P=kV
When P is doubled, V is doubled (say) from V to 2V.
To obtain work done, kV is integrated with respect to V within the limits V, 2V.
Work done W=1.5kV2.....(1)
The ideal gas equation is
PV=RT
But P=kV and T=273 K
Hence, kV2=R(273)....(2)
Substitute this in equation (1)
W=1.5(R)(273)
But Cv=1.5R
Hence, W=Cv(273)....(3)
In equation (2) when we substitute V with 2V and 273 with T, we get the final temperature T=4(273)
Change in internal energy ΔU=CV(T2−T1)U=CV(273)(4−1)=3CV(273)...(4)
Divide equation (4) with equation (3).
ΔUW=3CV(273)Cv(273)=3