Question

5.6 litre of helium gas at STP is adiabatically compressed to $0.7$ litre. Taking the initial temperature to be $T_1$, the work done on the gas is:

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

Answer 1

The correct option is A $\frac{9}{8}R{T}_1$

For an adiabatic process, $T{V}^{\gamma -1}=constant$
or $T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}$

where $V_1=5.6L$ and $V_2=0.7L$
also, $He$ is monoatomic so $\gamma =\frac{5}{3}$
so, $T_1(5.6{)}^{\frac{2}{3}}=T_2(0.7{)}^{\frac{2}{3}}$
this gives $T_2=4T_1$

$V_1=5.6L$ at STP so, moles of gas $n=\frac{5.6}{22.4}=\frac{1}{4}$

work done by the gas in adiabatic process is given by,
$w=\frac{P_2{V}_2-P_1{V}_1}{\gamma -1}$
$=\frac{nR{T}_2-nR{T}_1}{\gamma -1}$
$=\frac{nR\Delta T}{\gamma -1}$

Substituting these values to get
$w=\frac{R(4T_1-T_1)}{4\times 2/3}$
$w=\frac{9}{8}R{T}_1$