As a result of isobaric heating by $∆\mathrm{T}$ = 72 K, One mole of a certain ideal gas obtains an amount of heat 1.6 kJ. The value of y of the gas is

In isobaric heating, pressure remains constant. Hence work done is given by
$\begin{array}{l}\mathrm{W}=\mathrm{P}∫\mathrm{dV}=\mathrm{PV}=\mathrm{RdT}\\ ∴\mathrm{W}=8⋠314×72=598⋠6\mathrm{J}≈0⋠6\mathrm{kJ}\end{array}$
The increment in internal energy
$\begin{array}{l}\mathrm{dU}=\mathrm{Q}−\mathrm{W}=1⋠6−0⋠6=1\mathrm{kJ}\\ \mathrm{γ}=\frac{{\mathrm{C}}_{\mathrm{p}}}{{\mathrm{C}}_{\mathrm{v}}}=\frac{{\mathrm{C}}_{\mathrm{p}}}{{\mathrm{C}}_{\mathrm{p}}−\mathrm{R}}=\frac{{\mathrm{C}}_{\mathrm{p}}\mathrm{ΔT}}{\left({\mathrm{C}}_{\mathrm{p}}−\mathrm{R}\right)\mathrm{ΔT}}=\frac{\mathrm{Q}}{\mathrm{Q}−\mathrm{RΔT}}\\ =\frac{1.6}{1⋠6−0⋠6}=1⋠6\end{array}$