Question

5.6 L of helium gas at STP is adiabatically compressed to 0.7 L. Taking the initial temperature to be $T_1$, the work done in the process is

• A. $\frac{9}{8}RT_1$
• B. $\frac{3}{2}RT_1$
• C. $\frac{15}{8}RT_1$
• D. $\frac{9}{2}RT_1$

Answer 1

Answer: (A)

At STP, 22.4 L of any gas is 1 mole.
$\therefore \, \, \, \, 5.6L =\frac{5.6}{22.4}=\frac{1}{4} moles = n$
In adiabatic process
$ \, \, \, \, \, \, \, \, \, \, \, TV^{\gamma-1}=constant$
$\therefore \, \, \, \, \, T_2V_2^{\gamma -1}=T_1V_1^{\gamma-1} \, \, or \, \, T_2=T-1\bigg(\frac{V_1}{V_2}\bigg)^{\gamma-1}$
$ \, \, \, \, \, \, \, \gamma=\frac{C_p}{C_v}=\frac{5}{3}$ for monoatomic He gas
$\therefore \, \, \, T_2=T_1 \bigg(\frac{5.6}{0.7}\bigg)^{\frac{5}{3}-1} \, =4 T_1$
Further in adiabatic process, Q = 0
$\therefore \, \, \, \, \, \, \, \, \, W+\Delta U=0$
or $ \, \, \, \, w=-\Delta U=-nC_v \Delta T =-n\bigg(\frac{R}{\gamma-1}\bigg)(T_2-T_1)$
=-$\frac{1}{4}\bigg(\frac{R}{\frac{5}{3}-1}\bigg)(4T_1-T_1)=-\frac{9}{8}RT_1$
$\therefore \, $ Correct option is (a)