Question

One mole of an ideal gas at an initial temperature of $T \,K$ does $6R$ joules of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is $5/3$, the final temperature of gas will be

• A. $(T+2.4)K$
• B. $(T-2.4)K$
• C. $(T+4)K$
• D. $(T-4)K$

Answer 1

Answer: (D)

$(T-4)K$

Answer 2

In an adiabatic process, there is no heat transfer into or out of a system ie, $Q=0$. In an adiabatic process $Q=0$ So, from first law of thermodynamics. $W=-\Delta U=-n C_{v} \Delta T$ $=-n\left(\frac{R}{\gamma-1}\right)\left(T_{f}-T_{i}\right)$ $=\frac{n R}{\gamma-1}\left(T_{i}-T_{f}\right)$. . .(i) Here: $W=6 R J, n=1 \,mol$, $R=8.31\, J / mol-K, \gamma=\frac{5}{3}, T_{i}=T K$ Substituting given values in E (i), we get $\therefore 6 R=\frac{R}{(5 / 3-1)}\left(T-T_{f}\right)$ $\Rightarrow 6 R=\frac{3 R}{2}\left(T-T_{f}\right)$ $\Rightarrow T-T_{f}=4$ $\therefore T_{f}=(T-4) K$