- IIT–JEE Syllabus
Avogadro’ s law; equation of state for an ideal gas; kinetic theory of gases; root mean square velocity and its relation with temperature; average velocity; Gay Lussac’s law.
- The Gas Laws
2.1 Boyle’s Law
In 1662, Robert Boyle discovered that there existed a relation between the pressure and the volume of a fixed amount of gas at a fixed temperature. In his experiment, he discovered that the product of Pressure & Volume of a fixed amount of gas at a fixed temperature was approximately a constant.
∴V ∝ (T and n constant).
∴PV = Constant
It is important to understand here the various units of pressure and their relationships.
The SI unit of pressure is Nm−2, which is called ‘Pascal’ (Pa) The pressure at sea level due to the weight of the earth’s atmosphere is approximately 105 Pa. Since pascal is a small unit, we express pressures in bar units, where 1 bar = 105 Pa. Atmospheric pressure is about one bar. Atmospheric pressure can also be expressed in atmosphere units abbreviated as atm.
1 atm = 1.01325 bar
2.2 Charle’s – Gay- Lussac Law
In 1787, Jacques Charles and in 1808 Gay Lussac discovered that if the pressure is kept constant, the volume of a gas sample increases linearly with the temperature for a fixed amount of gas. This law led to the idea of temperature. The unit of temperature used is Kelvin.
T(K) = T(oC) + 273. ∴ V ∝ T (P and n are constant)
or = Constant
2.3 Avogadro’s Law
In 1812, Amadeo Avogadro stated that samples of different gases which contain the same number of molecules (any complexity, size, shape) occupy the same volume at the same temperature and pressure. It follows from Avogadro’s hypothesis that V ∝ n (when T and P are constant).
2.4 Ideal Gas Equation
Combining all these gas laws, a simple equation can be arrived at, which relates, P, V, n and T for a gas.
The equation is PV = nRT
This is called Ideal Gas Equation.
P is the pressure of the gas and can be expressed in atm or Pa. Correspondingly, the volume must be expressed in litres or m3 respectively. n is simply a number representing number of moles and T is the temperature in Kelvin. R is called the universal gas constant.
Numerical Values of R
- i) In litre atmosphere = 0.0821 litre atm deg–1 mole–1
- ii) In ergs = 8.314 × 107 erg deg–1 mole–1
iii) In calories = 1.987 cal deg–1 mole–1
Illustration 1: Calculate the pressure exerted on the walls of a 3 litre flask when 7 gms of N2 are introduced into the same at 27°C.
Solution: P =
= = 2.05 atm
Illustration 2: An open vessel at 27 oC is heated until 3/5th of the air in it has been expelled. Assuming that the volume of the vessel remains constant find
- a) the temperature at which vessel was heated?
- b) the air escaped out if vessel is heated to 900 K?
- c) temperature at which half of the air escapes out?
Solution: One should clearly note the fact that on heating a gas in a vessel there are the number of moles of gas which go out, the volume of vessel remains constant.
Let initial moles of gas at 300 K be ‘n’. On heating 3/5 moles of air are escaped out at temperature T.
∴Moles of air left at temperature T = =
- a) Under similar conditions of P and V
n1T1 = n2T2
n ×300 = ×T
T = 750 K
- b) On heating vessel to 900 K, let n1 moles be left again n1T1 = n2T2
n1×900 = 300 × n
⇒ n1 =
∴moles escaped out = n- moles
- c) Let n/2 moles are escaped out at temperature T then
n1T1 = n2T2
= n × 300
T = 600 K
Exercise 1: An open bulb containing air at 19°C was cooled to a certain temperature at which the no. of moles of gaseous molecules increased by 25%. What is the final temperature?
Exercise 2: A long rectangular box is filled with Cl2 (atomic weight 35.45) which is known to contain only 35Cl and 37Cl. If the box could be divided by a partition and the two types of Cl2 molecules put in the two compartments respectively, calculate where should the partition be made if pressure on both sides are to be equal. Is this pressure the same as the original pressure?
2.5 Relation between Molecular Mass and Gas Densities
Actual density: Density and Molecular weight
From the ideal gas equation P = ∴ M = |
Hence from a knowledge of d/P density under unit pressure molecular weight can be determined. At low pressure below 1 atm d/p has been found to decrease linearly with pressure. Extrapolation of the line to zero pressure
P = 0 gives the ideal value of (d/P) at P → 0
∴ M = RT
Vapour Density: For gases another term which is often used is vapour-density. Vapour density of a gas is defined as the ratio of the mass of the gas occupying a certain volume at a certain temperature and pressure to the mass of hydrogen occupying the same volume at the same temperature and pressure i.e. W(gas) =
and mol. wt.of Hydrogen is 2)
∴ = Vapour density of gas
Vapour density of a gas is same at any temperature, pressure and volume.
Molecular Weight and Effective Molecular Weight: Molecular wt. of a gas mixture or effective molecular weight:- Suppose we have to find the molecular weight of air and we are told that air contains 79% nitrogen and 21% oxygen (by mole or volume).
First of all let us understand what is meant by molecular wt. of a gas. Molecular wt. of a gas is the weight in gms of one mole of the gas (hence the unit gm/mole). Now if we take one mole of air it would contain 79/100 moles of Nitrogen and 21/100 moles of oxygen. The weight of one mole of air would be 0.79 × 28 + 0.21 × 2 = 22.54 gm/mole.
∴ Meffective = ΣXiMi = Σ mole fraction of a gas × mol. wt. of a gas.
Xi (Mole fraction) of a gas is =
2.6 Dalton’s Law of Partial Pressures
“The total pressure of a mixture of non-reacting gases is equal to the sum of their partial pressures”. By the partial pressure of a gas in a mixture is meant, the pressure that the gas will exert if it occupies alone the total volume of the mixture at the same temperature.
Consider n1, moles of gas 1 and n2 moles of gas 2 occupying a vessel of volume V at temperature T K and exerting a total pressure P.
Let n1 moles of gas 1 alone occupy the same vessel of capacity V at the same temperature TK and exert a pressure P1. Then by definition, P1 is partial pressure of gas 1 in the mixture.
Let n2 moles of gas 2 alone occupy the same volume V at the same temperature TK and exert a pressure P2. The P2 is the partial pressure of gas 2 in the mixture.
By Dalton’s Law P = P1 + P2
Derivation: n = n1 + n2 + …
+ +…
P = P1 + P2 + …
Assumption: Volume of all the gases is same.
Relationship between partial pressures and number of moles.
Important formula
- i) where x1 = mole fraction of gas (1)
- ii) Partial pressure of a gas in the mixture =
Partial pressure and aqueous tension
Dalton’s law is used to calculate the pressure of a dry gas when it is collected over water at atmospheric pressure. By Dalton’s law.
Pressure of dry gas = atmospheric pressure – aqueous tension Aqueous tension depends on temperature. It increases with temperature and becomes 760 mm at 100°C. |
Illustration 3: A mixture containing 1.6 g of O2 and 1.4 g of N2 and 0.4 g of He occupies a volume of 10 litre at 27 oC. Calculate the total pressure of the mixture and partial pressure of each component.
Solution: PHe = = 0.2463 atm
= 0.123 atm
= 0.123 atm
Ptotal = = 0.4925 atm
2.7 Graham’s Law of Diffusion
Diffusion is the tendency of any substance to spread throughout the space available to it. Diffusion will take place in all direction and even against gravity. So gases diffuse through firm substances and through small holes. The streaming of gas molecules through a small hole is called effusion.
According to Graham the rate of diffusion (effusion) of a gas at constant P&T is inversely proportional to square root of its molecular mass.
r ∝ at constant P & T
∴ at constant P & T
Since molecular mass of gas = 2 × vapour density, at constant P & T
The rate of diffusion (effusion) r of two gases under different pressure can be given by
at constant T only.
Therefore according to Graham’s law of diffusion (effusion) at constnt P & T.
; ρ1 & ρ2 are the respective densities
where V1 and V2 are volumes diffused (effused) in time t1 and t2.
where n1, n2 are moles diffused (effiused) in time t1 & t2.
where d1, d2 are distances travelled by molecules in narrow tube in time t1 and t2.
Illustration 4: 32 ml of H2 diffuses through a fine hole in 1 minute. What volume of CO2 will diffuse in 1 minute under same condition?
Solution:
= 6.82 ml
Exercise 3: A gaseous mixture of O2 and X containing 20% of X diffused through a small hole in 234 secs while pure O2 takes 124 secs to diffuse through the same hole. Find molecular weight of X?
- The Kinetics Theory of Gases
In order to derive the theoretical aspect of the various gas laws based on simple experimental facts, Maxwell proposed the following postulates under the heading of kinetic theory of gases.
3.1 The Postulates
- i) Each gas is made up of a large number of small (tiny) particles known as molecules.
- ii) The volume of a molecule is so small that it may be neglected in comparison to total volume of gas.
iii) The molecules are never in stationary state but they are believed to be in chaotic (random) motion. They travel in straight line in all possible directions with altogether different but constant velocities. The direction of motion is changed by the collision with container or with other molecules.
- iv) The molecules are perfectly elastic and bear no change in energy during collision.
- v) The effect of gravity on molecular motion is negligible.
- vi) The temperature of gas is a measure of its kinetic energy. KE ∝ absolute temperature.
vii) The pressure of the gas is thus due to the continuous collision of molecules on the walls of container. Pressure P = .
3.2 The Kinetic Equation
Maxwell also derived an equation on the basis of above assumption as
PV =
Where P = Pressure of gas
V = Volume of gas
m = mass of one molecule of gas
n = no. of molecules of gas
u = root mean square velocity of molecules
For 1 mole n = N is Avogadro number
m × N = Molecular mass M.
∴ PV = or u2 = or u = =
4. Velocities of gas molecules
4.1 Average Velocity
As per kinetic theory of gases, each molecule is moving with altogether different velocity. Let ‘n’ molecules be present in a given mass of gas, each one moving with velocity v1, v2, v3… vn. The average velocity or UaV = average of all such velocity terms.
Average velocity = =
Uav =
4.2 Root Mean Square Velocity
Maxwell proposed the term Urms as the square root of means of square of all such velocities.
= =
Also Urms =
4.3 Most probable velocity
It is the velocity which is possessed by maximum no. of molecules.
Vmp =
Furthermore UMP : UAV : Urms : : : :
: :
1 : 1.128 : 1.224
Also Uav = Urms × 0.9213
Exercise 4: The average speed at T1k and the most probable speed at T2k of CO2 gas is 9 ×104 cm sec–1. Calculate the value of T1 and T2.
4.4 Kinetic Energy of Gas
As per kinetic equation PV =
For 1 mole m× n = Molecular Mass (M)
∴ PV = or .
Also = kT
Where k is the Boltzmann constant
Illustration 5: Calculate rms speed of O2 at 273 K and 1 × 105Pa pressure. The density of O2 under these conditions is 1.42 kg m–3.
Solution: Data are given in SI units
= 459.63 m sec–1
Illustration 6: At what temperature will the r.m.s. velocity of oxygen be one and half times of its value at N.T.P.?
Solution:
Suppose the temperature required is T′ when the velocity will be C
or, T′ = × 273 = 614.25°K
Exercise 5: Calculate the rms speed of ozone kept in a closed vessel at 20°C and 82 cm Hg pressure.
5. Imperfect or Real Gases
So far the PVT behaviour of a gas has been presumed to obey the ideal gas equation PV = nRT. When measurements are extended to higher pressures, or even when very accurate measurements are made at ordinary pressures, the results do not confirm this equation. The ideal gas laws are not followed.
All gases exhibit, to some extent, deviations from the ideal-gas laws. When these deviations are recognized, the gas is said to behave as a real, non ideal or imperfect gas.
Compressibility factor Z: Real and ideal gases can be compared at various pressures and various temperatures by noting the extent to which the value of PV/RT deviates from 1. The quantity is given by the symbol Z and the name compressibility factor. That is
Z =
Greater is the departure of Z from unity more is the deviation from ideal behaviour.
- i) When Z < 1, this implies that gas is more compressible.
- ii) When Z > 1, this means that gas is less compressible.
iii) When Z = 1, the gas is ideal.
6. Van der Waals’ Equation of State for a Real Gas
An empirical equation of state generated by Van der Waals in 1873 reproduces the observed behaviour with moderate accuracy. For n moles of gas, the Van der Waals equation is
(V − nb) = nRT
where a and b are constants characteristic of a gas. This equation can be derived by considering a real gas and ‘converting ‘ it to an ideal gas.
Volume correction: We know that for an ideal gas P×V = nRT. Now in a real gas the molecular volume cannot be ignored and therefore let us assume that ‘b’ is the volume excluded (out of the volume of container) for the moving gas molecules per mole of a gas. Therefore due to n moles of a gas the volume excluded would be nb. ∴ a real gas in a container of volume V has only available volume of (V – nb) and this can be thought of as an ideal gas in container of volume (V – nb).
Pressure correction: Let us assume that the real gas exerts a pressure P. The molecules that exert the force on the container will get attracted by molecules of the immediate layer which are assumed not to be exerting pressure.
It can be seen that the pressure the real gas exerts would be less than the pressure an ideal gas would have exerted. The real gas experiences attractions by its molecules in the reverse direction. Therefore if a real gas exerts a pressure P, then an ideal gas would exert a pressure equal to P + p (p is the pressure lost by the gas molecules due to attractions). This small pressure p would be directly proportional to the extent of attraction between the molecules which are hitting the container wall and the molecules which are attracting these.
Therefore p ∝ (concentration of molecules which are hitting the container’s wall)
P ∝(concentration of molecules which are attracting these molecules) ⇒ p ∝
∴ p = where a is the constant of proportionality which depends on the nature of gas. A higher value of ‘a’ reflects the increased attraction between gas molecules.
Note: The Van der Waals constant b (the excluded volume) is actually 4 times the volume of a single molecule.
∴ b = 4 × 6.023 × 1023 ×, where r is the radius of a molecule.
The constants a & b: Vander Waals constant for attraction (a) and volume (b) are characteristic for a given gas. Some salient feature of a & b are:
- i) For a given gas Vander Waal’s constant of attraction ‘a’ is always greater than Vander Waals constant of volume (b).
- ii) The gas having higher value of ‘a’ can be liquefied easily and therefore H2 & He are not liquefied easily.
iii) The units of a = litre2 atm mole–2 & that of b = litre mole –1
- iv) The numerical values of a & b are in the order of 10–1 to 10–2 & 10–2 to 10–4
- v) Higher is the value of ‘a’ for a given gas, easier is the liquefaction.
Exercise 6: 2 moles of NH3 occupied a volume of 5 litres at 27°C. Calculate the pressure if the gas obeyed Vander Waals equation. Given a = 4.17 litre2 atm mole–2, b = 0.037 litre/mole.
6.1 Vander Waals equation, different forms
i) At low pressures: ‘V’ is large and ‘b’ is negligible in comparison with V. The Vander Waals equation reduces to:
; PV + = RT PV = RT – or PV < RT This accounts for the dip in PV vs P isotherm at low pressures. |
Deviation of gases from ideal behaviour with pressure. | ||
ii) At fairly high pressures
may be neglected in comparison with P. The Vander Waals equation becomes P (V – b) = RT PV – Pb = RT PV = RT + Pb or PV > RT This accounts for the rising parts of the PV vs P isotherm at high pressures. |
The plot of Z vs P for N2 gas at different temperature is shown here. |
iii) At very low pressures: V becomes so large that both b and become negligible and the Vander Waals equation reduces to PV = RT. This shows why gases approach ideal behaviour at very low pressures.
- iv) Hydrogen and Helium: These are two lightest gases known. Their molecules have very small masses. The attractive forces between such molecules will be extensively small. So is negligible even at ordinary temperatures. Thus PV > RT. Thus Vander Waals equation explains quantitatively the observed behaviour of real gases and so is an improvement over the ideal gas equation.
Vander Waals equation accounts for the behaviour of real gases. At low pressures, the gas equation can be written as,
or Z =
Where Z is known as compressibility factor. Its value at low pressure is less than 1 and it decreases with increase of P. For a given value of Vm, Z has more value at higher temperature.
At high pressures, the gas equation can be written as
P (Vm – b) = RT
Z = = 1 +
Here, the compressibility factor increases with increase of pressure at constant temperature and it decreases with increase of temperature at constant pressure. For the gases H2 and He, the above behaviour is observed even at low pressures, since for these gases, the value of ‘a’ is extremely small.
Illustration 7: One litre of a gas at 300 atm and 473 K is compressed to a pressure of 600 atm and 273 K. The compressibility factors found are 1.072 & 1.375 respectively at initial and final states. Calculate the final volume.
Solution: P1V1 = Z1nRT1 and P2V2 = Z2nRT2
or V2 =
= = 370.1 ml
- Some other important definitions
7.1 Mean Free Path (λ)
The average distance covered by a molecule between two successive collisions is called mean free path and is denoted by . ∴ = where n = no. of molecules/cc. Again, if P & T denote the pressure and temperature of the gas, from kinetic theory.
P × 10–3 = RT or n =
∴ =
Thus mean free path is directly proportional to temperature and inversely to pressure.
7.2 Collision Frequency (z)
It is the number of collisions taking place per second
z = πσ2Nuav
where σ = collision diameter. It is the distance between the centres of two molecule without collision.
N = number of molecules per unit volume,
And uav = average velocity
7.3 Relative Humidity (RH)
At a given temperature it is given by equation
RH =
7.4 Boyle’s Temperature (Tb)
Temperature at which real gas obeys the gas laws over a wide range of pressure is called Boyle’s Temperature. Gases which are easily liquefied have a high Boyle’s temperature [Tb(O2)] = 46 K] whereas the gases which are difficult to liquefy have a low Boyle’s temperature [Tb(He) = 26K].
Boyle’s temperature Tb =
where Ti is called Inversion Temperature and a, b are called van der Waals constant.
7.5 Critical Constants
- i) Critical Temperature (Tc): It (Tc) is the maximum temperature at which a gas can be liquefied i.e. the temperature above which a gas can’t exist as liquid.
- ii) Critical Pressure (Pc): It is the minimum pressure required to cause liquefaction at Tc
iii) Critical Volume: It is the volume occupied by one mol of a gas at Tc and Pc
Vc = 3b
Illustration 8: Calculate Vander Waals constants for ethylene
TC = 282.8 k; PC = 50 atm
Solution: b = = 0.057 litres/mole
a = = 4.47 lit2 atm mole–2
- Gas Eudiometry
The relationship amongst gases, when they react with one another, is governed by two laws, namely Gay-Lussac law and Avogadro’s law.
Gaseous reactions for investigation purposes are studied in a closed graduated tube open at one end and the other closed end of which is provided with platinum terminals for the passage of electricity through the mixture of gases. Such a tube is known as Eudiometer tube and hence the name Eudiometry also used for Gas analysis.
During Gas analysis, the Eudiometer tube filled with mercury is inverted over a trough containing mercury. A known volume of the gas or gaseous mixture to be studied is next introduced, which displaces an equivalent amount of mercury. Next a known excess of oxygen is introduced and the electric spark is passed, whereby the combustible material gets oxidised. The volumes of carbon dioxide, water vapour or other gaseous products of combustion are next determined by absorbing them in suitable reagents. For example, the volume of CO2 is determined by absorption in KOH solution and that of excess of oxygen in an alkaline solution of pyrogallol. Water vapour produced during the reaction can be determined by noting contraction in volume caused due to cooling, as by cooling the steam formed during combustion forms liquid (water) which occupies a negligible volume as compared to the volumes of the gases considered. The excess of oxygen left after the combustion is also determined by difference if other gases formed during combustion have already been determined. From the data thus collected a number of useful conclusions regarding reactions amongst gases can be drawn.
- a) Volume-volume relationship amongst Gases or simple Gaseous reactions.
- b) Composition of Gaseous mixtures.
- c) Molecular formulae of Gases.
- d) Molecular formulae of Gaseous Hydrocarbons.
The various reagents used for absorbing different gases are
O3 ⎯→ turpentine oil
O2⎯→ alkaline pyrogallol
NO ⎯→ FeSO4 solution
CO2,SO2 ⎯→alkali solution (NaOH, KOH, Ca(OH)2, HOCH2CH2NH2, etc.)
NH3 ⎯→ acid solution or CuSO4 solution
Cl2 ⎯→ water
Equation for combustion of hydrocarbons.
CxHy + O2 ⎯→ xCO2 + H2O
General Assumptions: In all problems, it is assumed that the sparking occurs at room temperature. This implies that water formed would be in liquid state and that nitrogen gas is inert towards oxidation.
Illustration 9: A gaseous hydrocarbon requires 6 times its own volume of O2 for complete oxidation and produces 4 times its volume of CO2. What is its formula?
Solution: The balanced equation for combustion
CxHy + O2 ⎯→ xCO2 + H2O
1 volume volume ∴ = 6 (by equation)
or 4x + y = 24 …(1)
Again x = 4 since evolved CO2 is 4 times that of hydrocarbon
∴ 16 + y = 24 or y = 8 ∴ formula of hydrocarbon C4H8
- Solution to Exercises
Exercise 1: Suppose the volume of bulb is V containing n moles at 19°C (i.e. 292 K) let the temperature be TK when n moles increases to 1.25 (i.e. by 25%). Since 1.25n moles at TK occupy a volume V.
∴ n moles at TK should occupy = ml
Applying
∴
or, T = = 233.6K = – 39.4°C
Exercise 2: Since 35.45 is the average atomic weight of 35Cl and 37Cl we have
= 35.45
∴ = 3.42. n1 : n2 = 3.44 : 1
∴ Ratio of lengths is 3.44 : 1
Since the pressure on both sides of the partition is equal the pressure before and after partition will be same. (no. of moles per unit volume is same)
Exercise 3: We have,
or, ∴ Mmix= 34.921
As the mixture contains 20% of X, the molar ratio of O2 and x may be represented as 0.8 n : 0.2 n
n being total no. of moles
Mmix = = 34.921
∴ Mx = 46.6
Exercise 4: We know for 1 mole of an ideal gas
r.m.s. speed =
r.m.s. speed : average speed : most probable speed = 1 : 0.9211 : 0.8165
Average speed at T1k = 0.9211 × = 9 ×104
∴ most probable speed at T2K = 0.8165 × = 9 × 104
Substituting R = 8.314 × 107 ergs K–1 mole–1
And M = 44 we get T1 = 168 K T2 = 2143 K
Exercise 5: Volume occupied by 1 mole of O3 at 20°C and 82 cm pressure
= 22400 × = 22282 ml
p = 82 × 13.6 × 981 dynes cm–2
Now we have,
= 3.9 × 104 cm sec–1
Exercise 6: Applying Vander Waal’s equation for n moles
(V – nb) = nRT
(5.2 – 2 × 0.037) = 2 × 0.082 × 300
or, p = 9.33 atm
- Solved Problems
10.1 Subjective
Problem 1: One litre flask contains air, water vapour and a small amount of liquid water at a pressure of 200 mmHg. If this is connected to another one litre evacuated flask, what will be the final pressure of the gas mixture at equilibrium? Assume the temperature to be 50 oC. Aqueous tension at 50 oC = 93 mmHg
Solution The aqueous tension remains same in both the flask. Also flask are at same temperature
∴ P1V1 = P2V2 where P1 = 200 – 93 = 107 V1 = 1 litre V2 = 2 litre ∴107 × 1 = P × 2 P = 53.5 mm Since aqueous tension is also present in flask, equivalent to 93 mm ∴Pressure of gaseous mixture = 93 + 53.5 = 146.5 mmHg |
Problem 2: One way of writing the equation of state for a real gas is
Where B is a constant. Derive an approximate expression for B in terms of vander waal’s constant ‘a’ and ‘b’
Solution:
or P =
Multiply by [V] then, PV =
or PV = RT
or PV = RT
Now
∴ PV =RT
or PV = RT
B = b –
Problem 3: Two containers A and B have the same volume. Container A contains 5 moles of O2 gas. Container B contains 3 moles of He and 2 moles of N2. Both the containers are separately kept in vacuum at the same temperature. Both the containers have very small orifices of the same area through which the gases leak out . Compare the rate of effusion of O2 with that of He gas mixture.
Solution: Since both the containers are in the same conditions of P,V and T,
=
As the mixture contains three moles of He and 2 moles of N2, the effective molecular weight of the mixture would be.
= 13.6
∴= = 0.652
Though this solution looks OK, there is one big flaw in it.
The error is that we have assumed that He and N2 from vessel B would effuse out with the same rate. This assumption was made in because we have taken the composition of the gas mixture coming out of the vessel to be same as that of the mixture that was inside the vessel . It should be duly noted that the two mixtures (inside and the are that effused out) have different compositions. Therefore first we must find the composition of the gas mixture coming out of the vessel B.
= 0.252
This means that initially the ratio of moles of N2 to the moles of He coming out of the vessel are in the molar ratio of 0.252 and not .
Let moles of the He coming out to be x
∴ moles of N2 coming out is 0.252 x
∴ = 0.2
= 0.8
⇒ Mmix = 0.2 ×28 + 0.8×4 = 8.8
∴ = 0.52
Problem 4: 50 ml. of a mixture of NH3 and H2 was completely decomposed by sparking into nitrogen and hydrogen. 40 ml of oxygen was then added and the mixture was sparked again. After cooling to the room temperature the mixture was shaken with alkaline pyro-gallol and a contraction of 6 ml. was observed. Calculate the % of NH3 in the original mixture.
Solution: Let the volume of NH3 be x ml
∴ volume of H2 = (50– x) ml
On sparking,
NH3 ⎯→ N2 + H2
Since 40 ml of O2 is added and sparked, it must have reacted with H2 to form liquid water. Moreover since 6 ml contraction occurs with alkaline pyrogallol, 34 ml is the volume of O2 used up.
∴ total volume of H2 is 68 (Q 2H2 + O2 ⎯→ 2H2O)
∴ (50–x) + x = 68
50 + = 68 ; x = 36
∴ NH3 = 72%
Problem 5: 20 ml of a mixture of methane and a gaseous compound of acetylene series were mixed with 100 ml of oxygen and exploded to complete combustion. The volume of the products after cooling to original room temperature and pressure, was 80 ml and on treatment with potash solution a further contraction of 40 ml was observed. Calculate (a) the molecular formula of the hydrocarbon, (b) the percentage composition of the mixture.
Solution: Oxygen is present in excess. This is because the combustion has been done completely (which means methane and the other gas are consumed completely) and on CO2 absorption 40 ml of a gas is still left over. This left over gas is oxygen.
∴ oxygen used up = 100–40=60 ml
CO2 produced = 40 ml
If the molecular formula of the gas of acetylene series is CnH2n–2 and its volume is a ml, then the volume of CH4 is (20–a) ml.
On combustion CH4(g) + 2O2(g) ⎯→ CO2(g) + 2H2O (l)
CnH2n–2 (g) + (g) ⎯→ nCO2 + (n–1) H2O (l)
Oxygen used up = 2(20–a) + a = 60
(from CH4) ( from CnH2n–2)
CO2 produced = (20–a) + na = 40
(from CH4) (from CnH2n–2)
Solving a = 10 ml
and n = 3
∴ formula of the gas is C3H4
% of C3H4 = 100 = 50%
% of CH4 = 50%
Problem 6: Calculate % of ‘free volume’ available in 1 mol gaseous water at 1.0 atm and 100 o C. Density of liquid water at 100 oC is 0.958 g/mol.
Solution: We know that PV = nRT
Volume occupied by 1 mol gaseous water at 100 oC and 1 atm pressure
= = 30.62 litre
volume of 1 mol liquid water = = = 1.88 × 10-2 litre
% of volume occupied by liquid water
= = 0.0613 %
% of free volume = 100 – 0.0613 = 99.93 %
Problem 7: A vertical hollow cylinder of height 1.52m is fitted with a movable piston of negligible mass and thickness. The lower half of the cylinder contains an ideal gas and the upper half is filled with mercury. The cylinder is initially at 300 K. When the temperature is raised half of the mercury comes out of the cylinder. Find the temperature assuming the thermal expansion of mercury to be negligible?
Solution: Initially at lower end
P = 76 cm of Hg + 76 cm of air = 152 cm, T = 300K, V = (where V1 is the volume of cylinder) Finally at lower end P = 76 cm of air + 38 cm of Hg = 114 cm, T = ?, V = ∴ ⇒ T = 337.5 K |
Problem 8: A mixture of propane and methane is contained in a vessel of unknown volume V at a temperature T and exerts a pressure of 320 mm Hg. The gas is burnt in excess O2 and all the carbon is recovered as CO2. The CO2 is found to have a pressure of 448 mm Hg in a volume V1 at the same temperature T. Calculate mole fraction of propane in mixture.
Solution: Let the moles of propane (C3H8) = n1
Moles of methane CH4 = n2
We know that PV = nRT
∴320 × V = (n1+n2) RT
after combustion
total moles of CO2 formed = (3n1+n2)
Once again we have PV = nRT
448 ×V = ….(2)
Dividing eqn (2) by (1) we have
⇒ = 0.25
mol fraction of C3H8 =
= = 0.2
Problem 9: What would be the final pressure of O2 in following experiment? A collapsed polethylene bag of 30 litre capacity is partially blown up by the addition of 10 litres of N2 at 0.965 atm at 298 K. Subsequently enough O2 is pumped into bag so that at 298 K and external pressure of 0.990 atm, the bag contains
30 litres N2.
Solution: Given pressure of N2 = 0.965 atm
Volume of N2 = 10 litre. Temperature of N2 = 298 K
∴ For N2 when bag is fully expanded,
Volume of N2 (alone) = 30 litre at 298 K
P1V1 = P2V2
0.965 × 10 = P2 × 30
∴ (alone) in 30 litre bag at 298 K = 0.322 atm
Now PM = +
0.990 = + 0.332
∴ = 0.668 atm
Problem 10: 1 mole of a gas is changed from is initial state (15 lit; 2 atm) to final state (4 lit, 10 aim) reversibly. If this change can be represented by a straight line in p-V curve, calculate maximum temperature, the gas attained.
Solution: Eqn. For the line is ; 11p + 8V = 142
For (pV)max, 11p =
or then apply
(pV)max = nRTmax
Problem 11: A mixture containing 1.12 litres of H2 and 1.12 litres of D2 (deuterium) at N.T.P., is taken inside a bulb connected to another bulb by a stop-cock with a small opening. The second bulb is fully evacuated, the stop-cock is opened for a certain time and then closed. The first bulb is now found to contain 0.05 g of H2 Determine the percentage composition by weight of the gases in the second bulb.
Solution: Volume of H2 left in the first bulb after diffusion =
= 0.56 litres at N.T.P.
Volume of H2 diffused = 1.12 -0.56 = 0.56 litres.
Let the volume of D2 diffused be v2.
According to Graham’s law of diffusion,
or (at. mass of D2 being 2)
hence, v2 = 0.396 litres at N.T.P.
Wt. of H2 in second bulb = 0.05 g, Wt. of D2 in the second bulb
= g.
Percentage of H2 = 41.67. Percentage of D2 = 100-41.67 = 58.33
Problem 12: Assume that air contains 79% N2 and 21% O2 by volume. Calculate the density of moist air at 25°C and 1 atm pressure when relative humidity is 60%. The vapour pressure of water at 25°C is 23.78 mm of Hg. Relative humidity is given by percentage relative humidity
=
Solution: First of all it is important to understand the distinction between vapour pressure and partial pressure of vapour.
When a liquid is in equilibrium with its vapours, the vapours exert a pressure on the surface of the liquid called vapour pressure. This is the maximum possible pressure exerted by the vapours of the liquid at that tempreature. If we assume that vapours of a liquid exist in a container without the liquid, then the vapours behave just like any other gas, obeying all the gas laws. The pressure exerted by the vapours now can be anything which is less than or equal to that of vapour pressure. This is called partial pressure of vapour. If the partial pressure somehow becomes more than vapour pressure the vapours start liquefying to form liquid till vapours would be in equilibrium with liquid, exerting the vapour pressure.
To this problem let us look at what we are supposed to calculate. We need to find the density of moist air. From the ideal gas equation, PV =
⇒ ρ =
Therefore, we need to find the molecular weight of moist air. To find the molecular weight of a mixture of gas, we need the molar composition or mole fraction of the gas mixture.
Partial pressure of water (vapour) = = 0.0187 atm.
∴ mole fraction of water vapour = = 0.0187
Pressue of (N2+O2) = (1–0.0187) atm = 0.9813 atm
Let the pressure of N2 be 79 x, then pressure of O2 is 21 x ( they are in that molar ratio)
∴ 79 x + 21 x = 0.9813 atm
x =
x = 0.009813 atm
∴ = 79 x = 0.7752 atm
∴ = 0.2061 atm.
Mole fraction of N2 = = 0.7752
Mole fraction of O2 = = 0.2061
∴Effective molecular weight of moist air
= (0.0187 ×18) + (0.7752 ×28) + (0.2061 ×32) = 28.63 g/mol
∴ ρ = = 1.1716 g / L
Problem 13: When a certain quantity of oxygen was ozonised in suitable apparatus, the volume decreases by 4 ml. On addition of turpentine oil, the volume decreases further by 8 ml. Find the formula of ozone, assuming all measurements are made at same temperature and pressure?
Solution: The reaction occurring in the ozoniser is the conversion of oxygen to ozone and the conversion is never 100%. Let the volume of oxygen in the beginning be ‘a’ ml.
nO2 2On
Initial volume a 0
Volume after attainment of equilibrium a–n x 2 x
The initial volume reduction is due to reaction alone.
∴ initial volume – volume at equilibrium = 4 ml
∴ a–(a–n x +2 x) = 4
n x – 2 x = 4
n x = 4+ 2 x ————– (i)
The second decrease in volume is due to absorption of ozone by turpentine oil.
∴ 2 x = 8 and x = 4
Putting the value of x in equation (i)
∴ n x = 4+8 = 12
n = = 3
∴ Molecular formula of ozone is O3.
Problem 14: A container holds 3L of N2(g) and H2O (l) at 29°C. The pressure is found to be
1 atm. The water in the container is instantaneously split into hydrogen and oxygen by electrolysis, according to the reaction, H2O(l) ⎯→H2(g) + . After the reaction is complete, the pressure is 1.86 atm. What mass of water was present in the container. The aqueous tension of water at 29°C is 0.04 atm.
Solution: This problem can be done in two ways. One is the right way and the other is wrong way. Let’s see the first method and guess whether it is right or wrong.
Total pressure in the beginning = = 1 atm
∴ = (1–0.04) = 0.96 atm.
When water is split into H2 + O2, the total pressure is 1.86 atm.
∴ = 1.86
0.96 + = 1.86
∴ = (1.86 – 0.96) × = 0.6 atm.
Now
= 0.0726 moles
∴ Mass of water = 0.0726 ×18 = 1.306 g
This method looks perfectly OK. But there is one small error made. The error is so small that it might be difficult to point out. Find the error on your own and only after several trials you should look at correct solution that follows now.
The error that has happened is that the pressure of 1 atm is no doubt the pressure of N2 and H2O vapours but the electrolysis is done only to the H2O (liquid) and that too instantaneously. This implies that the H2O (vapour) that was in equilibrium with the liquid initially would be still left after the complete electrolysis of liquid water. ∴ 1.86 atm is the pressure of H2,O2,N2 and H2O (vapour).
= 1.86
= 1.86
= = 0.0694 moles
∴= 0.0694 moles
∴ mass of H2O (l) = 0.0694 × 18 = 1.25 g
Problem 15: A spherical balloon of 21 cm diameter is to be filled up with hydrogen at STP from a cylinder containing the gas at 20 atm and 27°C. If the cylinder can hold 2.82 L of water, calculate the number of balloons that can be filled up.
Solution: We must understand first that the gas in the cylinder is at 20 atm and 27°C while its filling in the balloons requires STP conditions. If we calculate the volume of the gas at STP, it comes out to be,
V2 =
V2 = 51.324 L
This means as we fill the gas at STP, the volume of the gas expands to 51.324 L. It is this expansion that leads to filling of the gas in balloons. But since the cylinder’s volume is 2.82 L, this much gas would be inside the cylinder while the rest of the volume escapes out, filling the balloons.
Therefore the volume that is actually filled,
= (51.324 – 2.82) L = 48.504 L
Volume of one balloon = πr3 = = 4.8504 L
∴ Number of balloons filled up =
10.2 Objective
Problem 1: A gas can be liquefied by pressure alone when its temperature
(A) higher than its critical temperature
(B) lower than its critical temperature
(C) either of these
(D) none
Solution: A gas can be liquefied only if its temperature is lower than its critical temperature
∴(B)
Problem 2: Boyle’s law may be expressed as
(A) (B)
(C) (D) none
Solution: from Boyle’s law
PV = constant
PdV +VdP = 0
(PV = K)
∴ (B)
Problem 3: A vessel has N2 gas and water vapours at a total pressure of 1 atm. The partial pressure of water vapours is 0.3 atm. The contents of this vessel are transferred to another vessel having one third of the capacity of original volume, completely at the same temperature the total pressure of this system in the new vessel is
(A) 3.0 atm (B) 1 atm
(C) 3.33 atm (D) 2.4 atm
Solution:
Now new pressure of N2 in another vessel of volume V/3 at same temperature T is given by
∴
since aqueous tension remains constant, and thus total pressure in new vessel
= 2.1 + 0.3 = 2.4 atm
∴(D)
Problem 4: For two gases A and B with molecular weights MA and MB, it is observed that at a certain temperature T1 the mean velocity of A is equal to the root mean square velocity of B. thus the mean velocity of A can be made equal to the mean velocity of B if
(A) A is at temperature T and B at T′, T > T′
(B) A is lowered to a temperature T2 , T2 < T while B is at T
(C) Both A and B are raised to a higher temperature
(D) Both A and B are placed at lower temperature
Solution: (UAV)A = and (Urms)B =
∴
for A (UAV) = for B VAV =
∴T2 = or T2 < T
∴ (B)
Problem 5: The circulation of blood in human body supplies O2 and releases CO2. the concentration of O2 and CO2 is variable but on an average, 100 ml blood contains 0.02 g of O2 and 0.08 g of CO2. The volume of O2 and CO2 at 1 atm and at body temperature 37oC, assuming 10 lt blood in human body, is
(A) 2 lt, 4 lt (B) 1.5 lt, 4.5 lt
(C) 1.59 lt, 4.62 lt (D) 3.82 lt, 4.62 lt
Solution: 100 ml blood has 0.02 g P2 and 0.08 g CO2
∴ 10,000 ml blood has 2 g O2 and 8 g CO2
using PV = nRT
for O2,
⇒ = 1.59 litre
for CO2, ⇒ = 4.62 litre
∴ (C)
Problem 6: At 100oC and 1 atm, if the density of liquid water is 1.0 g/cc and that of water vapour is 0.0006 g/cc, then the volume occupied by water molecule in one litre of steam at that temperature is
(A) 6 cc (B) 60 cc
(C) 0.6 cc (D) 0.06 cc
Solution: Mass of 1 lt water vapour = V ×d = 1000 × 0.0006 = 0.6g
∴volume of liquid water = = 0.6cc
∴(C)
Problem 7: The K.E. of N molecule of O2 is x Joules at –123°C. Another sample of O2 at 27°C has a KE of 2x Joules. The latter sample contains.
(A) N molecules of O2 (B) 2N molecules of O2
(C) N/2 molecules of O2 (D) N/4 molecule of O2
Solution: KE = ; T = – 123 + 273 = + 150 K
= xJ = 225 × 8.314 = xJ
At 27°C = 27+ 223 = 300K
KE for = 2x Joule =
N molecules
∴ x Joule = 3 × 8.314 × 75
In both the cases x Joules correspond to N molecules.
∴ (A)
Problem 8: If for two gases of molecular weights MA and MB at temperature TA and TB,
TAMB = TBMA, then which property has the same magnitude for both the gases.
(A) density (B) pressure
(C) KE per mol (D) Vrms
Solution: i) density of a gas (ρ) =
Since , ∴ at the same pressure = .
But if pressure is different then ≠.
- ii) Pressure of the gases would be equal if their densities are equal other wise not.
- KE per mol =
∴It will be different for the two gases.
- Vrms = , since ; Vrms of A = Vrms of B
∴ (D)
Problem 9: Helium atom is two times heavier than a hydrogen molecule. At 298 K, the average kinetic energy of a Helium atom is
(A) two times that of hydrogen molecule
(B) same as that of a hydrogen molecule
(C) four times that of a hydrogen molecule
(D) half that of a hydrogen molecule
Solution: The average kinetic energy of an atom is given as kT.
∴ It does not depend on mass of the atom.
∴ (B)
Problem 10: Dalton’s law of partial pressure is not applicable to, at normal conditions
(A) H2 and N2 mixture (B) H2 and Cl2 mixture
(C) H2 and CO2 mixture (D) H2 and O2 mixture
Solution: H2 and Cl2 reacts to form HCl; Dalton’s law of partial pressure is valid only for the gases which don’t react at ordinary conditions
∴(B)
Problem- 11: The ratio between the rms velocity of H2 at 50 K and that of O2 at 800 K is
(A) 4 (B) 2
(C) 1 (D) 1/4 [IIT–JEE ’96]
Solution: Vrms (H2 at 50 K) =
Vrms (O2 at 800K) =
∴(C)
Problem 12: Which of the following curve does not represent Boyle’s law?
Solution: P = Where C is a constant. We can see that (c) is true as the graph of
P vs would be a straight line.
(A) is true because log P = log C – log V.
(B) is true because
which means that as V increases the slope decreases and is always negative
∴ (D)
Problem 13: The temperature of an ideal gas is increased from 140 K to 560 K. If at 140 K the root mean square velocity of the gas molecule is V, at 560 K it becomes
(A) 5V (B) 2V
(C) V/2 (D) V/4
Solution: The Vrms at 140K is V
∴ V =
At 540 K = V′ = = = 2 = 2V
∴ (B)
Problem 14: The behaviour of a real gas is usually depicted by plotting compressibility factor Z versus P at a constant temperature. At high temperature and high pressure, Z is usually more than one. This fact can be explained by van der Waals equation when
- the constant ‘a’ is negligible and not ‘b’
- the constant ‘b’ is negligible and not ‘a’
- both constants ‘a’ and ‘b’ are negligible
- both the constants ‘a’ and ‘b’ are not negligible.
Solution: (V – nb) = nRT
At low pressures, ‘b’ can be ignored as the volume of the gas is very high. At high temperatures ‘a’ can be ignored as the pressure of the gas is high.
∴P (V–b) = RT
PV – Pb = RT
PV = RT + Pb
∴ (A)
Problem 15: X ml of H2 gas effuses through a hole in a container in 5 seconds. The time taken for the effusion of the same volume of the gas specified below under identical conditions is :
(A) 10 seconds : He (B) 20 seconds : O2
(C) 25 seconds : CO (D) 35 seconds : CO2
Solution:
∴ (A) is incorrect
(B) is correct
(C) is incorrect
=
(D) is incorrect
∴ (B)
- Assignments (Subjective)
LEVEL – I
- a) How large a balloon could you fill with 4.0 g of He gas at 22 oC and 720 mm
of Hg? - b) Calculate the density of CO2 at 100 0C and 800 mm Hg pressure.
- a) A container has 3.2 g of a certain gas at NTP. What would be the mass of the same gas contained in the same vessel at 200°C and 16 atm. Pressure?
- b) A certain quantity of a gas measured 500 mL at a temperature of 15°C and 750 mm Hg. What pressure is required to compress this quantity of gas into a 400 mL vessel at a temperature of 50°C?
- a) Three footballs are respectively filled with N2, H2 and He. In what order are these footballs to be reinflated?
- b) The ratio of rates of diffusion of two gases A and B under same pressure is 1:4. If the ratio of their masses present in the mixture is 2:3. What is the ratio of their mole fraction in mixture?
- a) Calculate the rms speed in cm/sec at 25°C of a free electron and a molecule of UF6.
- b) The flask (A) and (B) have equal volumes. Flask (A) contains H2 gas at 300K, while (B) contains equal mass of CH4 at 900K. Calculate the ratio of average speed of molecules in flask (A) and (B)/
5. a) Calculate total energy of one mole of an ideal monatomic gas at 27°C?
- b) How much thermal energy should be added to 3.45 g Ne in a 10 litre flask to raise the temperature from 0°C to 100°C. (Atomic weight Ne = 20.18)?
- The critical temperature and pressure of CO2 gas are 304.2K and 72.9 atm respectively. What is the radius of O2 molecule assuming it to behave as Vander Waal’s gas?
- Argon has TC= –122°C PC = 48 atm. What is the radius of the argon atm?
*8. 0.22g of a sample of a volatile compound containing C, H and Cl only yielded on combustion in O2 0.195g of CO2 and 0.0804g of H2O. 0.12g of the compound occupied a volume of 37.24 ml at 105°C and 768 mm Hg pressure. Calculate the molecular formula of the compound.
*9. 10 ml of mixture of CH4 and CO2 are exploded with excess O2. A contraction of 17 ml was observed after explosion. After the treatment with KOH solution there was a 2nd contraction of 14 ml. Find the % of gases in mixture?
*10. a) 3040 ml of a gas mixture at 27°C and 750 mm pressure contain 20% CH4, 60% CO and 20% H2. What weight of KClO3 will be required to liberate just sufficient oxygen to oxidise completely the gas mixture?
- b) 25cc of a gaseous mixture containing N2 and N2O is passed through a hot Cu tube. The resulting gas has a volume of 20 ml. find the percentage of N2 and N2O in the original gas mixture (The volumes have been measured under same conditions of temperature & pressure)?
LEVEL – II
- a) Two flasks of equal volume connected by a narrow tube (of negligible volume) are at 27°C and contain 0.7 moles of H2 at 0.5 atm. One of the flask is then immersed into a bath kept at 127°C, while the other remains at 27°C. Calculate the final pressure and the number of moles of H2 is each flask.
- b) 1 gm of an alloy of Al and Mg reacts with excess HCl to form AlCl3, MgCl2 and H2. The evolved H2 collected over mercury at 0°C occupied 1200 ml at 699 mm Hg. What is the composition of alloy?
*2. a) Ladenburg found that a sample of ozonised oxygen containing 86.16% of O3 by weight required 430 secs to diffuse under conditions where pure O2 required 369.5 secs. Determine the vapour density of O3.
- b) The normal density of PH3 at 0°C and at different pressure are
P (atm) 1 3/4 1/2 1/4
d/P (gm/lit) 1.5307 1.5272 1.5238 1.5205
Calculate the molecular weight of PH3 and atomic weight of P
- Assuming O2 molecule spherical in shape and occupying the radius 150 pm. Calculate
- i) the volume of single molecule of gas
- ii) the percentage of empty space in one mole of O2 at NTP
*4. 9 volumes of gaseous mixture consisting of gaseous organic compound A and just sufficient amount of O2 required for complete combustion yielded on burning 4 volumes of CO2, 6 volumes of water vapour and 2 volumes of N2 all volumes measured at the same temperature and pressure. If the compound A contained only C, H and N
- i) how many volumes of O2 required for complete combustion?
- ii) What is the molecular formula of the compound A?
- Two gases in adjoining vessels are brought into contact by opening a stop cock between them. The one vessel measured 0.25 lit and contained NO at 800 torr and 220 K, the other measured 0.1 lit and contained O2 at 600 torr and 220 K. The reaction to form N2O4 (solid) exhausts the limiting reactant completely.
- a) Neglecting the vapour pressure of N2O4 what is the pressure of the gas remaining at 220 K after completion of the reaction?
- b) What weight of N2O4 is formed?
- a) The compressibility factor for CO2 at 273K and 100atm pressure is 0.2005. Calculate the volume occupied by 0.2 mole of CO2 gas at 100 atm using (i) ideal gas (ii) real gas nature
- b) Calculate the pressure exerted by 1 mole of methane (CH4) in a 250 ml container at 300 K using van der Waal’s equation. What pressure will be predicted by ideal gas equation ?
a = 2.253L2 atm. mol-2,
b = 0.04281 lit mol-1
R = 0.0821 L atm mol-1K.
- a) To what minimum degree centigrade the temperature of earth will have to be raised to get atmosphere free earth like moon?
Given Vesp = When Re = 6.37 ×106 m
Assume : atmospheric air chiefly contains N2 and O2 only. - b) A straight glass tube has two inlets X and Y at the two ends. The length of the tube is 200 cm. HCl gas through inlet X and NH3 gas through inlet Y are allowed to enter the tube at the same time. White fumes first appear at a point P inside the tube. Find the distance of P from X?
- a) A 10 litre box contained 41.4 gm of a mixture of CxH8 and CxH12. At 44°C the total pressure is 1.56 atm. Analysis of the gas mixture shows 86% of C & 14% of hydrogen by weight of hydrocarbon sample.
What gases are in the box?
How many moles of each gas are in the box?
*b) 24 ml. of water gas containing only hydrogen and carbon monoxide in equal proportions by volume, are exploded with 80 ml. of air, 20% by volume of which is oxygen. If all gaseous volumes are measured at room temperature and pressure, calculate the composition by volume of the resulting gaseous mixture.
- An open vessel at 27°C is heated until three-fifths of the air in it has been expelled. Assuming the volume of the vessel remains constant, find the temperature to which the vessel has to be heated.
*10. At 60°C the density of N2O4 gas was found to be 30.2. Calculate the percentage of NO2 molecules by weight and by volume.
LEVEL – III
*1. The solubility of oxygen, nitrogen and CO at 15°C, 760 mm pressure are 0.048, 0.022 and 0.03 per litre of water respectively. A mixture containing 10% O2, 70% N2 and 20% CO at a pressure of 740 mm is bubbled through 200 ml water at 15°C. What weight of each gas is dissolved?
*2. A vessel contains 10 g of I2(s) and N2 at a pressure of 10 atm at 25°C. The volume of the vessel is one litre. If this vessel is connected to a 40 litre empty vessel and the temperature of the first vessel is increased to 250°C and of the second to 200°C. Calculate the final pressure in vessels.
- A column of Hg of 10 cm in length is contained in the middle of a narrow 1 m long tube which is closed at both ends. Both the halves of the tube contained air at a pressure 76 cm of Hg. By what distance will the column of Hg be displaced if the tube is held vertical?
- A thin tube of uniform cross section is sealed at both ends. It lies horizontally. The middle 5 cm containing Hg and the two equal ends containing air at the same pressure P0. When the tube is held at an angle 60° with the vertical, the length of air column above and below the mercury are 46 and 44.5 cm respectively. Calculate the pressure P0 in cm of Hg (the temperature of the system is kept at 30°C).
*5. The respiration of the suspension of yeast cells was measured by observing the decrease in pressure of gas above the cell suspension. The apparatus show that the gas was confined to a constant volume, 16 ml and the entire pressure change was caused by uptake of O2 by the cells. The pressure was measured in a manometer, the fluid of which had a density of 1.034 gm/ml. The entire apparatus was immersed in a thermostat at 37°C. In a 30 min. observation period the fluid in open side of manometer dropped 37 mm. neglecting the solubility of O2 in yeast suspension, compute rate of O2 consumed by the cells in m3 of O2 (S.T.P.) per hour.
*6. Inside the spherical glass flask A of radius 1 m containing 300g of H2 there was a rubber balloon B containing some N2. Inside the balloon B was another rubber balloon C containing some O2. At 27°C it was found that the balloon B had a radius of 60 cm and balloon C had a radius of 30 cm. Calculate total weight of gases inside the flask. If another 50 gm of H2 was not introduced in flask A, what would be the volume of B & C. If the flask is now dipped in a bath containing acetone and dry ice (–78°C) for sufficient time the temperature of the entire contents would reacts the bath temperature. Assuming that at this temperature rubber cracks up and both the balloons burst what would be pressure inside flask.
- Assuming ideal gas behaviour, how many atoms of Ar are contained in a typical human breath of 0.5L at 1.0 bar and 27°C? Air consists of 1% Ar atoms. Assuming that the argon atoms from last breath of Plato have been distributed randomly throughout the atmosphere ( 5 × 1018 m3). How long would it take to breath one of the atoms? A typical adult breath rate is 10 min–1.
- In a space shuttle the CO2 output per astronaut has been estimated as 44g/hr. An experimental catalytic converter reduces CO2 at a rate of 600 ml STP per minute into H2O. What fraction of time would such a converter have to operate in order to keep up the CO2 output of one astronaut?
- A silent electric discharge was passed through 100 ml. air when 95 ml. of ozonized air was formed. This ozonized air took 48.7 sec. to diffuse from a very small hole. If 100 ml. of air diffused through the same hole under identical conditions, it took 50 sec. Find the mol. wt. of ozone assuming air to contain 79% N2 and 21% O2.
- 10 ml of ammonia were enclosed in an eudiometer and subjected to electric sparks. The sparks were continued till there was no further increase in volume. The volume after sparking measured 20 ml. Now 30 ml. of O2 were added and sparking was continued again. The new volume then measured 27.5 ml. All volumes were measured under identical conditions of temperature and pressure. V.D. of ammonia is 8.5. Calculate the molecular formula of ammonia. Nitrogen and Hydrogen are diatomic.
*11. To maintain the body temperature at 310 K a Russian Sanyasi breathing his lung capacity once in 5 minutes while meditating in a Himalayan cave where the atmospheric pressure is 70 cm. of Hg. After having learnt the art of meditation he goes back home and meditates on the Baltic sea shore where the pressure is 760 torr. Assuming that the atmosphere temperature is 283 K at both the places and that air is composed of 79 % N2 and 21 % O2 , Calculate his breathing rate required to maintain the body temperature at 310 K. If his O2 requirement is 3.24 × 1022 molecules per hour, find the capacity of his lungs?
- A compound exists in the gaseous state both as a monomer (A) and dimer (A2). The molecular weight of the monomer is 48. In an experiment, 96 g of the compound was confined in a vessel of volume 33.6 litres and heated to 273°C. Calculate the pressure developed, if the compound exists as a dimer to the extent of 50 per cent by weight, under these conditions ( R = 0.082)
- A mixture of H2O vapour, CO2 and N2 was trapped in a glass apparatus with a volume of 0.731 ml. The pressure of total mixture was 1.74 mm of Hg at 23°C. The sample was transferred to a bulb in contact with dry ice (–75°C) so that (H2O)v are frozen out. When the sample returned to normal value of temperature, pressure was 1.32 mm of Hg. The sample was then transferred to a bulb in contact with liquid N2 (–95°C) to freeze out CO2. The measured, pressure was 0.53 mm of Hg at normal temperature. How many moles of each constituent are in mixture?
- Calculate the number of binary collisions per cc of of N2 gas at 1 atm and 25°C. The diameter of N2 molecule = 3.74Å.
*15. Find out the percentage of molecule in a gas at 1 atm and 27°C whose kinetic energy lines in the range of E and E + 0.01E.
[Where E = KT]
- Assignments (Objective Problems)
LEVEL – I
- If the pressure of a given mass of gas is reduced to half and temperature is doubled simultaneous the volume will be
(A) Same as before (B) Twice as before
(C) ¼ the as before (D) None
- The molecules of a gas A travel four times faster than the molecules of gas B at same temperature. The ratio of molecular weights (MA/MB) is
(A) 1/16 (B) 4
(C) 1/4 (D) 16
- The ratio of rms velocity to average velocity of gas molecules at a particular temperature is
(A) 1.086 : 1 (B) 1 : 1.086
(C) 2 : 1.086 (D) 1.086 : 2
- X ml of H2 gas effuses through a hole in a container in 5 secs. The time taken for effusion of same volume of gas specified below under identical condition is
(A) 10 secs : He (B) 20 secs : O2
(C) 25 secs : CO (D) 55 secs : CO2
- The compressibility factor for an ideal gas is
(A) 1.5 (B) 1
(C) 2 (D) ∞
- On increasing temperature, the fraction of total gas molecule which has acquired most probable velocity will
(A) increase (B) decrease
(C) remains constant (D) cant say without knowing pressure
- Which of the following pair will diffuse at the same rate?
(A) CO2 and N2O (B) CO2 and NO
(C) CO2 and CO (D) N2O and NO
- The value of Vander Waals constant ‘a’ is maximum for
(A) helium (B) nitrogen
(C) CH4 (D) NH3
- A mixture of H2 and O2 in 2:1 volume is allowed to diffuse through a porous partition what is the composition of gas coming out initially
(A) 1:2 (B) 4:1
(C) 8:1 (D) 1:4
- The temperature at which a real gas obeys the ideal gas laws at fairly wide range of pressure is
(A) Critical temperature (B) Inversion temperature
(C) Boyel’s temperature (D) Reduced temperature
- For non-zero value of force of attraction between gas molecules gas equation will be
(A) PV = nRT – (B) PV = nRT + nbP
(C) PV = nRT (D) P =
12. Distribution of molecules with velocity is represented by the curve. Point A in the curve shifts to the higher value of velocity if
(A) T is increased (B) V is increased (C) P is increased (D) All |
- According to Charle’s law
(A) (B)
(C) (D) none
- For a given mass of gas, if pressure is reduced to half and temperature is increased two times, then the volume would become
(A) (B) 2v2
(C) 6v (D) 4v
- The density of O2 gas at 25oC is 1.458 mg/lt at one atm pressure. At what pressure will O2 have the density twice the value?
(A) 0.5 atm (B) 2atm/25oC
(C) 4atm/25oC (D) none
LEVEL – II
1. I, II, III are three isotherm respectively at T1, T2 & T3 temperatures will be in order
(A) T1 = T2 = T3 (B) T1 < T2 < T3 (C) T1 > T2 > T3 (D) T1 > T2 = T3 |
- NH3 is liquefied more easily than N2. Hence
(A) a and b of NH3 > that of N2 (B) a(NH3) > a(N2) but b (NH3) < b(N2)
(C) a(NH3) < a(N2) but b (NH3) > b(N2) (D) None
- 0.2 mole sample of hydrocarbon CxHy yields after complete combustion with excess O2 gas, 0.8 mole of CO2, 1.1 mole of H2O. Hence hydrocarbon is
(A) C4H10 (B) C4H8
(C) C4H5 (D) C8H16
- When 2 gm A gas is introduced into an evacuated flask kept at 25°C, the pressure is found to be 1 atm. If 3g of another gas B is then added to same flask, the total pressure becomes 0.5 atm. The ratio of molecular weights is
(A) 1:1 (B) 1:2
(C) 2:3 (D) 1:4
- Air open vessel at 127°C is heated until 1/5th of air in it has been expelled. Assuming that the volume of vessel remains constant the temperature to which the vessel has been heated is
(A) 177°C (B) 277°C
(C) 377°C (D) 477°C
- 3.2g S is heated if occupy a volume of 780 ml at 450°C and 723 mm pressure. Formula of sulphur is
(A) S2 (B) S
(C) S4 (D) S8
- A gas cylinder containing cooking gas can withstand a pressure of 14.9 atm. The pressure gauge of cylinder indicates 12 atm at 27°C. Due to sudden fire in building the temperature starts rising. The temperature at which the cylinder will explode is.
(A) 42.5°C (B) 67.8°C
(C) 99.5°C (D) 25.7°C
- A sample of gas is at 0oC. The temperature at which its rms speed of the molecule will be doubled is
(A) 103oC (B) 273oC
(C) 723oC (D) 818oC
- 6 g each of the following gases at 87oC and 750 mm pressure are taken. Which of them will have the least volume
(A) HF (B) HCl
(C) HBr (D) HI
- The temperature at which H2 has same rms speed (at 1 atm) as that of O2 at NTP is
(A) 37 K (B) 17 K
(C) 512 K (D) 27 K
- In a closed vessel, a gas is heated from 300 K to 600 K the kinetic energy becomes/remain
(A) half (B) double
(C) same (D) four times
- Air contains 79 % N2 and 21 % O2 by volume. If the barometric pressure is 750 mmHg. The partial pressure of oxygen is
(A) 157.7 mmHg (B) 175.5 mmHg
(C) 315.0 mmHg (D) none
- The gaseous mixture contains 1g of H2, 4g of He, 7g of N2 and 8g of O2. The gas having the highest partial pressure is
(A) H2 (B) O2
(C) He (D) N2
- Which of the following gases would have the highest rms speed at 0 oC
(A) O2 (B) CO2
(C) SO3 (D) CO
- Two gases A and B present separately in two vessels X & Y at the same temperature with molecular weights M & 2 M respectively are effused out. The orifice in vessel X is circular while that in Y is a square. If the radius of the circular orifice is equal to that of the length of the square orifice, the ration of rates of effusion of gas A to that of gas B is.
(A) (B)
(C) 2π (D)
- Answers (Subjective Problems)
LEVEL – I
- a) 25.65 litre 2. a) 1.3184g
- b) 5124 g/litre b) 1051 mm
- a) H2 > He > N2 4. a) 1.16 × 107cm/sec
- b) 1:24 b) 632
- a) 900 cals 6. 1.62Å
- b) 29 cal 7. 1.47 × 10–8 cm
- C2H4Cl2 9. 45% CH4, 40% C2H4, 15% CO2
- a) 7.96g
- b) 60% N2, 40% NO
LEVEL – II
- a) 0.571 atm, 0.3 and 0.4 moles 2. a) 23.3
- b) 25% Mg, 54.75% Al b) 30.98
- 1.41 × 1023 cc/molecule, 99.96% 4. 7 volume, C2H6N2
- 0.30 atm, 0.402g 6. a) 0.448 litre, 8.987 × 10–3 lit
- a) 1.6 × 105K 8. a) C5H8, C5H12 0.35, 0.244 moles
- b) 12 cm b) O2= 4ml, N2 = 64 ml, CO2 = 12 ml
- 750 K 10. 52.3%, 68.7%
LEVEL – III
- 0.0093, 0.00426, 0.0058g/ml 2. 0.4254 atm
- 3 cm 4. 75.4 cm
- 102 mm3 6. 0.73 atm
- 8.47 min 8. 0.622
- 48.26 10. NH3
- 5.42 per min per breath 12. 2 atm
- 2.1 × 10–8, 3.1 × 10–8, 1.7 × 10–8 14. 8.98 × 1028
- 46%
- Answers to Objective Assignments
LEVEL – I
- D 2. A
- A 4. B
- B 6. B
- A 8. D
- C 10. C
- A 12. D
- D 14. D 15. B
LEVEL – II
- B 2. B
- A 4. C
- D 6. D
- C 8. D
- D 10. B 11. B 12. A 13. A 14. D
- A