2 mole of an ideal gas is expanded adiabatically and reversibly from 10 L to 80 L. The initial temperature of gas was 47°C. $∆\mathrm{U}$ for this process is ____ cal (C_v, of gas =  3 R/2).

For adiabatic process, q = 0

$\begin{array}{l}\text{ For gas, }{C}_{\mathrm{V}}=\frac{3}{2}\mathrm{R},{\mathrm{C}}_{\mathrm{P}}={\mathrm{C}}_{\mathrm{V}}+\mathrm{R}=\frac{3}{2}\mathrm{R}+\mathrm{R}=\frac{5}{2}\mathrm{R}\\ \text{ and }\gamma ={\mathrm{C}}_{\mathrm{P}}/{\mathrm{C}}_{\mathrm{V}}=\frac{5}{3}\end{array}$

$\begin{array}{l}\text{ Now for adiabatic reversible process, }{\mathrm{TV}}^{\mathrm{\gamma }-1}\\ \text{ or, }{\mathrm{T}}_2={\mathrm{T}}_1\cdot {\left(\frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}\right)}^{\mathrm{\gamma }-1}\\ {\mathrm{T}}_2={\mathrm{T}}_1{\left(\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}\right)}^{\mathrm{\gamma }-1}=320{\left(\frac{10}{80}\right)}^{\frac{5}{3}-1} = 80 \mathrm{K}\\ \text{ Now, }\mathrm{\Delta U}=\mathrm{W}={\mathrm{nC}}_{\mathrm{V}}\left({\mathrm{T}}_2-{\mathrm{T}}_1\right)\\ =2\times \frac{3}{2}\mathrm{R}\times (80-320)=-1440 \mathrm{cal}\end{array}$