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Physical Chemistry — ThermodynamicsIChO

Physical Chemistry — Thermodynamics Chemistry Question

Expansion of Ideal Gas and Thermodynamics of Liquid Mixing

3.1.

A quantity of 0.10 mol of an ideal gas A initially at 22.2 o C is expanded from 0.200 dm 3 to 2.42 dm 3 . Calculate the values of work (w), heat (q), internal energy change (U), entropy change of the system (Ssys), entropy change of the surroundings (Ssurr), and total entropy change (Suniv) if the process is carried out isothermally and irreversibly against an external pressure of 1.00 atm.

Model Answer

U = 0, q = -w
w = – (1.00 atm) × (2.42 – 0.200 dm 3 ) × (101.325 J dm -3 atm -1 )
= – 225 J
q = 225 J
Ssurr = – 225 J/295.4 K = – 0.762 J K -1
Ssys = 0.100 mol × 8.3145 J mol -1 K -1 × 295.4 K × / (295.4 K)
= 2.07 J K -1
Suniv = 2.07 + (– 0.762) = 1.31 J K -1

3.2.

If 3.00 mol of A is condensed into liquid state and is mixed with 5.00 mol of liquid B, calculate the changes in entropy and Gibbs free energy upon such mixing at 25.0 C. This mixture can be assumed to be ideal.

Model Answer

H mix = V mix = 0
The other functions are given by these equations:
The mole fraction of A is 3.00 / (3.00 + 5.00) = 0.375.
The mole fraction of B is 1.000 – 0.375 = 0.625.
ΔG mix = 8.314 J mol -1 K -1 × 298.0 K × (0.375×ln 0.375 + 0.625×ln 0.625) =
= – 1639 J mol -1
ΔS mix = – 8.314 J mol -1 K -1 × (0.375×ln0.375 + 0.625×ln0.625) = 5.50 J K -1 mol -1

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