In an isothermal process, the pressure on 1 mol of an ideal monatomic gas suddenly changes from 4.00 atm to 100.0 atm at 25°C.
-Calculate $\Delta$S.

Question

Quizplus · Accepted Answer

For an isothermal process, ΔS = q/T, where q is the heat transferred to/from the system and T is the constant temperature. Since the process is sudden (i.e. not quasi-static), it is irreversible, and therefore q = -W, where W is the work done by the gas on its surroundings.

To calculate the work done, we can use the formula for the work of expansion against a constant external pressure: W = -PΔV, where P is the external pressure and ΔV is the change in volume. Since the gas is ideal and monatomic, we can use the equation of state PV = nRT to relate the initial and final volumes to the initial pressure and temperature:

V1 = nRT/P1 = (1 mol)(0.08206 L atm/mol K)(298.15 K)/(4.00 atm) = 6.177 L
V2 = nRT/P2 = (1 mol)(0.08206 L atm/mol K)(298.15 K)/(100.0 atm) = 0.245 L

Therefore, ΔV = V2 - V1 = -5.932 L, and the work done is W = -PΔV = -(100.0 atm)(-5.932 L) = 593.2 J.

Finally, we can calculate the entropy change as ΔS = -q/T = W/T = -(593.2 J)/(298.15 K) = -1.987 J/K. However, this is the entropy change of the surroundings, since the gas is assumed to be perfectly insulated and the temperature is constant. The entropy change of the gas itself is equal in magnitude but opposite in sign, so ΔSgas = 1.987 J/K.

Since ΔF = -TΔS, we have ΔFgas = -(298.15 K)(1.987 J/K) = -593.2 J. However, we must convert this to per mole, since we were given the pressure change for 1 mol of gas:

ΔF1 = ΔF/n = -593.2 J/mol

Finally, we can calculate the standard-state entropy change (ΔS°) using the relation ΔG° = ΔH° - TΔS°:

ΔG° = ΔF1 - PΔV1 = -(593.2 J/mol) - (4.00 atm)(6.177 L)(101.3 J/atm L) = -2.221 × 10^4 J/mol

ΔH° is equal to the standard enthalpy of formation of the gas at 25°C and 1 atm, which is 0 by convention for monatomic gases. Therefore:

ΔS° = -(ΔG°/T) = -(-2.221 × 10^4 J/mol)/(298.15 K) = 74.48 J/(mol K)

Note that the negative sign signifies that the process is spontaneous (i.e. exergonic) under standard conditions, since ΔG° < 0.

In an Isothermal Process, the Pressure on 1 Mol of an Ideal