JEE MAIN - Chemistry (2022 - 27th July Morning Shift - No. 17)
The molar heat capacity for an ideal gas at constant pressure is $$20.785 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$$. The change in internal energy is $$5000 \mathrm{~J}$$ upon heating it from $$300 \mathrm{~K}$$ to $$500 \mathrm{~K}$$. The number of moles of the gas at constant volume is ____________. [Nearest integer] (Given: $$\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$$)
Answer
2
Explanation
$\mathrm{C}_{\mathrm{p}}=20.785 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$
$$ \begin{aligned} &\text { and } \Delta \mathrm{U}=\mathrm{nC} \mathrm{v} \Delta \mathrm{T} \\\\ &\therefore \quad \mathrm{nC}_{\mathrm{v}}=\frac{5000}{200}=25 \end{aligned} $$
and we know that
$$ \begin{aligned} &C_{p}-C_{v}=R \\\\ &20.785-\frac{25}{n}=8.314 \\\\ &n=\frac{25}{(20.785-8.314)}=2 \end{aligned} $$
$$ \begin{aligned} &\text { and } \Delta \mathrm{U}=\mathrm{nC} \mathrm{v} \Delta \mathrm{T} \\\\ &\therefore \quad \mathrm{nC}_{\mathrm{v}}=\frac{5000}{200}=25 \end{aligned} $$
and we know that
$$ \begin{aligned} &C_{p}-C_{v}=R \\\\ &20.785-\frac{25}{n}=8.314 \\\\ &n=\frac{25}{(20.785-8.314)}=2 \end{aligned} $$
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