Chemistry - Thermodynamics - 1 Question with Solution | TestHub

To calculate the work done, we neet to know the final temperature.

${\mathrm{T}}_1=\mathrm{300 }\mathrm{K},{\mathrm{P}}_1=1\mathrm{atm},\mathrm{n}=2\mathrm{mole},{\mathrm{P}}_2=2\mathrm{atm},{\mathrm{V}}_1=\frac{{\mathrm{nRT}}_1}{{\mathrm{P}}_1}$

$\mathrm{w}=\Delta \mathrm{Y}={\mathrm{nC}}_{\mathrm{v}}\left({\mathrm{T}}_2-{\mathrm{T}}_1\right)=-{\mathrm{P}}_2\left({\mathrm{V}}_2-{\mathrm{V}}_1\right)$

For a monatomic gas, ${\mathrm{C}}_{\mathrm{v}}=\frac{3}{2}\mathrm{R}$

$\Rightarrow \frac{3}{2}\mathrm{nR}\left({\mathrm{T}}_2-{\mathrm{T}}_1\right)=-{\mathrm{P}}_2\left({\mathrm{V}}_2-{\mathrm{V}}_1\right)$

$\frac{3}{2}\mathrm{nR}\left({\mathrm{T}}_2-{\mathrm{T}}_1\right)=-{\mathrm{P}}_2\left(\mathrm{nR}\frac{{\mathrm{T}}_2}{{\mathrm{P}}_2}-\mathrm{nR}\frac{{\mathrm{T}}_1}{{\mathrm{P}}_1}\right)$

$\frac{3}{2}\left({\mathrm{T}}_2-{\mathrm{T}}_1\right)=-{\mathrm{T}}_2+\frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}\Rightarrow {\mathrm{T}}_2=\frac{2}{3}\left(\frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}+\frac{3}{2}\right){\mathrm{T}}_1$

${\mathrm{T}}_2=0.4\left(2+1.5\right)300=420 \mathrm{K}$

Work done, $\mathrm{W}={\mathrm{nC}}_{\mathrm{v}}\mathrm{\Delta {\rm T}}=2\times \frac{3}{2}\mathrm{R}\times \left(420-300\right)$

$\Rightarrow \mathrm{W}=2\times \frac{3}{2}\times 2\times 120=720\mathrm{cal}$