UPSC Combined Geo-Scientist, Mains · 2017 · Mains Paper II

Q1(b). Adiabatic expansion of helium

Question

Ideal He at 500 K expands adiabatically and reversibly to twice its volume. Calculate the final temperature in degrees Celsius.

T₂=314.98 K=41.83° C.

Monatomic ideal helium has C_(V,m)=3R/2, C_(p,m)=5R/2 and γ=5/3.
For reversible adiabatic expansion, TV^γ-1=constant. Hence T₂=T₁(V₁/V₂)^γ-1=500(1/2)^2/3=314.98 K.
Convert only after using absolute temperatures: t₂=T₂-273.15=41.83° C.

The full model answer

As written for the answer booklet: the diagram carries the working, the lines stay short. Read here only; it is not offered as a download.

CoversAdiabatic expansion of heliumThermodynamics