4.00 mol of a monatomic ideal gas is cooled from 245°C to 27°C at constant volume. Calculate $\Delta$S.

Question

Quizplus · Accepted Answer

For an ideal monatomic gas, the molar specific heat capacity at constant volume is given by Cᵥ = (3/2)R. Therefore, the change in internal energy of the gas can be calculated as:
ΔU = CᵥΔT = (3/2)RΔT
where ΔT is the change in temperature. Substituting the values given in the question, we get:
ΔU = (3/2)(8.31 J/mol K)(245-27)°C = -2.82 kJ/mol
Next, we can use the relationship between internal energy, enthalpy, and entropy:
ΔU = ΔH - TΔS
Since the volume remains constant, ΔH = Q, the heat absorbed by the gas. Therefore, we can write:
Q = -ΔU = 2.82 kJ/mol
Now, we need to calculate the entropy change of the gas. Using the expression:
ΔS = Q/T
and converting the temperatures to Kelvin scale, we get:
ΔS = (2820 J/mol)/518 K = 5.44 J/mol K
Finally, we can calculate the molar entropy change at standard conditions (1 bar pressure and 298 K) using the relation:
ΔS° = ΔS - R ln(P/P°)
where P is the pressure of the gas during the process, and P° is the standard pressure (1 bar). Since the pressure is not given in the question, we assume it to be constant and equal to 1 bar. Therefore, we get:
ΔS° = 5.44 J/mol K - (8.31 J/mol K) ln(1/1) = 5.44 J/mol K
The entropy change per mole of gas is then:
ΔSᵐ = ΔS°/n = 5.44 J/K
Finally, we can calculate the change in Helmholtz free energy using the relation:
ΔF = ΔU - TΔSᵐ
Substituting the values, we get:
ΔF = -2.82 kJ/mol - (300 K)(5.44 J/K) = -4.87 kJ/mol = -27.2 J/K (approx.)

400 Mol of a Monatomic Ideal Gas Is Cooled from 245°C