Question

The volume $V$ of a monoatomic gas varies with its temperature $T$ as shown in the graph. The ratio of work done by the gas to the heat absorbed by it when it undergoes a change from state 'A' to state 'B' is

• A. $\frac{2}{5}$
• B. $\frac{2}{3}$
• C. $\frac{1}{3}$
• D. $\frac{2}{7}$

Answer 1

The correct option is A $\frac{2}{5}$

$V\propto T$
isobaric process
$W=P\Delta V=nR\Delta T\Delta Q=n{C}_p\Delta T$
For monoatomic gas, $f=3C_p=\left(\frac{f}{2}R+R\right)=\frac{5}{2}RSo,\frac{W}{\Delta Q}=\frac{nR\Delta T}{n{C}_p\Delta T}=\frac{2}{5}$