(a) (i) Calculate the coefficient of viscosity of air at temperatures : (I) 298 K and (II) 0 K. Assume that the collision cross-section (πσ²) of air is 0·28 (nm)² and average molar mass of air is 29 g mol⁻¹. (10 marks)

(ii) Arrange Boyle's temperatu

Question

(a)

(i)

Calculate the coefficient of viscosity of air at temperatures : (I) 298 K and (II) 0 K. Assume that the collision cross-section (πσ²) of air is 0·28 (nm)² and average molar mass of air is 29 g mol⁻¹. 10 marks

(ii)

Arrange Boyle's temperature of the gases Ar, CH₄ and C₆H₆ in increasing order. Give reason(s) for the answer. 5 marks

(ii) Arrange Boyle's temperature of the gases Ar, CH₄ and C₆H₆ in increasing order. Give reason(s) for the answer. (5 marks)

(b) (i) Which of the following liquids has greater surface tension : Ethanol or Dimethyl ether. Explain the answer with reasons. (5 marks)

(ii) Calculate the difference in pressure across the liquid-air interface for a water droplet of radius 150 nm. (5 marks)

(c) (i) Calculate the change in Helmholtz energy for a reversible isothermal compression of 1 mole of an ideal gas whose volume decreases from 100·0 L to 22·4 L. Assume that temperature is 298 K. (10 marks)

(ii) Why does a tyre get hot when air is pumped into it ? Can a tyre be inflated without a rise in temperature ? (5 marks)

(iii) Calculate the pressure of O₂ (in atm) over a sample of NiO at 25°C if ΔG° = 212 kJ/mole for the following reaction : NiO (s) ⇌ Ni (s) + ½ O₂ (g) (5 marks)

(iv) Estimate the final temperature of one mole of gas at 200·00 atm and 19·0°C as it is forced through a porous plug to a final pressure of 0·95 atm. Given : The Joule-Thomson coefficient (μJT) of the gas is 0·150 K/atm. (5 marks)

UPSC Answer Check · Accepted Answer

(a)(i)
Using kinetic theory of a hard-sphere gas:
η = (1/3) ρ v_avg λ, with λ = 1/(√2 n πσ²). Since ρ = n m,
η = m v_avg/(3√2 πσ²), where v_avg = √(8RT/πM).

Given πσ² = 0.28 nm² = 2.8×10⁻¹⁹ m², M = 29 g mol⁻¹ = 0.029 kg mol⁻¹.
m = M/N_A = 0.029/(6.022×10²³) = 4.816×10⁻²⁶ kg.

At T = 298 K:
v_avg = √(8×8.314×298/(π×0.029)) = 466.4 m s⁻¹.
η = (4.816×10⁻²⁶ × 466.4)/(3√2 × 2.8×10⁻¹⁹)
= 1.89×10⁻⁵ Pa s.

At T = 0 K:
v_avg ∝ √T, so v_avg = 0 and hence η = 0 Pa s by this model. Real air would not remain gaseous at 0 K.

Final: η(298 K) = 1.89×10⁻⁵ Pa s; η(0 K) = 0 Pa s (ideal hard-sphere extrapolation).

(a)(ii)
For a van der Waals gas, B = b − a/(RT). Boyle temperature is where B = 0:
T_B = a/(Rb).

T_B increases with stronger attractive forces relative to molecular size. Ar has only weak dispersion forces. CH₄ has moderate dispersion forces. C₆H₆ has a large, highly polarisable π-electron cloud, giving much stronger dispersion forces. Hence:
Ar < CH₄ < C₆H₆.

Typical values also support this: T_B(Ar) ≈ 510 K, T_B(CH₄) ≈ 650 K, T_B(C₆H₆) ≈ 1900 K.

(b)(i)
Ethanol has greater surface tension than dimethyl ether. Ethanol contains an −OH group and forms strong intermolecular hydrogen bonds. Dimethyl ether has no O−H bond and cannot act as a hydrogen-bond donor; it has only weaker dipole–dipole and dispersion forces. Stronger cohesive forces at the liquid–air interface give ethanol higher surface tension.

Final: Ethanol > dimethyl ether.

(b)(ii)
By the Young–Laplace equation for a spherical droplet:
Δp = 2γ/r.

Taking γ(water–air, 25°C) = 0.072 N m⁻¹ and r = 150 nm = 1.50×10⁻⁷ m:
Δp = (2 × 0.072)/(1.50×10⁻⁷)
= 9.6×10⁵ Pa.

Thus Δp = 9.6×10⁵ Pa = 9.6 bar ≈ 9.5 atm. The pressure inside the droplet is greater than outside.

(c)(i)
For an ideal gas under isothermal conditions, ΔU = 0. Therefore:
ΔA = ΔU − TΔS = −TΔS.

For 1 mole ideal gas:
ΔS = R ln(V₂/V₁).

So:
ΔA = −RT ln(V₂/V₁) = RT ln(V₁/V₂).

V₁ = 100.0 L, V₂ = 22.4 L:
V₁/V₂ = 100.0/22.4 = 4.4643.
ln(4.4643) = 1.4961.

RT = 8.314 × 298 = 2477.71 J mol⁻¹.
ΔA = 2477.71 × 1.4961 = 3.707×10³ J mol⁻¹.

Final: ΔA = +3.71 kJ mol⁻¹.

(c)(ii)
When air is pumped into a tyre, the pump does work on the air. If compression is rapid, heat exchange with surroundings is small, so the process is nearly adiabatic. Work done increases the internal energy of the air, hence its temperature rises. Friction in the pump also contributes heat.

Yes, a tyre can in principle be inflated without a net rise in temperature if the compression is done slowly and isothermally, allowing heat to escape to the surroundings. Using pre-cooled air or cooling the tyre during inflation also prevents temperature rise.

(c)(iii)
For the reaction:
NiO(s) ⇌ Ni(s) + ½ O₂(g).

Solids have unit activity, so:
K = P_O₂^(1/2).

Using ΔG° = −RT ln K:
212000 J mol⁻¹ = −RT ln(P_O₂^(1/2))
= −(RT/2) ln P_O₂.

Thus:
ln P_O₂ = −(2 × 212000)/(8.314 × 298)
= −424000/2477.71
= −171.126.

P_O₂ = e^(−171.126) = 4.8×10⁻⁷⁵ atm.

Final: P_O₂ = 4.8×10⁻⁷⁵ atm.

(c)(iv)
Using the Joule–Thomson coefficient:
μ_JT = (∂T/∂P)_H.

For approximately constant μ_JT:
ΔT = μ_JT ΔP.

ΔP = 0.95 − 200.00 = −199.05 atm.
ΔT = 0.150 × (−199.05) = −29.8575 K.

Initial T = 19.0°C = 292.15 K.
Final T = 292.15 − 29.8575 = 262.2925 K.

Final: T_final = 262.29 K = −10.86°C.

Model answer

What "Calculate" is asking you to do

Structure that answers it

Where marks are lost

How this answer will be evaluated

Approach

Key points expected

Evaluation rubric

Must cover

Loses marks

Earns more

Extra mark

Must cover

Loses marks

Earns more

Extra mark

Must cover

Loses marks

Earns more

Extra mark

Must cover

Loses marks

Earns more

Extra mark

Must cover

Loses marks

Earns more

Extra mark

Must cover

Loses marks

Earns more

Extra mark

Must cover

Loses marks

Earns more

Extra mark

Must cover

Loses marks

Earns more

Extra mark

Practice this exact question

More from Chemistry 2024 Paper I