|Density and Molecular weight
From the ideal gas equation
∴ 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
Mean Free Path
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.
Gases are usually liquefied by lowering or temperature and increase in pressure. Of these two factors the influence of temperature is important. For every gas there is a temperature above which liquefaction is impossible whatever the pressure be.
The limiting temperature above which liquefaction is impossible is called critical temperature (TC) of the gas. The minimum pressure necessary to liquefy a gas at critical temperature is called its critical pressure. The volume occupied by a 1 gm mole of substance at critical temperature and pressure is called critical volume.
At critical point
For Vander Waal’s gas
VC = 3b, PC = TC =
∴ = 2.66
a = 3PCVC2, b = VC =
The temperature at which real gas Obeys Boyle’s Law. The minimum of PV – P curve is on the PV axis (i.e. P = 0).
Hence for Boyle point at P = 0 = 0
For Vander Waal’s gas = TB =
Reduced Equation State
Since critical constants are definite quantities for a gas, it is possible to express its P, V, T as multiples or sub-multiples of critical values let us say.
P = πPC; V = φVC, T = θTC
Where π, φ, θ are called reduced pressure, reduced volume and reduced temperature.
The Vander Waal’s equation may be changed as
(V – b) = RT [for 1 mole of gas]
(θφVC – b) = RθTC
(3φb – b) =
(3φ + 1) = 8θ
This is called law of corresponding states.
Relative Humidity (RH)
At a given temperature it is given
If isotopic enrichment is to be increased the number of operations can be calculated by overall separation factor f.
Overall separation factor f =
Where n1, n2, n′1, n′2 are the concentration of two isotopes before and after processing.
Theoretical separation factor for a single step f′ =
If required enrichment of species in attained after x times then (f′)x = = f
or x logf′ =
or x =
Illustration 1: Calculate the pressure exerted on the walls of a 3 lit flask when 7 gms of N2 are introduced into the same at 27°C.
Solution: P =
= = 2.05 atm
Illustration 2: At what temperature will the r.m.s. velocity of oxygen be one and half times of its value at N.T.P.
Suppose the temperature required is T′ when the velocity will be C
or, T′ = × 273 = 614.25°K
Illustration 3: 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 final volume.
Solution: P1V1 = Z1nRT1 and P2V2 = Z2nRT2
or V2 =
= = 370.1 ml
Illustration 4: Calculate Vander Waals constant for ethylene
TC = 282.8 k; PC = 50 atm
Solution: b = = 0.057 litres/mole
= 4.47 lit2 atm mole–2
Illustration 5: Argon has TC = –122°C; PC = 48 atm
What is the radius of Argon atom
Solution: b =
or r =
= 1.47 × 10–8 cm
Illustration 6: 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
Illustration 7: 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 8: 32 ml of H2 diffuses through a fine hole in 1 minute. What volume of CO2 will diffuse in 1 minute under same condition?
= 6.82 ml
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?
Exercise 3: Calculate the rms speed of ozone kept in a closed vessel at 20°C and 82 cm Hg pressure.
Exercise 4: 2 moles of NH3 occupied a volume of 5 litres at 27°C. Calculate the pressure if the gas obeyed Vander Waals equation.
Exercise 5: 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.
Exercise 6: 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.
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
or, T = = 233.6K = – 39.4°C
Exercise 2: Since 35.45 is the average atomic weight of 35Cl and 37Cl we have
∴ = 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: 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 4: 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
Exercise 5: 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 6: 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
- 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) ∞
- A monoatomic, diatomic and triatomic gas are mixed taking one mole of each CP/CV for the mixture is
(A) 2/5 (B) 1.67
(C) 1.33 (D) 1.428
- 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 density of a mixture O2 and N2 at NTP is 1.3 gm lit–1. Partial pressure of O2 is
(A) 0.12 atm (B) 0.19 atm
(C) 0.28 atm (D) 0.42 atm
- 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 are 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
- Level – II Q.No. 15
- Level – I Q. No. 15
- Level – I Q. No. 13
|16. See the figure. The valves of X & Y are opened simultaneously. The white fumes of NH4Cl will first from at
(D) A, B & C simultaneously.
- N2 + 3H2 ⎯→ 2NH3. 1 mole N2, 4 mole N2 are taken in 15 lit flask at 27°C. After complete conversion of N2 into NH3, 5 lit of H2O is added. Pressure set up in the
(A) atm (B) atm
(C) atm (D) atm
- If V is the volume of 1 molecule of gas under given condition then Vander Waals constant b is
(A) 4V (B) 4V/N0
(C) N0/4V (D) 4VN0
- The compressibility of a gas is less than 1 at STP. Therefore
(A) Vm (molar volume) > 22.4 lit (B) Vm < 22.4 lit
(C) Vm = 22.4 lit (D) None
- Level – II Q.NO.1
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, 0.1 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) 1:3 (D) 1:4
- Air open vessel at 27°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
- Assume centre of sun to consist of gases whose average molecular weight is 2. The density and pressure of the gas are 1.3 g cc–1 and 1.12 × 109 atm respectively. The temperature of sun is
(A) 2 × 103 K (B) 2 × 105K
(C) 2 × 107 K (D) 2 × 109 K
- 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 1which the cylinder will explode is.
(A) 42.5°C (B) 67.8°C
(C) 99.5°C (D) 25.7°C
- Level – II Q. 7
- Level – II Q. 8
- Level – II Q. 14
- A mixture of 2g of H2 and 8 gm O2 is kept in a vessel of 20 lit capacity at 27°C. Total pressure inside the vessel is
(A) 0.5 atm (B) 1 atm
(C) 1.5 atm (D) 2 atm
- 1 mole of N2O4 (g) at 300 K is kept in dosed container under 1 atm. It is heated to 600K when 20% mass of N2O4(g) decomposes to NO2(g). The resultant pressure is
(A) 1.2 atm (B) 2.4 atm
(C) 2 atm (D) 1 atm
- A closed vessel contains equal volume of N2 & O2 at a pressure of pmm. If N2 is remover from the system then the pressure will be
(A) P (B) 2P
(C) P/2 (D) P2
- Three rubber tubes are respectively filled with H2, O2, N2 and He. The tube which will be required to be reinflated first is
(A) H2 filled tube (B) O2 filled tube
(C) N2 filled tube (D) He filled tube
- At what temperature will H2 molecules have same kinetic energy as N2 molecules have at 35°C.
(C) (D) 35°C
- 432g of gas A takes 36 minutes to diffuse through a porous hole. 288 ml of another gas B takes 48 minutes to diffuse through porous hole under same condition of temperature and pressure. If molecular weight of B is 64, molecular weight of A is
(A) 16 (B) 32
(C) 64 (D) 128
- The mole of water present in 1 lit of air at 20°C and 255% relative humidity. The vapour pressure of water at 20°C is 17.5 torr is
(A) 4 × 10–2 mole (B) 4 × 10–4 mole
(C) 4 × 10–6 mole (D) 4 mole
[Given relative humidity = ]
LEVEL – I
- Calculate the rms speed in cm/sec at 25°C of a free electron and a molecule of UF6.
- 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)
- 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 mole of H2 is each flask.
- Level – I Q. 15
- Level – I Q.13
- Level – I Q. 8
- Level – I Q. 2
- Level – 1Q. 15
- A mixture of HCO2H & (CO2H)2 is heated with concentrated H2SO4. The gas produced is collected and on its treatment with KOH solution the volume of the gas decreased by one sixth. Calculate the molar ratio of the two acids in the original mixture.
- 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.
- 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.
- 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.
- 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?
- 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 – III Q. NO.7
LEVEL – II
- Argon has TC= –122°C PC = 48 atm. What is the radius of the argon atm?
- For Vander Waal’s gas, show Boyle temperature (where a real gas tends to obey Boyle’s Law) TB= .
- 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.
- 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.
- 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
- Calculate the mean free path of the molecules of hydrogen (σ = 2Å) at very high altitudes where the pressure is 10–4 mm and temperature 0°C.
- A mixture containing 1.12 lit D2 and 2.24 lit H2 at N.T.P is taken inside a bulb connected to another bulb through a stop cock with a small opening. The 2nd bulb is fully evacuated. The stop cock is opened for a certain time and then closed. The first bulb is now found to contain 0.1g of H2. Determine the % of weight of the gases in 2nd bulb.
- Assuming O2 molecule spherical in shape and occupying the radius 150 pm. Calculate
- a) the volume of single molecule of gas
- b) the percentage of empty space in one mole of O2 at NTP
- c) Comment on percentage of empty space.
- 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.
- Level – II Q. No. 3
- Level – III Q. No.9
- Level – III Q. No.13
- 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?
- The densities of graphite and diamond are 2.25 and 3.51 gms/ml. The ΔGf0 values 0J mole–1 and 2900 J mole–1 for graphite and diamond respectively. Calculate the equilibrium pressure for conversion of graphite into diamond at 25°C.
- Level –III Q.4.
LEVEL – III
- 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?
- 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.
- A half litre bomb at 27°C is filled with C2H4 at 1 atm and oxygen at 5 atm and the mixture is exploded. Calculate the final pressure if final temperature is (a) 27°C; (b) 127°C. The aqueous tension at 27°C is 27 mm.
- Calculate the relative rates of diffusion for 235UF6 & 236UF6 in gaseous form. Also if naturally occurring uranium ore having U235 & U238 in the ratio 0.l72 & 99.28% and if it is desired to enrich the U235 to 10% of the sample making use of relative rates of diffusion of UF6 having U235 and U238 isotopes, how many diffusion stages are required?
- 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).
- An organic compound CxH2yOy was burnt with twice the amount of O2 needed for complete combustion to CO2 and H2O. The hot gases when cooled 0°C and 1 atm measured 22.4 litre, the vapour pressure of pure water at 20°C is 17.5 mm of Hg is lowered by 0.104 mm when 50g of organic compound is dissolved in 1000g of water. Give the molecular formula of organic compound.
- 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 monometer, 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.
- 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.
- If ideal gas equation is (V – b) = RT.
What will be the ratio of ?
- Level – III Q. No. 8
- Level – III Q. No. 11
- In phosgene gas reaction at 400°C the initial pressure are pCO= 342 mm &
= 352 mm and the total pressure at equilibrium is 440 nm CO + CO2 COCl2. Calculate percentage dissociation of N2 is present in the system with a pressure of 0.4 atm, the total pressure being the same 1 atm.
- 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?
- 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]
Answers to Objective
Level – I
- D 2. A
- A 4. B
- B 6. D
- A 8. C
- C 10. C
- A 12. D
- D 14. D
- D 16. C
- D 18. D
- B 20. C
LEVEL – II
- B 2. B
- A 4. C
- D 6. C
- D 8. C
- C 10. B
- B 12. C
- B 14. C
- A 16. D
- A 18. A
- A 20. C
Answers to Subjective Problems
LEVEL – I
- Mass of electron on atomic weight scale is amu
Mass of 1 mole of electron =
We have C = =
∴ r.m.s. speed of an electron = 1.16 × 107 cm/sec
To calculate rms speed of UF6 put M = 238 + 6 × 19 = 352
- Let the volume of each bulb be V lit
For two connected bulbs P = 0.5 atm. n = 0.7 mole
T = 300 K. Volume = 2V
We have PV = nRT
0.5 (2V) = 0.7 × 0.082 × 300
or V = 17.22 lit
When one of the bulbs is maintained at 127°C i.e. 400K & other at 300 K let the moles of H2 in these bulbs be n1 & n2 respectively.
n1 + n2 = 0.7 …(1)
Since stopcock is open, the pressure in each bulb
Will be same let it be p atm. Thus for bulb at 400K
pV = n1 × 0.082 × 400
or 17.22 p = 32.8 n1 or p = 1.9 n1 …(2)
For 2nd bulb at 300 K
p × 17.22 = n1 × 0.082 × 300
p = 1.42 n2 …(3)
From equation (1), (2) and (3) p = 0.571 atm
n1 = 0.3 mole n2 = 0.4 mole
- HCO2H ⎯→ H2O + CO
a mole a mole
a mole a mole
Decomposition of HCO2H (a mole) and H2C2O4 (b mole)
H2C2O4 ⎯→ H2O + CO + CO2
b b b mole no.
b b b volume after decomposition
H2O is absorbed by H2SO4 and CO2 is absorbed by KOH
Thus as given,
or = 4
The molar ratio of HCO2H & H2C2O4 = 4 :1
- On combustion in O2 all carbon and hydrogen atms of the compound are converted into CO2 and H2O respectively.
44 gm CO2 having 12 gm C
∴ 0.195 gm CO2 having 0.0531 gm C
Similarly 0.0804 gm H2O contain 0.0089 gm H2.
Hence 0.22 gm compound contain 0.0531 gm C, 0.0089 gm H2 & 0.158 gm H2
∴ % of C = 24.13%; H = 4.04% Cl = 71.8%
or V weights = 12 gm
22400 ml weights 98.5 gm
The empirical formula is CH2Cl
Molecular formula (CH2Cl)n
Let volume of 0.12g compound occupies V ml at N.T.P. then
Molecular weight = 98.5
∴ 12 n + 2n + 35.5 n = 49.5n = 98 or n = 2
∴ molecular formula = C2H4Cl2
- Reactions are CH4 + 2O2 ⎯→ CO2 + 2H2O
C2H4 + 3O2 ⎯→ 2CO2 + 2H2O
Let volume of CH4, C2H4 x and y ml
∴ volume of CO2 = 10 – (x + y)ml
1st volume contraction = (x + 2y + y + 3y) – (x + 2y) = 2x + 2y
By question, 2x + 2y = 17
or 2x = 17 – 8 = 9 or x = 4.5
Because y = 4 [total volume of CO2 = x + 2y + 10 – x – y = 14]
∴ y = 4
Hence volume of C2H4 = 4.5 ml
Volume of CH4 = 4 ml
Volume of CO2 = 1.5 ml
% of CH4 = = 45%
% of C2H4 = = 40%
% of CO2 = = 15%
- Let the volume of gas mixture at N.T.P. be V ml
V ml = 2730 ml
According to problem
100 ml mixture contain 20 ml methane 60 ml CO and 20 ml H2
∴ Volume of methane in 2730 ml of mixture = = 546 ml
Volume of CO= = 1638 ml
Volume of Hg = (2730 – 546 – 1638) = 546 ml
The reactions are
CH4 + 2O2 ⎯→ CO2 + H2O …(1)
2CO + O2 ⎯→ 2CO2 …(2)
2H2 + O2 ⎯→ 2H2O …(3)
∴ Total volume of O2 required for oxidation of
546 ml CH4, 1638 ml CO & 546 ml H2 are
= ml = 2184 ml
Again 2KClO3 ⎯→ 2KCl + 3O2
3 × 22.4 lit of O2 at NTP are obtained from 245 gm of KClO3
2.184 lit of O2 at NTP are obtained from of KClO3 = 7.96g of KClO3
- Mg + 2HCl ⎯→ MgCl2 + H2
24g 22.4 lit at N.T.P.
2Al + 6HCl ⎯→ 2AlCl3 + 3H2
2 × 27 = 54g 3 × 22.4 lit at N.T.P.
Let V be volume of liberated H2 at N.T.P.
∴ or V = 1103.7 ml
Let weight of Mg in alloy be x gm ∴ wt. of Al = (1 – x)g
∴ By problem, (1 – x) = 1103.7
∴ x = 0.4525, % of Al = 54.75%
∴ Wt. of Mg = 0.4525g
Wt. of Al = (1 – 0.4525)g = 0.5475gm
% of Mg = 45.25
- When the mixture of gases is passed over hot copper, N2 will remain unchanged but NO will be reduced by Cu to N2
2Cu + 2NO ⎯→ 2CuO + N2
Let x cc of N2 be present in 25cc mixture. Hence the volume of NO in the mixture is (25 – x)cc
From the equation under same conditions of temperature and pressure 2 volume of NO gives 1 volume of N2
(25 – x)cc NO will give cc of N2
Total volume after the reaction = cc
By question, = 20 or x = 15cc
So the mixture contains 15cc N2 and (25 – 15) or 10cc NO
Hence percentage of N2 = = 60%
Percentage of NO = = 40%
LEVEL – II
- b =
r = = 1.47 × 10–8 cm
- N2O4 2NO2
1 – α 2α Total no. of moles = 1 + α
The calculated density of N2O4 = 46
Observed density ρ 30.2
Hence α = = = 0.523
The fraction of gas molecules decomposed = 0.523
% of NO2 molecules by weight = 52.3%
Again in the mixture the ratio of gm moles of NO2 and N2O4 is
∴ % of NO2 by volume = × 100 = 68.7%
- At Boyle point TB at P = 0
= 0 …(1)
Vander Waals equation P =
Using equation (1)
Since when P → 0 the volume V will be infinity large
i.e. V – b ≈ V ∴ TB =
- Let V be the volume diffusing out in each case
Let dm.dm and d be the densities of the mixture of pure oxygen and ozone respectively.
∴ dm = 16 ≈ 21.91
Again for 100 gms the volume =
∴ d = 23.3
- From the extrapolation of vs P curve to P = 0 the value of is found to be 1.517 gm lit–1.
∴ = 1.517 × 0.082 × 273 = 34
∴ The atomic weight of P = 34 – 3 × 1.008 = 30.98
- = 3.54 × 1012 molecules/cc
= mean free path
= = 5.9 × 102 cm
|7.||At STP before diffusion D2 = 1.12 lit = 0.2 g
H2 = 2.24 lit = 0.2g in I bulb
After diffusion H2 = 0.1g
H2 diffuses from I = 0.1g
Now for diffusion D2 + H2
∴ Wt. of gases in II bulb = Wt. of H2 + Wt. of O2
= (0.1 + 0.14)g = 0.24g
% of D2 by weight = = 58.33%
% of H2 in bulb II = 41.67%
- a) Volume of 1 molecule = πr3
( r = 150 pm = 150 × 10–10 cm)
Volume of 1 molecule =
V1 = 1.41 × 10–23 cc/molecule
- b) The volume occupied by N molecules = N × V1
= 6.023 × 1023 × 1.41 × 10–23
= 8.49 cc per mole
∴ Volume of 1 mole of N2 = 22400 ml of STP
Thus empty space = (22400 – 8.49) = 22391.51cc
% empty space = × 100 = 99.96%
- i) CxHyNz + O2 ⎯→ CO2+ H2O (Vapour) + N2
(suppose the compound is CxHyNz)
∴ (9 – V) volume V volume 4 volume 6 volume 2 volumes
(9 – V) mole V moles 4 moles 6 moles 2 moles
By 2V = 2 × 4 + 1× 6 = 14 or, V = 7 volume
- ii) CxHyNz + O2 ⎯→ CO2 + H2O (vapour) N2
2 moles 7 moles 4 moles 6 moles 2 moles
∴x × moles of CxHyNz = V moles of CO2
x + 2 = 1 × 4 x = 2
xyz moles of CxHyNz = 2 × moles of H2O (vapour)
x moles of CxHyNz = 2 × moles of N2
z × 2 = 2 × 2 , z = 2
Here the compound is C2H6N2
- Applying PV = nRT
for NO, × 0.25 = n1RT ∴n1 =
for O2, = n2RT n2 =
Since NO & O2 react in 2 : 1 molar ratio
2NO + O2 ⎯→ N2O4
Here O2 is the limiting reagent
Mole of NO remain after completion of reaction
The pressure due to remaining NO is
p = or p = 0.30 atm
Wt. of N2O4 = mole of N2O4 × 92
= = 0.402g
- Where, Cgraphite ⎯→ Cdiamond
ΔG0 = ΔG(diamond) – ΔG(Graphite)
= 2900 mole–1
or ΔG2 – ΔG1 = ΔV (p2 – p1)
or p2 =
or p2 = pa
= 1.5 × 109pa
LEVEL – III
- Applying PV = nRT =
or P = where S = solubility
∴ = constant for a gas at a given temperature
Given that P1 = 760 mm, P2 = 740 mm
S2 = = 0.97S1
For oxygen solubility at 740 mm = 0.97 × 0.048 = 0.0465 g ml–1
= 0.0465 g ml–1
For nitrogen solubility at 740 mm = 0.97 × (0.02)2 = 0.0213 g ml–1
For CO solubility at 740 mm = 0.97 × 0.03 = 0.029 g ml–1
∴ Amount of O2 dissolved in 200 ml of water = (0.0465 × 0.2) = 0.0093 g ml–1
Amount of N2 dissolved in 200 ml of water = 0.0213 × 0.2 = 0.00426 g ml–1
Amount of CO dissolved in 200 ml of water = 0.029 × 0.2 = 0.0058 g ml–1
- The number of molecules per cc at 25°C & 1 atm
n = = 2.46 × 1019
The average velocity CAV = = 4.8 ×104 cm/sec
Hence no. of binary collisions per cc per sec
= × (3.74 × 10–8)2 × 4.8 ×104 × (2.462 × 1019)2
= 8.98 × 1028
- Suppose the number of moles of CxH2yOy is m & total volume of O2 used is 2V litre at 0°C and 1 atm pressure
∴Volume of O2 reacted = v lit
Volume of O2 remained = v lit
Volume of O2 unreacted + Volume of CO2 formed = 2.24 lit
Volume of CO2 formed = (2.24 – V) lit
CxH2yOy + O2 ⎯→ H2O + CO2
m moles V lit 0.9g (2.24 – V) mole
m moles mole 0.05 mole moles
From question, x × m = …(1)
2y × m = 2 × 0.05 = 0.1 …(2)
y × m + 2 × = 0.05 + 2 × …(3)
from (1) & (3) v = 1.12
Putting the value of V, x × m = 0.05 …(4)
From (4) & (2) = 1 …(5)
From Raoult’s Law
or or M = 150
∴ 12x + 18y = 150 …(6)
∴ from equation (5) & (6) x = 5, y = 5
- Observed pressure drop to mm Hg 37 mm = 2.8 mm Hg in 30 min, 56 mm Hg drop per hour with 5.6 mm Hg as the oxygen partial pressure change to STP.
V2 = = 0.1 cm3 = 1 × 102 mm3
- Volume of flask A =
Volume of balloon B =
Volume occupied by H2 in A =
= 3.284 m3 = 3284 lit
Pressure in A = = 1.123 atm
Pressure in B = 1.123 atm
Volume of N2 in B =
= 0.7917 m3 = 791.7 lit
Moles of N2 = = 36.14
Mass of N2 = (36.14 × 28) = 1011.92g
Mass of O2 in C = = 165.12g
Total weight of all the gases (165.2 + 300 + 1011.92)g = 1477.12g
Total pressure inside the flask of –78°C after the balloon cracks up
= = 0.73 atm
- (V – b) = RT
PV = RT + Pb –
≈ neglecting last term as very small
or, PV = RT
Differentiating this equation w.r.t. pressure and equating the same to zero
Hence, = 1 or = 2.45
- a) Let x be the partial pressure of phosgene gas at equilibrium. At equilibrium then
pCO = (342 – x) mm = (352 – x) nm = x mm
∴ (352 – x) + (342 – x) + x = 440
or, x = 254 mm = 0.334 atm
∴ pCO = 342 – 254 = 88 mm = 0.116 atm
= 352 – 254 = 98 mm = 0.128 atm
The equilibrium constant for dissociation of COCl2
COCl2 CO + Cl2
1 – α α α
kp = = 4.4 × 10–2, Kp = = 4.4 × 10–2
Since P = 1, α = 0.206 ∴ % dissociation = 20.6
- b) When inert N2 is present with = 0.4 atm
The total pressure of reactants R resultants at equilibrium = 0.6 atm
Kp = α1 = degree of dissociation in presence of N2
∴ = 4.4 × 10–2
α1 = 0.26, α1 = 26%
The presence of N2 has enhanced extent of dissociation
- For CO2 output rate = 44 g per hour
= 1 mole/hr
= 22.4 lit/hr at STP
CO2 reduction rate = 600 ml/min
600 × 60 ml/hr
= 36 lit/hr
Fraction of time by which converter has to be operated =
= = 0.622
- We have
= 6.2 × 10–14 ergs
dE = 0.01 E = 6.2 × 10–16 ergs
= × e–3/2 × (6.2 × 10–16)
= 0.0046 ∴ 46%
level – I
- At what temperature the volume of a given mass of an ideal gas will be thrice of that at 27°C
- Calculate the weight of CH4 in a 9 litre cylinder at 16 atm and 27 oC
- Calculate the average speed of CO at 100 oC
- 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.
- 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.
- 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.
- 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.
- 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
- 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
LEVEL – III
- 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.
- 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.
- 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?
- 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?
- 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
- 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
- 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
- B 10. B
- A 12. D
- A 14. B
- C 16. B
- A 18. A
- D 20. C