JEE MAIN - Chemistry (2024 - 8th April Morning Shift - No. 25)
Consider the figure provided.
$$1 \mathrm{~mol}$$ of an ideal gas is kept in a cylinder, fitted with a piston, at the position A, at $$18^{\circ} \mathrm{C}$$. If the piston is moved to position $$\mathrm{B}$$, keeping the temperature unchanged, then '$$\mathrm{x}$$' $$\mathrm{L}$$ atm work is done in this reversible process.
$$\mathrm{x}=$$ ________ $$\mathrm{L}$$ atm. (nearest integer)
[Given : Absolute temperature $$={ }^{\circ} \mathrm{C}+273.15, \mathrm{R}=0.08206 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}{ }^{-1} \mathrm{~K}^{-1}$$]
Answer
55
Explanation
$$\begin{aligned}
& \mathrm{V}_1=100 \mathrm{~L} \\
& \mathrm{~V}_2=10 \mathrm{~L} \\
& \mathrm{~W}=-\mathrm{nR} \operatorname{Tl} \frac{\mathrm{V}_2}{\mathrm{~V}_1} \\
& =-1 \times 0.08206 \times 291.15 \times 2.303 \log \frac{10}{100} \\
& =55 \mathrm{~L} \text { atm } \\
&
\end{aligned}$$
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