MCQ
For $2\ moles$ of $He$ (ideal gas), find the work done in a process where it is heated from $200\ K$ to $400\ K$ such that the enthalpy of the gas varies as : $H = 10\ V^2$
- A$-100\ R$
- ✓$-200\ R$
- C$-300\ R$
- D$-400\ R$
✓
Answer
Correct option: B.
$-200\ R$
b
${\text{M}}{\mkern 1mu} \alpha \,T \Rightarrow {\text{T}}{\mkern 1mu} \alpha {\mkern 1mu} {{\text{V}}^2} \Rightarrow {\text{T}}{\mkern 1mu} {{\text{V}}^{ - 2}} = constant \Rightarrow {\text{P}}{{\text{V}}^{ - 1}}$
${\text{M}}{\mkern 1mu} \alpha \,T \Rightarrow {\text{T}}{\mkern 1mu} \alpha {\mkern 1mu} {{\text{V}}^2} \Rightarrow {\text{T}}{\mkern 1mu} {{\text{V}}^{ - 2}} = constant \Rightarrow {\text{P}}{{\text{V}}^{ - 1}}$
cosntant $=(\mathrm{K})$
$\therefore w=-\int P d V=-\int K V d V=\frac{-K}{2}\left[V_{2}^{2}-V_{1}^{2}\right]$
$=-\frac{\left[\mathrm{P}_{2} \mathrm{V}_{2}^{-1}\left(\mathrm{V}_{2}\right)^{2}-\mathrm{P}_{1} \mathrm{V}_{1}^{-1}\left(\mathrm{V}_{1}\right)^{2}\right]}{2}$
$=-\frac{\left[\mathrm{P}_{2} \mathrm{V}_{2}-\mathrm{P}_{1} \mathrm{V}_{1}\right]}{2}=-\frac{\mathrm{n} \mathrm{R}}{2}\left[\mathrm{T}_{2}-\mathrm{T}_{1}\right]$
$-\frac{2 \mathrm{R}}{2}[400-200]=-200 \,\mathrm{R}$
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