- 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
Illustration-1: A gas is present at a pressure of 2 atm. What should be the increase in pressure so that the volume of the gas can be decreased to ¼ th of the initial value. If the temperature is maintained constant
Solution: PV = Constant for a given mass of gas at constant pressure
P1V1 = P2V2
P1 = 2 atm
V1 = V
V2 = V/4
2 × V = P2 ×
P2 = 8atm
∴pressure should be increased from 2 to 8 atm
∴ total increase = 8-2 = 6 atm
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
Exercise-1: 5g of ethane is confined in a bulb of one litre capacity. The bulb is so weak that it will burst if the pressure exceeds 10 atm. At what temperature will the pressure of the gas reach the bursting value?
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
2.5. Relation between Molecular Mass and Gas Densities
2.5.1 Actual density
For an ideal gas, PV = nRT or PV = , where w = mass of the gas in gms. and
M = Molecular wt. in g/mole.
∴ PM = RT
or PM = ρRT, where ρ is the density of the gas.
∴ ρ =
Illustration 3 The density of an unknown gas at 98 oC and 0.974 atm is 2.5 × 10-3 g/ml. What is the mol wt of gas?
Solution density = 2.5 × 10-3 g/ml = 2.5 g/lt
PM = dRT
0.974 × M = 2.5 × 0.0821 × 371
⇒ M = 78.18
2.5.2 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.
Exercise-2: When 3.2 g S is vapourised at 450 oC and 723 mm pressure, the vapours occupy a volume of 780 ml. What is the mol. Formula of S vapours under these conditions? Calculate the vapour density also.
2.5.3 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.
- 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 -4: The density of a mixture of O2 and N2 at STP is 1.3 g/litre. Calculate the partial pressure of O2
Solution: Let n1 and n2 be moles of O2 and N2 respectively in mixture
∴ Meff = ∑xiMi = ….(1)
and Meff =
= 29.137 ….(2)
equating (1) and (2)
or = 29.137
mole fraction of O2 = 0.28
= PM × mole fraction
= 1 × 0.28 = 0.28 atm
Exercise- 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.
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 constnt P & T
Since molecular mass of gas = 2 × vapour density, at constant P & T
The rate of diffusion (effusion) r of two gases under diffferent 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- 5: Pure O2 diffuses through an aperture in 224 seconds, whereas mixture of O2 and another gas containing 80% O2 diffuses from the same in 234 sec. What is molecular wt. of gas?
Solution: The gaseous mixture contains 80% O2 ad 20% gas.
∴ Average molecular weight of mixture (Mm) = ——– (i)
Now for diffusion of gaseous mixture and pure O2.
or ———— (ii)
∴ Mm = 34.92
By (i) and (ii) mol. weight of gas (m) = 46.6
- 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 moelcules 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 collison.
- 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
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 = =
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.
Furthermore UMP : UAV : Urms : : : :
1 : 1.128 : 1.224
Also Uav = Urms × 0.9213
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- 6: Calculate Vrms of O2 at 273K and 1 Pa pressure. The density of O2 under these conditions is 1.42 kg/m3?
Solution: = = 460 m/s
Also Vrms = = 461.28 m/s.
Note: In calculation of velocities pressure should be taken in Pascal’s or N/m2 units, density in kg/m3 units, molar mass of gas in kg units, R = 8.314 J/mole K to get velocity in m/s units.
Exercise -4: Calculate the kinetic energy of one mole of CH4 at -73 oC also calculate Vmp for the same under the same conditions
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
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- 5: There are four gases A,B,C and D with the Vander Waal’s constant a1,a2,a3 and a4 respectively such that a3 > a2 > a1 > a4 which of the gas is most easily liquefiable.
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: Using Vander Waal’s equation calculate the constant ‘a’ when two moles of gas contained in a four litre flask exerts a pressure of 11.0 atm at a temperature of 300 K. The volume of ‘b’ is 0.05 lit/mol-1
Solution: Vander Waal’s equation for n moles of gas is
(V-nb) = nRT
given V = 4 lt P=11.0 atm T = 300 K, b = 0.05 lt/mol n = 2
∴ [4 -2×0.05] = 2 × 0.082 × 300 ⇒ a = 6.46 atm litre2 mol-2
- Some other important definitions
7.1 Mean Free Path (λ)
It is the average distance travelled by a particle between collisions
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
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
Exercise- 6: Calculate Boyle temperature for N2 if its ‘a’ and ‘b’ values ar 1.39 litre2 atm mol-2 and 39.0 ml/mol
- 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
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- 8: A 20 c.c. mixture of CO, CH4 and He gases is exploded by an electric discharge at room temperature with excess of oxygen. The decrease in volume is found to be 13 c.c. A further contraction of 14 c.c. occurs when the residual gas is treated with KOH solution. Find out the composition of the gaseous mixture in terms of volume percentage.
Solution: Let the volume of CO be ‘a’ c.c. and CH4 be ‘b’ c.c
∴ volume of He = (20–a–b)
On explosion with oxygen
CO (g) + ⎯→ CO2(g)
CH4(g) + 2O2(g) ⎯→ CO2(g) + 2H2O (l)
‘a’ c.c. of CO give ‘a’ c.c. of CO2 and ‘b’ c.c. of CH4 gives ‘b’ c.c. of CO2. Therefore the volume decrease is due to the consumption of O2. O2 consumed for ‘a’ c.c. of CO is c.c. and O2 consumed for ‘b’ c.c. of CH4 is ‘2b’ c.c.
The further contraction occurs because of the absorption of CO2 by
KOH, a+b = 14
∴ b = 4 c.c.
∴ a = 10 c.c.
∴ percentage composition of CO = = 50%
percentage composition of CH4 = = 20%
percentage composition of He = = = 30%
Exercise- 7: Three volumes of a gaseous compound containing C, H and S only, reacts with O2 to yield 3 volume CO2, 3 volume SO2 and 6 volume of H2O. what is the formula of compound. How much O2 is needed for complete combustion of this molecule?
- Solutions to Exercises
Exercise-1: PV = nRT
10 × 1 = 5/30 × 0.082 ×T
T = 731.7K
Exerise-2: PV =
M = 256
Hence atomaticity = = 8
Hence formula = S8
Vapour density = = 128
Exercise-3: PHe = = 0.2463 atm
= 0.123 atm
= 0.123 atm
Ptotal = = 0.4925 atm
Exercise-4: K.E. per mole = = = 24.92 kJ
Vmp = = 4.56 × 102 m/s
Exercise-5: Greater is the value of ‘a’ greater is the force of attraction among the molecules hence easier is the liquification. Hence C gas will be most easily liquefiable.
Eercise-6: Tb = = 434.11 K
Exercise-7: Let the mol formula be CxHySz
CxHySz + O2 ⎯→ xCO2 + + zSO2
Given that 3x = 3, x = 1
= 6. y = 4 and 3z = 3, z = 1
Hence molecular formula = CH4S
O2 required = = (1+1+1) = 3 mole
- Solved Problems
Problem-1: Two container 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.
∴= = 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 happened 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.
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
⇒ Mmix = 0.2 ×28 + 0.8×4 = 8.8
∴ = 0.52
Problem -2: The pressure exerted by 12g of an ideal gas at temperature t°C in a vessel of volume V litre is one atm. When the temperature is increased by 10 degrees at the same volume the pressure increases by 10%. Calculate the temperature t and volume V (Molecular mass of gas = 120).
Solution: Given P = 1 atm, w = 12g, T = (t + 273) K, V = V litre (I case)
If T = (t + 283); P = 1 + = 1.1 atm, w = 12g, V = V litre (II case)
Using as equation:
I case 1 × V = × R (t + 273) …(i)
II Case 1.1 × V = × R (t + 283) …(ii)
By equations (i) and (ii),
∴ 1⋅ 1t + 300.3 = t + 283
∴ 0.1t = –17.3
∴ t = –173°C = 100 K
Also from Case (i) 1 × V = × 0.082 × 100 ( m = 120)
V = 0.82 litre
Problem- 3: 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 -4: 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 occuring 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.
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- 5: 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
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- 6: 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.
= 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).
= = 0.0694 moles
∴= 0.0694 moles
∴ mass of H2O (l) = 0.0694 × 18 = 1.25 g
Problem- 7: 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’
or P =
Multiply by [V] then, PV =
or PV = RT
or PV = RT
∴ PV =RT
or PV = RT
B = b –
Problem- 8: 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 = 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 =
Problem- 9: 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 = 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- 10: 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-1: 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
Problem-2: 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
Solution: A gas can be liquefied only if its temperature is lower than its critical temperature
Problem -3: Boyle’s law may be expressed as
(C) (D) none
Solution: from Boyle’s law
PV = constant
PdV +VdP = 0
(PV = K)
Problem- 4: 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
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
Problem-5: 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
Problem- 6: 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
⇒ = 1.59 litre
for CO2, ⇒ = 4.62 litre
Problem-7: 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
Problem-8: 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 =
∴ x Joule = 3 × 8.314 × 75
In both the cases x Joules correspond to N molecules.
Problem-9. 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 ≠.
- 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
Problem- 10. 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.
Problem- 11. 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
Problem- 12. 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
Problem- 13: 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
∴ (A) is incorrect
(B) is correct
(C) is incorrect
(D) is incorrect
Problem- 14. 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
Problem- 15: 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) =
- Assignments (Subjective)
LEVEL – I
- At what temperature the volume of a given mass of an ideal gas will be thrice of that at 27°C
- 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).
- How large a balloon could you fill with 4.0 g of He gas at 22 oC and 720 mm of Hg
- Calculate the weight of CH4 in a 9 litre cylinder at 16 atm and 27 oC
5. Calculate total energy of one mole of an ideal monatomic gas at 27°C?
- Calculate the density of CO2 at 100 0C and 800 mm Hg pressure
- Three footballs are respectively filled with N2, H2 and He. In what order are these footballs to be reinflated
- Calculate the average speed of CO at 100 oC
- 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.
- By how many folds the temperature of a gas would increase when the root mean square velocity of gas molecules in a container of fixed volume is increased from
5 × 104 cm s–1 to 15 × 104 cms–1.
- A carbon dioxide gas sample occupies 480 ml at 1 atm and 275 K. The pressure of the gas is now lowered and temp raised until its volume is 1.2 lit. Find the density of the gas under new condition.
- 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
- 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.
- 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
- What weight of air does an automobile tyre hold under following condition
Atmospheric pressure = 0.95 atm
Atmospheric temp = 25.8°C
Internal volume of the inflated tyre = 5.2 litre
Assume air is composed of 79% N2 and 21% O2 by volume
LEVEL – II
- How much thermal energy should be added to 3.45 g Neon in a 10 litre flask to raise the temperature from 0 oC to 100 oC. Atomic weight of Neon 20.18
- A gas cylinder contains 370g of O2 at 30atm and 25 oC. What mass of O2 would escape if first the cylinder was heated to 75 oC and then the valve were held open until the gas pressure was 1 atm, the temperature maintained at 75 oC.
- A meteorological balloon has a radius of1m when released from sea level at normal pressure and 25 oC and expanded to a radius of 3m when it had risen to its maximum attitude where the temperature was – 20 oC. What is the pressure inside the balloon at that height?
- 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
- 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.
- 50 ml of hydrocarbon having 85.7% carbon at S.T.P. on combustion gave 0.37 gm of CO2, 0.142 gm of H2O and an unknown quantity of methane. Deduce the molecular formula of hydrocarbon.
- 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.
- What fraction of total volume does helium atoms actually occupy at S.T.P. What will be its van der Waal’s equation under these circumstances. Given van der Waal constant ‘b’ for He= 24 cm3 mole–1.
- 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 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
- 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 (a) ideal gas (b) real gas nature
- A 20g chunk of dry ice is placed in an empty 0.75 litre wine bottle tightly closed. What would be the final pressure in the bottle after all CO2 has been evaporated and temperature reaches to 25 oC.
- At room temp. NO gas at 1.6 atm and O2 gas at 1 atm pressure were allowed to enter simultaneously through two ends of 1 m long straight glass tube of uniform cross sectional area. At what distance from the end which was used to introduce NO gas, brown vapour will first appear.
If we want to tap out only pure NO2 gas where should hole be made in the tube.
- A gas occupied 0.418 litre at 740 mm of Hg at 27 oC. Calculate
- a) its volume at STP
- b) molecular weight if gas weighs 3.0 g
- c) new pressure of gas if the weight of gas is increased to 7.5 g and temperature becomes 280 K
- d) the volume of vessel at 300 K
- Vander Waal’s constant b for a gas is 4.2 × 10-2 litre/mol. How close the nuclei of the two molecules come together?
LEVEL – III
- 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.
- A vessel of volume 5 litre contains 1.4 g of N2 at a temperature 1800 K. Find the pressure of the gas if 30 % of its molecules are dissociated into atoms at this temperature.
- 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.
- Two flask of equal volume have been joined by a narrow tube of negligible volume. Initially both flasks are at 300 K containing 0.60 mole of O2 gas at 0.5 atm pressure. One of the flask is then placed in a thermostat at 600 K. Calculate final pressure and the number of O2 gas in each flask.
- 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.
- 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.
- 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.
- 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.
- A long rectangular box is filled with chlorine (at. wt.: 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 chlorine molecules put in the two compartments respectively, calculate where should the partition be made if the pressure on both sides are to be equal. Is this pressure the same as the original pressure?
- 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.
- 1 mole of a gas is changed from is initial state (15 lit; 2 atm) to final state (4 lit, 10 atim) reversibly. If this change can be represented by a straight line in p-V curve, calculate maximum temperature, the gas attained.
- 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.
- 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 H2Ov 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?
- The composition of the equilibrium mixture for the equilibrium Cl2 2Cl at 1470 K may be determined by the rate of diffusion of the mixture through a pin hole. It was found that at 1470 K, the mixture diffuses 1.16 times as fast as Krypton (atomic weight = 83.8) diffuses under the same conditions. Find the degree of dissociation of Cl2 at equilibrium?
- Assignments (Objective Problems)
LEVEL – I
- According to charle’s law
(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
- Which sample contains least no. of molecules
(A) 1g of CO2 (B) 1 g of N2
(C) 1 g of H2 (D) 1 g of CH4
- A gas ‘A’ having mol. wt 4 diffuses thrice as fast as the gas B at a given T. The mol wt. of gas B is
(A) 36 (B) 12
(C) 18 (D) 24
- In the equation of state of an ideal gas PV = nRt, the value of the universal gas constant would depend only on.
(A) the nature of gas (B) the pressure of gas
(C) the temperature of the gas (D) the units of measurement
- 380 ml of a gas at 27oC, 800 mmHg weighs 0.455g the mol wt. of gas is
(A) 27 (B) 28
(C) 29 (D) 30
- The value of Vander Waals constant ‘a’ is maximum for
(A) helium (B) nitrogen
(C) CH4 (D) NH3
- A balloon filled with ethyne is pricked with a sharp point and quickly dropped in a tank of hydrogen gas under identical conditions. After a while the balloon will have:
(A) shrunk (B) enlarged
(C) completely collapsed (D) remains unchanged in size
- The temperature at which a real gas obeys the ideal gas laws over a fairly wide range of pressure is
(A) critical temperature (B) inversion temperature
(C) Boyle’s temperature (D) reduced temperature
- At Boyle’s temperature, compressibility factor ‘Z’ for a real gas is
(A) Z = 1 (B) Z = 0
(C) Z > 1 (D) Z < 1
- Under same condition of temp. and pressure, a cycloalkene was found to diffuse 3 times slower than hydrogen. Cycloalkene is
(A) Cyclopropene (B) Cyclobutene
(C) Cyclopentene (D) Cyclohexene
- 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
- Average molecular distance of gaseous molecules at 27°C at 1 atm is
(A) 3 ×10–5 cm (B) 3 ×10–8 cm
(C) 3 ×10–6 cm (D) 4.26 ×10–7 cm
- If 8 g methane be placed in 5 Lt container at 27°C. Its Boyle’s constant will be
(A) 12.3 litre atm (B) 2.46 atm
(C) 5 litre atm (D) 1.11 atm litre
- A gas in an open container is heated from 27°C to 127°C. The fraction of the original amount of gas remaining in the container will b
- A flask of methane (CH4) was weighed, methane was then pushed out and that flask again weighed when filled with oxygen at the same temperature and pressure. The mass of oxygen would be
(A) the same as that of methane
(B) half of that of methane
(C) double of that of methane
(D) negligible in comparison to that of methane.
- An ideal gas Law PV = nRT, is a relation between the four variables that describes the state of any gas. Which of the following is/are intensive variables?
(A) V (B) P
(C) n (D) T
- Under identical experimental condition which of the following pair of gas will be most easy to separate by diffusion process?
(A) H2 and D2 (B) and
(C) CO2 and C3H8 (D) O2 and N2
- The unit of van der Waal’s constant ‘a’ is
(A) litre (B) atmosphere
(C) atmosphere litre2 mole–1 (D) atmosphere (litre)2 (mole)–2
LEVEL – II
- 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) 819oC
- 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.
(C) 2π (D)
- X ml of Hydrogen 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) 55 seconds : CO2
- 1 Litre CO and 1.75 litre CH4 at the same temp and pressure were mixed together. What is the relation between the masses of two gases in the mixture
- A spherical air bubble is rising from the depth of a lake where pressure is ‘P’ atm and temperature is ‘T’ Kelvin. The percentage increase in its radius when it comes to the free surface of lake will be. (Assume temperature and pressure at the surface be respectively P/4 and 2T Kelvin)
(A) 100% (B) 50%
(C) 40% (D) 200%
- Let the most probable velocity of hydrogen molecules at a temp t°C is Vo. Suppose all the molecules dissociate into atoms when temp is raised to (2 t + 273)°C then the new r.m.s velocity is
(A) VO (B) VO
(C) VO (D) VO
- An ideal gas molecule is present at 27°C. By how many degree centigrade its temperature should be raised so that its Vrms , Vmp and Vav all may double.
(A) 900°C (B) 108°C
(C) 927°C (D) 81°C
- The mole fraction of nitrogen in mixture of nitrogen and oxygen in which the partial pressure of oxygen is 63 cm and the total pressure is 90 cm comes as
(A) 0.7 (B) 0.3
(C) 0.6 (D) none of these
- A flask contains 10g of a gas (relative molecular mass 100) at a pressure of 100 KPa was evacuated to a pressure of 0.01 KPa at constant temperature. Which one of the following is the best estimate of the number of molecules left in the flask.
(A) 6.02 × 1016 (B) 6.02 × 1017
(C) 6.02 × 1018 (D) 6.02 × 1019
- A compound exists in the gaseous state both as monomer (F) and dimer (A2). The molecular weight of the monomer is 48. In an experiment, 96g of the compound was confined in vessel of volume 33.6 L and heated to 273°C. Calculate the pressure developed, if the compound exists as a dimer to extent of 50% by weight under these conditions.
(A) 7.5 atm (B) 2.0 atm
(C) 0.9 atm (D) 5.4 atm
- 5.40 gm of an unknown gas at 27°C occupies the same volume as 0.14 gm of hydrogen at 17°C and same pressure. The molecular weight of unknown gas is
(A) 79.8 (B) 81
(C) 79.2 (D) 83
- If 8 g methane be placed in 5 Lt container at 27°C. Its Boyle’s constant will be
(A) 12.3 litre atm (B) 2.46 atm
(C) 5 litre atm (D) 1.11 atm litre
- Two containers A and B contain the same gas. If the pressure, volume and absolute temperature of the gas A is twice as compared to that of B, and if the mass of the gas B is xg, the mass of gas in A is
(A) xg (B) 4xg
(C) 2/xg (D) 2xg
- The number of moles of Hydrogen in 0.224 L of hydrogen gas at STP (273 K, 1 atm) (assuming ideal gas behaviour) is
(A) 1 (B) 0.1
(C) 0.01 (D) 0.001
- Answers (Subjective Problems)
LEVEL – I
627°C 2. 1.632
3. 25.56 lt. 4. 96g
5. 900 cal 6. 1.5124g/lt
7. H2 > He > N2 8. 1.502 × 103 m/s
9. 750 K 10. T1 = 9T2
- 0.780 12. 29.534 gm
13. 1051 mm 14. 1 : 24
LEVEL – II
- 51.29 2. 359.7
- 0.031 atm 4. 1.62
- 1.6 × 105 6. 146.5 mm Hg
- 81.12 cm 8. 2.68 × 10–4
- 98.52 atm 10. i) C5H8 and C5H12,
- ii) C5H8 = 0.35 mole C2H12 = 0.244 mole
- a) 0.0448 lt, (b) 8.987 × 10–3 lt 12. 15.87 atm
- a) 0.623 ml, b) no where 14. a) 0.37 lt, b) 181.54,
- c) 2.27 atm, d) 0.418lt
LEVEL – III
- 99.93% 2. 1.92 atm
- 337.5K 4. 0.66atm
- 0.2 6. 0.4254 atm
- 0.668 atm 8. O2 = 4ml, N2 = 64ml, CO2 = 12ml
- 3.44 : 1 10. NH3
- 698 K 12. H2 41.67%, D2 58.33%
- 2 atm 14. 1.7 × 10–8
- Answers to Objective Assignments
LEVEL – I
- D 2. D
- B 4. A
- A 6. D
- B 8. D
- B 10. C
- A 12. B
- B 14. D
- A 16. A
- C 18. D
- A 20. D
LEVEL – II
- D 2. D
- B 4. B
- A 6. A
- D 8. A
- B 10. B
- A 12. D
- A 14. B
- C 16. B
- A 18. A
- D 20. C