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If $1.5 \mathrm{~L}$ of an ideal gas at a pressure of $20 \mathrm{~atm}$ expands isothermally and reversibly to a final volume of $15 \mathrm{~L}$, the work done by the gas in $\mathrm{L}$ atm is
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Verified Answer
The correct answer is:
−69.09
Given,
$$
\begin{aligned}
& p_1=20 \mathrm{~atm} \\
& V_1=1.5 \mathrm{~L} \\
& V_2=15 \mathrm{~L}
\end{aligned}
$$
At constant temperature,
$$
\begin{aligned}
p_1 V_1 & =p_2 V_2=n R T \\
\because \quad W & =-n R T \text { in }\left(\frac{V_2}{V_1}\right) \\
& =-2.303 \times 20 \times 1.5 \times \log \left(\frac{15}{1.5}\right) \\
& =-69.09 \mathrm{~L} \mathrm{~atm}
\end{aligned}
$$
$$
\begin{aligned}
& p_1=20 \mathrm{~atm} \\
& V_1=1.5 \mathrm{~L} \\
& V_2=15 \mathrm{~L}
\end{aligned}
$$
At constant temperature,
$$
\begin{aligned}
p_1 V_1 & =p_2 V_2=n R T \\
\because \quad W & =-n R T \text { in }\left(\frac{V_2}{V_1}\right) \\
& =-2.303 \times 20 \times 1.5 \times \log \left(\frac{15}{1.5}\right) \\
& =-69.09 \mathrm{~L} \mathrm{~atm}
\end{aligned}
$$
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