Question

$1$ mole of an ideal gas at initial temperature of $T\text{K}$ does $6R$ joules of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is $\frac{5}{3},$ then final temperature of the gas will be

• A. $(T-4)\text{K}$
• B. $(T+4)\text{K}$
• C. $(T+2.4)\text{K}$
• D. $(T-2.4)\text{K}$

Answer 1

The correct option is A $(T-4)\text{K}$

Given, $\gamma =\frac{5}{3}$
Specific heat at constant volume is given by,
$C_v=\frac{R}{\gamma -1}=\frac{3R}{2}$
From first law of thermodynamics,
$\Delta Q=\Delta U+\Delta W$
For an adiabatic process, $\Delta Q=0$
$\Rightarrow \Delta U=-\Delta W$
From the data given in the question, we can write that
$C_V\Delta T=-6R$
$\Rightarrow \Delta T=-\frac{6R}{C_V}$
$\Rightarrow \Delta T=-4\text{K}$
This gives the final temperature $T_f$ as $(T-4)\text{K}.$
Thus, option (a) is the correct answer.

Alternate method:
Work done by gas during an adiabatic process is given by
$W=\frac{R}{\gamma -1}[T_1-T_2]$
$\Rightarrow 6R=\frac{R}{\frac{5}{3}-1}[T-T_f]$
$\Rightarrow T-T_f=4\text{K}$
$\Rightarrow T_f=(T-4)\text{K}$
Hence, option (a) is the correct answer.