The volume (V) of a monatomic gas varies with its temperature (T), as shown in the graph. The ratio of work do

Question

The volume (V) of a monatomic 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

Accepted Answer

Given process is isobaric. $dQ = n C_p dT$. For a monatomic gas, $C_p = \frac{5}{2}R$. So, $dQ = n(\frac{5}{2}R)dT$. Work done $dW = P dV = nR dT$. The required ratio is $\frac{dW}{dQ} = \frac{nRdT}{n(\frac{5}{2}R)dT} = \frac{2}{5}$.

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