Chemistry - Thermodynamics - II Question with Solution | TestHub
A 2.4 g sample of helium is expanded reversibly and adiabatically to 8.0 times its volume, then compressed reversibly at constant pressure back to the initial volume, and finally heated at constant volume until its absolute temperature is twice the initial value. Assume ideal gas behaviour. The entropy change of the gas over the whole process is (R = 2 cal/K-mol, ln 2 = 0.7)
Options:
Answer:
Solution:
n = 2.4/4 = 0.6 mol, CV = 3, CP = 5 cal/K-mol. Step 1 (reversible adiabatic): Δ S = 0, temperature falls to T/4 (82/3 = 4). Step 2 (isobaric, 8V → V): Δ S = nCPln(1/8) = 0.6 × 5 × (-2.1) = -6.3 cal/K, temperature falls to T/32. Step 3 (isochoric, T/32 → 2T, ratio 64): Δ S = nCVln 64 = 0.6 × 3 × 4.2 = +7.56 cal/K. Total = -6.3 + 7.56 = +1.26 cal/K. Check by state function: initial (V, T), final (V, 2T), so Δ S = nCVln 2 = 0.6 × 3 × 0.7 = 1.26 cal/K. Matches (d).
