Question

5.6 litre of helium gas at STP is adiabatically compressed to 0.7 litre. Taking the initial temperature to be T, the work done in the process is:

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

Answer 1

The correct option is A $\frac{9}{8}RT$

$T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}\Rightarrow \frac{T_2}{T_1}={\left(\frac{V_1}{V_2}\right)}^{\gamma -1}$

$\frac{T_2}{T}={\left[\frac{V}{\left(\frac{V}{8}\right)}\right]}^{\frac{5}{3}-1}\Rightarrow T_2=4T$

At STP, 1 mole of gas occupies 22.4 litre, so no. of moles of 5.6 litre gas is equal to $\frac{1}{4}$. i.e., $n=\frac{1}{4}$

Work done, $W=\frac{nR(T_1-T_2)}{\gamma -1}=\frac{\frac{1}{4}\times R\times (T-4T)}{(\frac{5}{3}-1)}=-\frac{9}{8}RT$