Chemistry - Thermodynamics - I Question with Solution | TestHub
Match the column
Column I | Column II |
|---|---|
P) Reversible isothermal process | i) w = nRTln(P2)/(P1) |
Q) Reversible adiabatic process | ii) w = nCV,m(T2 - T1) |
R) Irreversible adiabatic process | iii) PV = nRT |
S) Irreversible isothermal process | iv) w = -∫V1V2 Pext· dV |
(Column I = Ideal gas; Column II = Related equations)
Options:
Answer:
Solution:
For a reversible isothermal change of an ideal gas w = -nRTln(V2/V1) = nRTln(P2/P1); PV = nRT always holds, and w = -∫ Pext dV is the general definition of work (with Pext = P for a reversible path) — P matches i,iii,iv. For any adiabatic process q = 0, so w = Δ U = nCV,m(T2 - T1) whether the path is reversible or not; the gas law and the general work integral also apply — Q and R match ii,iii,iv. For an irreversible isothermal process the reversible logarithmic formula does not apply, but PV = nRT (at the equilibrium end states) and w = -∫ Pext dV do — S matches iii,iv.
