Chemistry - Thermodynamics - I Question with Solution | TestHub
Match Column I (Process for an ideal gas) with Column II (Work expression)
Column I | Column II |
|---|---|
P) Reversible isothermal expansion from P1 to P2 | i) w = nRTln(P2)/(P1) |
Q) Reversible adiabatic expansion | ii) w = (P2V2 - P1V1)/(γ - 1) |
R) Irreversible isothermal expansion against constant external pressure Pext | iii) w = -Pext(V2 - V1) |
S) Free expansion into vacuum | iv) w = 0 |
| v) w = nR(T2 - T1) |
Options:
Answer:
Solution:
Reversible isothermal: w = -nRTln(V2/V1) = nRTln(P2/P1) (P – i). Reversible adiabatic: w = Δ U = nCV,m(T2 - T1) = (nR(T2-T1))/(γ - 1) = (P2V2 - P1V1)/(γ - 1), using CV,m = R/(γ - 1) and PV = nRT (Q – ii). Against a constant external pressure, w = -PextΔ V (R – iii). Into a vacuum there is no opposing pressure, so w = 0 (S – iv). Trap: nRΔ T (v) uses R in place of CV,m and is never the adiabatic work.
