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A total of $48 \mathrm{~J}$ heat is given to one mole of helium kept in a cylinder. The temperature of helium increases by $2^{\circ} \mathrm{C}$. The work done by the gas is:
Given, $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$.
Options:
Given, $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$.
Solution:
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Verified Answer
The correct answer is:
$23.1 \mathrm{~J}$
$1^{\text {st }}$ law of thermodynamics
$\begin{aligned}
& \Delta \mathrm{Q}=\Delta \mathrm{U}+\mathrm{W} \\
& \Rightarrow+48=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}+\mathrm{W} \\
& \Rightarrow 48=(1)\left(\frac{3 \mathrm{R}}{2}\right)(2)+\mathrm{W} \\
& \Rightarrow \mathrm{W}=48-3 \times \mathrm{R} \\
& \Rightarrow \mathrm{W}=48-3 \times(8.3) \\
& \Rightarrow \mathrm{W}=23.1 \text { Joule }
\end{aligned}$
$\begin{aligned}
& \Delta \mathrm{Q}=\Delta \mathrm{U}+\mathrm{W} \\
& \Rightarrow+48=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}+\mathrm{W} \\
& \Rightarrow 48=(1)\left(\frac{3 \mathrm{R}}{2}\right)(2)+\mathrm{W} \\
& \Rightarrow \mathrm{W}=48-3 \times \mathrm{R} \\
& \Rightarrow \mathrm{W}=48-3 \times(8.3) \\
& \Rightarrow \mathrm{W}=23.1 \text { Joule }
\end{aligned}$
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