**1) 5g of an unknown gas has a pressure P at temperature TK in a vessel. On increasing the temperature by 50 ^{0}C, 1g of the gas was given out to maintain the pressure P. The original temperature was**

(a) **1132 K** (b) **100 K** (c) **200 K** (d) **273 K**

(c) i) PV = (5/M) RT

ii) PV = (4/M) R(T+50)

∴ T = 200 K

**2) The van der Waals’ equation for 0.5 mol of gas is **

(a) **(p + a/4V ^{2})(V-b)/2 = 2RT/2** (b)

**(p + a/4V**

^{2})(2V-b) = RT(c) **(p + a/4V ^{2})(2V-4b) = RT** (d)

**(p + a/4V**

^{2}) = 2RT/2(V-b)(b) (P + an^{2}/V^{2})(V-nb) = nRT

For n = 1/2 , (P + a/4V^{2}) (V – b/2) = 1/2 RT

(P + a/4V^{2}) (2V-b) = RT

**3) The compressibility factor for a real gas at high pressure is:**

(a) **1** (b) **1+(Pb/RT)** (c) **1-(Pb/RT)** (d) **1+(RT/Pb)**

(b) Van der Waals’ equation for one mole of a real gas is (P + a/V^{2})(V-b) = RT

Or PV = RT + Pb + ab/V^{2} – a/V

At high pressures, the van der Waals’ gas equation reduces to PV = RT + Pb

Or PV/RT = 1 + (Pb/RT)

∴ Compressibility factor Z = PV/RT = 1 + Pb/RT

**4) Consider the arrangement of bulbs shown below at constant temperature.What is the pressure of the system when all the stop cocks are opened? Temperature remains constant**

(a) **0.56 atm** (b) **0.65 atm** (c) **1 atm** (d) **1.5 atm**

(a) Total number of moles = (635/760)(1/RT)+(212/760)(1/RT)+(418/760)(0.5/RT)

Let the pressure of gases be P atm when all the stop cocks are open

P(1+1+0.5) = [(635/760)(1/RT)+(212/760)(1/RT)+(418/760)(0.5/RT)] RT

P = 0.56 atm

**5) A mixture of N _{2(g)} , Ne_{(g)} ,and water vapour are taken in a vessel of volume V. The molar ratio of N_{2} and Ne is 1:1. The pressure due to water vapour is found to be 0.2 atm and total pressure of the mixture is 1.0 atm. If all the contents of the mixture are transferred to another vessel whose volume is v/3 at same temperature, then the total pressure in the second vessel is**

(a) **3.0 atm** (b) **1.3 atm** (c) **2.6 atm** (d) **2.0 atm**

(c) P_{H2O} = 0.2 atm

P_{N2} = P_{Ne} = 0.4 atm

P_{1}V_{1} = P_{2}V_{2}

P_{1}V_{1} = P_{2}V_{2 }(V_{1}/3)

∴ P_{2} = 3P_{1}

∴(P_{2})N_{2} = (P_{Ne}) = 1.2

P_{total} = 1.2+1.2+0.2

= 2.6 atm

**6) One mole of nitrogen gas at 0.8 atm takes 38 second to diffuse through a pin hole whereas one mole of an unknown compound of xenon with fluorine at 1.6 atm takes 57 second to diffuse through the same hole. The molecular formula of the compound is **

(a) **XeF _{6}** (b)

**XeF**(c)

_{4}**XeF**(d)

_{2}**XeF**

_{8}(a) r_{N2}/r_{XeFn} = PN_{2}/P_{XeFn }= √M_{XeFn}/√M_{N2}

r_{N2}/r_{XeFn} = (Molar volume/t_{N2})/(Molar volume/t_{XeFn})

= t_{XeFn}/t_{N2} = 57/38

∴ 57/38 = 0.8/1.6 (√M_{XeFn}/√28)

M_{XeFn} = (57/38)^{2}×(1.6/0.8)^{2}×28 = 252

Xe+nF = 252

131+n×19 = 252

n = 6

Molecular formula = XeF_{6}

**7) Pure O _{2} diffuses through an aperture in 224 seconds, whereas mixture of O_{2} and another gas containing 80% O_{2} diffuses from the same in 234 sec under similar condition of pressure and temperature. The molecular mass of gas will be **

(a) **46.6** (b) **34.9** (c) **60.0** (d) **none of these**

(a) M_{mix} = (M_{1}n_{1} + M_{2}n_{2})/(n_{1}+n_{2})

r_{1}/r_{2} = (V/t_{1})/(V/t_{2}) = √M_{2}/√M_{1}

⇒ t_{2}/t_{1} = √M_{2}/√M_{1}

234/224 = √M_{mix}/√32

⇒ M_{mix }= 34.92 gm

34.92 = [M_{1}(20)+32×(80)]/100

⇒ M_{1} = 46.6

**8) If volume occupied by CO _{2} molecules is negligible, then the pressure exerted by one mole of CO_{2} gas in terms of temperature (T), assuming V to be single valued, is**

(a) **P = RT/4a** (b) **P = RT/(4×a-b)** (c) **P = R ^{2}T^{2}/4a** (d)

**R**

^{2}T^{2}/4ab(c) (P + a/V^{2})(V) = RT

PV^{2}-RTV+a = 0

V = [+RT±√(RT)^{2}-4aP]/2P

V is single valued, (RT)^{2}-4aP = 0

P = R^{2}T^{2}/4a

**9) The heat capacity ratio ‘γ’ can be determined by a method based on adiabatic expansion of the gas. For a gas initial temperature, pressure and volume are T _{1}, P_{1} and V_{1}, while the same in final state are T_{2}, P_{2} and V_{2}. Which of the following is correct expression for the determination of γ. **

(a) **ϒ = (logT _{1}-logT_{2})/(logV_{2}-logV_{1})** (b)

**ϒ = (logV**

_{2}-logV_{1})/(logT_{1}-logT_{2})(c) **ϒ = (logP _{1}-logP_{2})/(logV_{2}-logV_{1})** (d)

**ϒ = 1+(logP**

_{1}-logP_{2})/(logV_{2}-logV_{1})(c) PV^{γ} = constant

P_{1}V_{1}^{γ} = P_{2}V_{2}^{γ}

P_{1}/P_{2} = (V_{2}/V_{1})^{γ}

⇒ V_{2}/V_{1} . T_{1}/T_{2} = (V_{2}/V_{1})^{γ}

log P_{1}/P_{2} = γ log (V_{2}/V_{1})

⇒ T_{1}/T_{2} = (V_{2}/V_{1})^{γ-1}

γ = (log P_{1} – log P_{2})/(log V_{2} – log V_{1})

γ-1 = (log T_{1} – log T_{2})/(log V_{2} – log V_{1})

γ = [(log V_{2} – log V_{1})+(log T_{1} – log T_{2})]/(log V_{2} – log V_{1})

**10) J.H. Vander Waals introduced an approximate equation of state. This equation was proposed on the basis of experimental evidence available to him in conjunction with rigorous thermodynamic arguments. Which of the following is not an expression of Vandeer Waals equation of state? (a and b are constant and V _{m} is the molar volume). **

(a) **P = (nRT)/(V -nb) – a(n/V) ^{2}** (b)

**P = RT/(V**

_{m}-b) – a/(TV_{m}^{2})(c) **Z = V _{m}/(V_{m}-b**

**) – a/(V**(d)

_{m}RT)**V**

_{m}^{3}-(b +**RT/P)V**_{m}^{2}+(a/P)V_{m}– ab/P = 0(b) Vander Waals equation

(P + n^{2}a/V^{2})(V-nb) = nRT for n mols

(P + a/V_{m}^{2})(V_{m}-b) = RT for 1 mol

⇒ P = RT/(V_{m}-b) – (a/V_{m}^{2})

Multiplying by m V_{m}/RT

PV_{m}/RT = V_{m}/(V_{m}-b) – a/V_{m}RT = Z

Equation (B) is not the same expression (In fact, it is the Berthelot’s equation of state).

**11) For a gas which of the following statements are correct**

(a) **Inversion temperature is double for Boyle temperature**

(b) **Boyle temperature is double for inversion temperature**

(c) **Critical temperature is less than inversion and Boyle temperature**

(d) **Inversion temperature is more than Boyle temperature and critical temperature**

(a, c, d) In version temperature T_{i} = 2a/Rb

Boyle’s temperature T_{B} = a/Rb = T_{i }= 2×T_{B}

Critical temperature T_{C} = 8a/27Rb

Inversion temperature T_{i} = 2a/Rb

Critical temperature is less than Inversion temperature Inversion tempature more than critical temperature and Inversion temperature.

**12) Two contains have the same volume and have ideal gases A and B in them. The rms velocity of A is twice of B. Which of these conditions could justify it? **

(a) **Molar mass of A is half of B. Temperature and pressure same**

(b) **Temperature of A is four times of B. Molar mass and pressure same **

(c) **Temperature of A twice of B, molar mass of A half of B and pressure unknown**

(d) **Molar mass of B is four times of A and Pressure of A is twice of B with temperature same**

(b, c, d) C_{rms} = √3RT/√M

C_{1} = 2C_{2}

√3RT_{1}/√M_{1} = 2(√3RT_{2}/√M_{2})

3RT_{1}/M_{1} = (4×3RT_{2})/√M_{2})

⇒ T_{1}/M_{1 }= 4T_{2}/M_{2}

Hence if T_{1 }= 4T_{2} or T_{1 }= 2T_{2} and M_{1} = 1/2 M_{2} or M_{2} = 4M_{1} the equation would be satisfied.

**13) You have two gases, A and B, in two separate containers of equal volume and at equal temperature and pressure. Therefore the correct statements are **

(a) **You must have the same number of moles of each gas**

(b) **Because the two temperatures are equal, the average kinetic energies of the two samples are equal**

(c) **Since the energy of such a system corresponds to translational motion, the root mean square velocities of the two are equal and thus the particles in each sample move, on average, with the same relative speed**

(d) **Since A and B are different gases, each must have different molar mass. If A has a higher molar mass than B, the particles of A must be hitting the sides of the container with more force. Thus the pressure in the container of gas A must be higher than that in the container with gas B. Hence our assumption that the pressure is equal cannot be true. **

(a, b)

(a) Since temperature, pressure and volume are same the number of moles are same.

(b) The average kinetic energies depend only on temperature hence are same.

(c) Same kinetic energies do not mean same velocity. Mass is different hence velocity will be different.

(d) The above statement is correct as velocities are not same. Mass and velocity compensate each other.

**14) 0.7 g of Na _{2}CO_{3} . xH_{2}O is dissolved in water and the volume made up to 100 mL, 20 mL of it requires 19.8 mL of N/10 HCl for complete neutralisation. the value of x should not be is : **

(a) **7** (b) **3** (c) **2** (d) **5**

(a, b, d) No. of meq of Na_{2}CO_{3}. xH_{2}O in 20 ml

= 19.8×(1/10 =1.98

No. of meq in 100 ml = 1.98 × 5 = 9.9

i.e. (mass/eq. mass)×1000 = 9.9

⇒ 0.7/(M/2) = 9.9

M = 141.40 (molar mass of Na_{2}CO_{3 }. xH_{2}O)

i.e. 106 + 18 x = 141.40

x = 2

**15) Consider two gases, A and B each in a 1.0 L container with both gases at the same temperature and pressure. The mass of gas A in the container is 0.34 g and the mass of gas B in the container is 0.48 g. Choose the correct statements**

(a) **Both samples have the same number of molecules**

(b) **Both samples have same average kinetic energy**

(c) **Both samples have the same average velocity**

(d) **Both samples have the same total kinetic energy**

(a, b, d)

(a) Since the volume, pressure and temperature are same, the number of moles are same.

(b) The average kinetic energy depends only on temperature hence remains the same

(c) Molar mass of the two gases are different hence the velocity will not be the same

(d) Since the moles and temperature are same, the total kinetic energy is same.

**16) An equimolar binary/ternary solution of A, B and C (all are volatile liquids) is made at 25°C. Assuming all solutions to be ideal in nature, which of the following value of vapour pressure is possible for their solutions if vapour pressure of A, B and C respectively are 50, 25 and 10 at 25°C. **

(a) **37.5** (b) **17.5** (c) **40** (d) **28.3**

(a, b, d)

Possible Binary solutions are solutions of A & B, B & C and A & C

Possible Ternary solution is solution of A, B & C.Since the solutions are equimolar so mole fraction of each component in binary solution is 1/2 and in Ternary solution is 1/3 for

For solution of A & B, P = 1/2(50)+1/2(25) = 37.5 22

For solution of B & C, P = 1/2(25)+1/2(10) = 17.5

For solution of A & C, P = 1/2(50)+1/2(10) = 30

For solution of A, B & C, P = 1/2(50+25+10) = 28.3

**17) AT 55°C, ethanol has a vapour pressure of 168 mm of Hg and that of methyl cyclohexane is 280 mm of Hg. A solution of the two with mole fraction of ethanol 0.68 has a total vapour pressure of 376 mm of Hg. Which of the following is true if this solution forms from its components with? **

(a) **Absorption of heat** (b) **The solution shows an appreciable positive deviation**

(c) **ΔV _{mixing} = +ve** (d)

**ΔV**

_{mixing}= -ve(a, b, c)

P_{total} = P_{E}ºX_{E}+P_{H}ºX_{M} =(168) (0.68) + (280) (0.32)= 203.84 mm of Hg

(P_{total})_{expected} = 203.84 mm of HG

(P_{total})_{actual }= 376 mm of Hg

_{(Ptotal)actual }> (P_{total})_{expected}

Solution shows positive deviation from Raoult’s law

Thereby, ΔH_{solution} > 0 & ΔV_{mixing} > 0

**18) A graph is plotted with temperature of a solution containing benzene and toluene as a function of mole fraction. Choose the correct options. **

(a) **a → b represents evaporation** (b) **b → c represents condensation**

(c) **c → d represents evaporation** (d) **c → d represents condensation **

(a, b, c)

In graph ‘a’ & ‘b’ represent composition of benzene-toluene mixture in liquid phase and vapor phase respectively which are in equilibrium with each other at a given temperature.

If vapors with composition ‘b’ are taken and condensed in another beaker then in liquid phase benzene toluene mixture will have composition ‘c’ which is same as that of ‘b’.

Composition ‘c’ & ‘d’ represent another composition of benzene toluene mixture in liquid phase and vapor phase respectively which are in equilibrium with each other at a given temperature.

**19) Consider following solutions: **

**I. 1M aqueous glucose solution **

**II. 1M aqueous sodium chloride solution **

**III. 1M aqueous ammonium phosphate solution **

**IV. 1M benzoic acid in benzene Select correct statements for the above solutions:**

(a) **All are isotonic solutions** (b) **III is hypertonic of I, II and IV**

(c) **IV is hypotonic of I, II and III** (d) **II is hypotonic of III but hypertonic of I and IV **

(b, c, d)

π = iCRT

π_{I} = (1) (1) (RT) = RT

π_{II} = (2) (1) (RT) = 2RT

π_{III} =(4) (1) RT) = 4RT

π_{IV} = (1) (RT) = RT/2

(a) All solutions have different Osmotic pressure and hence are not isotonic.

π_{III }> π_{II} > π_{I} > π_{IV}

**20) Freezing point lowering expression is ΔT _{f} = K_{f}m (molality)**

**Which of the following assumptions are considered for the validity of above equation?**

(a) **The solution is dilute**

(b) **The ΔH _{f }(latent heat of fusion of solvent) is independent of temperature between the actual and normal freezing point.**

(c) **The solid phase consists of pure solvent when solution is allowed to cool.**

(d) **ΔT _{f} is not equal to 3K_{f} for 3 mol L^{-1} solution**

(a, b, d)

(a) Raoult’s law for solution of non volatile solute in volatile solvent is valid for ideal solutions which are infinitely dilute solutions.

(b) K_{f} = (MRT_{f}^{2})/(ΔH_{f}×1000)

Cryoscopic constant is considered to be constant for a range of temperature between actual and normal freezing point.

(c) This is not an assumption. It is a fact that only solvent solidifies on freezing.

(d) ΔT_{f} ≠ 3K_{f}

as solution cannot be 3 molal (it is a concentrated solution)

**21) In an insulated container 1 mole of a liquid, molar volume 100 ml is at 1 bar. Liquid is steeply taken to 100 bar, when volume of liquid decreases by 1 ml then value of ΔH is found to be 1100x bar ml then find x? **

Volume of 1 mole of liquid = 100ml at 1 bar

Volume of 1 mole of liquid = 99ml at 100 bar

The process is irreversible as it is steeply changed from 1 bar to 100 bar

Therefore w = -p(v_{2} –v_{1}) = -100(99-100) = 100 bar ml

Also ΔU = q + w

q = 0 since container is insulated

⇒ ΔU = 100 bar ml

⇒ ΔH = ΔU + Δ(PV) = ΔU + (P2V2 – P1V1) = 9900 bar ml = 1100x bar ml

⇒ x = 9

**22) A hydrocarbon contains 14.3% of hydrogen. 10 ml of same gaseous hydrocarbon requires exactly 75 ml of oxygen for complete combustion. What is the number of carbon atoms in the hydrocarbon. **

H – 14.3%, C = 85.7%

∴ Emperical formula is CH_{2}

(CH_{2})_{n} + (3n)/2 O_{2} → nCO_{2}+nH_{2}O

1 mL reacts with 3n/2 mL

10 mL reacts with (3n/2)10

∴ 3n .10 = 75

n = 150/30 = 5

**23) At 77°C and one atmospheric pressure, N _{2}O_{4} is 70% dissociated into NO_{2}. What will be the volume occupied by the mixture under these conditions if we start with 10 g of N_{2}O_{4}. **

Molar mass of N_{2}O_{4} = 28 + 64 = 92 g mol^{–1}

N_{2}O_{4} ↔ 2NO_{2}

= (10/92)-(70/100)×(10/92) , 2×0.076

= 0.109 – 0.076 , = 0.152

= 0.033

Total mass after dissociation = 0.033 + 0.152 = 0.185

T = 77°C

T =77 + 273 K = 350 K

PV = nRT

V = nRT/P = (0.18 mole×0.0821 L atm K^{-1} mol^{-1}×350 K)/1 atm = 5.32 L

**24) A mixture of 20 ml of CO, CH _{4} and N_{2} was burnt in excess of O_{2} resulting in contraction of 13 ml of volume. The residual gas was then treated with KOH solution to show a contraction of 14 ml in volume. Calculate the volume of N_{2} ?**

CO = a ml

CH_{4} = b ml

N_{2} = c ml

a + b + c = 20 ml . . . . .(i)

CO + 1/2 O_{ 2} → CO_{2}

Vol. of CO_{2} = a,

CH_{4} + 2O_{2} → CO_{2} + 2H_{2}O(l)

vol. of CH_{4} = b

vol. of CO_{2} = b

N_{2} + O_{2} → No reaction

Volume of CO_{2} formed = volume of CO_{2} absorbed by KOH a + b = 14 ml . . . . . (ii)

∴ a + b + c + vol. of O_{2} taken – vol. of O_{2} left – (a + b) – c = 13

Vol. of O_{2} used = 13

(a/2)+2b = 13 . . . . . . (iii)

Solving (i), (ii) and (iii)

a = 10 ml, b = 4 ml, c = 6 ml

**25) Freon-12 is used as a refrigerant in central home air conditioners. The rate of effusion of Freon-12 to Freon-11 (molar mass =137.4 g/mol) is 1.07 : 1. The formula of Freon-12 is one of the following : CF _{4}, CF_{3} Cl, CF_{2}Cl_{2}, CFCl_{3}, or CCl_{4}. Find the number of fluorine atoms in a molecules of Freon-12. **

Let the formula of freon-12 be CF_{x}Cl_{y}

x + y = 4

Molar mass of freon-12 = 12 + 19x + 35.5y

r_{Freon11}/r_{Freon12} = 1/1.07

= √(12+19x+35.5y)/√(137.4)

Squaring 1/1.1449 = (12+19x+35.5y)/(137.4)

= 19x+35.5y = (137.4/1.449) – 12

Substituting y = 4-x

19x+35.5(4-x) = 108

19x+142-35.5x = 108

-16.5x = -34

x ∼ 2

**26) The stopcock, connecting the two bulbs of volumes 5 litres and 10 litres containing an ideal gas at 9 atm and 6 atm respectively, is opened. what is the final pressure in the two bulbs if the temperature remained the same ? **

For the first bulb, p_{1}V_{1} = n_{1}RT

for the second bulb, p_{2}V_{2} = n_{2}RT

on adding : p_{1}V_{1 }+ p_{2}V_{2 }= (n_{1}+n_{2})RT…(i)

When the stopcock is opened, the total volume is (V_{1}+V_{2}) and total no. of moles is (n_{1}+n_{2}).

Suppose the equilibrium pressure at each bulb is p atm. Then,

p(V_{1 }+ V_{2}) = (n_{1}+n_{2})RT …(ii)

From equations, (i) and (ii) we have

p_{1}V_{1}+p_{2}V_{2 }= p(V_{1 }+ V_{2})

Given that, p_{1} = 9atm,V_{1} = 5lit, p_{2} = 6atm,V_{2} = 10lit.

substituting : 9×5+6×10 = p×15

p = 7 atm

**27) 0.789 g of crystatlline barium hydroxide is dissolved in water. For the neutralization of this solution, 20 mL of N/4 HNO _{3} is required. How many molecules of water are present in one mole of this base ? (Ba = 137.4, O = 16, N = 14) **

Let the molecular formula be Ba(OH)_{2}. xH_{2}O

Mol. mass of Ba(OH)_{2}.xH_{2}O = 171.4 + 18x

equivalent mass of Ba(OH)_{2}. xH_{2}O = (171.4 18x)/2

20 mL N/4 HNO_{3} ≡ 20mL N/4 Ba(OH)_{2} . xH_{2}O

Amount of Ba(OH)_{2}.xH_{2}O = [(171.4 18x)/2×4]×20/1000 = (171.4+18x)/400 g

Amount of Ba(OH)_{2}.xH_{2}O = 0.789 g

Hence, (171.4+18x)/400 = 0.789g

or 171.4+18x = 0.789×400

x = 144.2 /18 = 8.01 ≈ 8

Thus, 8 g moles of water molecules are present in one g mole of the base.

**28) 4 g of an ideal gas was introduced into a bulb of volume of 0.821 dm ^{3} at certain pressure, P and temperature T. The above bulb was placed in a thermostat mentioned at temperature (T+125)K of 0.8 gm of the gas was left off to keep the original pressure. Calculate the pressure in atmospheres. [Molecular weight of the gas is 40 g mole^{–1} and R value is 0.0821 lit-atm-k^{–1}mol^{–1}].**

Initial no. of moles = 4/10 = 0.1

Final no. of moles = 3.2/40 = 0.08

since P and V are constant n_{1}T_{1} = n_{2}T_{2}

0.1T = 0.08(T_{1}+125)

T_{1} = 500K

from PV = nRT

P = (0.1×0.082×500)/0.821 = 5 atm

**29) Two flasks A and B have equal volumes. A is maintained at 300 K and B at 600 K. while A contains H _{2} gas, B has an equal mass of CH_{4} gas. Assuming ideal behaviour for the both gases find the ratio of (u_{av})_{A} : (u_{av})_{B}. **

Since (U_{av})_{A} = √8RT_{A}/πM_{A}

= √(8R(300K))/π(0.002kg mol^{-1})

(U_{av})_{B} = √8RT_{B}/πM_{B }= √8RT_{B}/πM_{B}

= √(8R(300K))/π(0.016kg mol^{-1})

Therefore, (U_{av})_{A}/(U_{av})_{B} = (√300/√2)×(√16/√600) = 2

**30) 5 mL of a gaseous hydrocarbon was exposed to 30 mL of O _{2}. The resultant gas, on cooling is found to measure 25 mL, of which 10 mL was absorbed by NaOH and the remainder by pyrogallol. All measurements are made at constant pressure and temperature. The sum of number of carbon and hydrogen atom in the hydrocarbon is **

Let the formula of hydrocarbon be C_{x}H_{y}.Its combustion reaction is given as

C_{x}H_{y}(g) + (x + y/4)O_{2}(g) → xCO_{2}(g)+y/2 H_{2}O(l)

Initial vol. in ml 5 30 0 0

Final volume 0 30-(x + y/4)5 5x 0

Out of 25 ml of resultant gas 10 ml was CO_{2} (absorbed by NaOH) and the remaining 15ml was oxygen (absorbed by pyrogallol).

5x = 10 ; x = 2

30-(x + y/4)5 = 15

On solving , y = 4

formula of gaseous hydrocarbon is C_{2}H_{4}