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. $\frac{9}{8}R{T}_1$
• B. $\frac{3}{2}R{T}_1$
• C. $\frac{15}{8}R{T}_1$
• D. $\frac{9}{2}R{T}_1$

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
$T{V}^{\gamma -1}=constant$
$\therefore T_2{V}_2^{\gamma -1}=T_1{V}_1^{\gamma -1}or{T}_2=T-1(\frac{V_1}{V_2}{)}^{\gamma -1}$
$\gamma =\frac{C_p}{C_v}=\frac{5}{3}$ for monoatomic He gas
$\therefore T_2=T_1(\frac{5.6}{0.7}{)}^{\frac{5}{3}-1}=4T_1$
Further in adiabatic process, Q = 0
$\therefore W+\Delta U=0$
or $w=-\Delta U=-n{C}_v\Delta T=-n\left(\frac{R}{\gamma -1}\right)(T_2-T_1)$
=-$\frac{1}{4}\left(\frac{R}{\frac{5}{3}-1}\right)(4T_1-T_1)=-\frac{9}{8}R{T}_1$
$\therefore$ Correct option is (a)