Question

In an isobaric reversible process, the ratio of heat supplied to the system ($dq$) and work done by the system ($dW$) for a ideal monoatomic gas is:

Answer 1

At constant pressure, (isobaric process) heat supplied is given as:
$q_p=n{C}_{p}dT$
$C_p$ for a monoatomic gas = $\frac{5}{2}R$
So, $q_p=\frac{5}{2}nRdT$
Work done by system is given as:
$dw=nRdT$
Hence, the ratio is: $\frac{dq}{dw}=\frac{\left(5/2\right)nRdT}{nRdT}$
$=\frac{5}{2}=2.5$