$5.6 \mathrm{~L}$ of helium gas at STP is adiabatically compressed to $0.7 \mathrm{~L}$. Taking the initial…
$5.6 \mathrm{~L}$ of helium gas at STP is adiabatically compressed to $0.7 \mathrm{~L}$. Taking the initial temperature to be $T_1$, the work done in the process is
$\frac{9}{8} R T_1$
$\frac{3}{2} R T_1$
$\frac{15}{8} R T_1$
$\frac{9}{2} R T_1$
Solution
At STP, 22.4 L of any gas is 1 mole.
$
\therefore \quad 5.6 \mathrm{~L}=\frac{5.6}{22.4}=\frac{1}{4} \text { moles }=n
$
In adiabatic process,
$
\begin{array}{rlrl}
& T V^{\gamma-1} & =\text { constant } \\
& \therefore & T_2 V_2^{\gamma-1} & =T_1 V_1^{\gamma-1} \\
& \text { or } & T_2 & =T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}
\end{array}
$
$
\begin{aligned}
& \gamma=\frac{C_p}{C_V}=\frac{5}{3} \text { for monoatomic He gas. } \\
& \therefore \quad T_2=T_1\left(\frac{5.6}{0.7}\right)^{\frac{5}{3}-1}=4 T_1 \\
& \because \text { Further in adiabatic process, } \\
& Q=0 \\
& \therefore \quad W+\Delta U=0 \\
& \text { or } \quad W=-\Delta U \\
& =-n C_V \Delta T \\
& =-n\left(\frac{R}{\gamma-1}\right)\left(T_2-T_1\right) \\
& =-\frac{1}{4}\left(\frac{R}{\frac{5}{3}-1}\right)\left(4 T_1-T_1\right) \\
& =-\frac{9}{8} R T_1 \\
&
\end{aligned}
$
$\therefore$ Correct option is (a).
Analysis of Question
(i) From calculation point of view question is moderately tough.
(ii) Volume of gas is decreasing. Therefore, work done by the gas should be negative