Answer:
Option A
Explanation:
At STP ,22.4 L of any gas is 1 mole
$\therefore$ $5.6 L= \frac{5.6}{22.4}=\frac{1}{4} moles =n$
In adiabatic process
$ T V ^{\gamma-1}= constant$
$\therefore$ $T_{2} V_{2}^{\gamma-1}= T_{1} V_{1}^{\gamma-1}$
or $T_{2}=T_{1}\left(\frac{V_{1}}{V_{2}}\right)^{\gamma-1}$
$\gamma= \frac{C_{p}}{C_{V}}=\frac{5}{3}$ for monoatomic He gas
$\therefore$ $T_{2}= T_{1}\left(\frac{5.6}{0.7}\right)^{5/3-1}=4T_{1}$
$\because$ Further in adiabatic process
$Q=0$
$\therefore$ $W+ \triangle U=0$
or $W= -\triangle U$
= $-n C_{v} \triangle T$
=$-n\left(\frac{R}{\gamma-1}\right)(T_{2}-T_{1})$
=$-\frac{1}{4}\left(\frac{R}{\frac{5}{3}-1}\right)(4T_{1}-T_{1})$
= $-\frac{9}{8} RT_{1}$
$\therefore$ correct option is (a)
Analysis of Question
(i) From a calculation point of view question is moderately tough.
(ii) Volume of gas is decreasing. Therefore, work done by the gas should be negative.
