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. $\frac{2}{7}$
• D. $\frac{1}{3}$

Answer 1

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

From graph, we can deduce that the given process is isobaric, $\Delta Q=n{C}_P\Delta T$
Given, gas is ideal and monoatomic in nature.
so, $\Delta Q=n\left(\frac{5}{2}R\right)\Delta T$ and $\Delta W=PdV=nR\Delta T$
Ratio of work done by the gas to the heat absorbed by the gas is given by $\frac{dW}{dQ}=\frac{nR\Delta T}{n\left(\frac{5}{2}R\right)\Delta T}=\frac{2}{5}$
Thus. option (a) is the correct answer.