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 :
Answer & explanation
Correct answer: option 1
The process is isobaric :
dQ = n Cp dT = n \(\frac{5}{2}R\) dT
dW = P dV = n R dT
\(\frac{dW}{dQ}\) = \(\frac{n R dT}{n (\frac{5}{2} R) dT}\) = \(\frac{2}{5}\)