Question

One mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of 27°C. If the work done during the process is 3kJ, then final temperature of the gas is ($C_v$ = 20 J/K):
(a) 100 K
(b) 150 K
(c) 195 K
(d) 255 K

Answer 1

Answer is B.

∆U=q+w. (1)For adiabatic process, q=0. So, ∆U=W.Now, Cv=dU/dT. So, dU=Cv*dT and hence W=Cv*dT. (2)Now, W=3kJ=3000J.So, Change in temperature=dT= W/Cv = 150 K.Initial temperature=27°C=300 K.So, Final temperature= 300-150 (since,dT=150) = 150 K.