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An ideal gas expands isothermally from $10^{-3} \mathrm{~m}^3$ to $10^{-2} \mathrm{~m}^3$ at $300 \mathrm{~K}$ against a constant pressure of $10^5 \mathrm{Nm}^{-2}$. The work done on the gas is
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Verified Answer
The correct answer is:
$-900 \mathrm{~J}$
For an isothermal irreversible expansion, Work done $(\mathrm{W})=-\mathrm{p}_{\mathrm{ext}}\left(\mathrm{V}_2-\mathrm{V}_1\right)$ where, $\mathrm{V}_1=$ initial volume
$$
\mathrm{V}_2=\text { final volume }
$$
Given, $\mathrm{p}_{\mathrm{ext}}=10^5 \mathrm{Nm}^{-2}, \mathrm{~V}_1=10^{-3} \mathrm{~m}^3, \mathrm{~V}_2=10^{-2} \mathrm{~m}^3$
On substituting the given values in Eq, (i),
We get,
$$
\begin{aligned}
W & =-10^5 \mathrm{Nm}^{-2}\left(10^{-2} \mathrm{~m}^3-10^{-3} \mathrm{~m}^3\right) \\
& =-10^5 \mathrm{Nm}^{-2} \times 10^{-3}(10-1) \mathrm{m}^3 \\
& =-900 \mathrm{Nm}=-900 \mathrm{~J}
\end{aligned}
$$
$$
\mathrm{V}_2=\text { final volume }
$$
Given, $\mathrm{p}_{\mathrm{ext}}=10^5 \mathrm{Nm}^{-2}, \mathrm{~V}_1=10^{-3} \mathrm{~m}^3, \mathrm{~V}_2=10^{-2} \mathrm{~m}^3$
On substituting the given values in Eq, (i),
We get,
$$
\begin{aligned}
W & =-10^5 \mathrm{Nm}^{-2}\left(10^{-2} \mathrm{~m}^3-10^{-3} \mathrm{~m}^3\right) \\
& =-10^5 \mathrm{Nm}^{-2} \times 10^{-3}(10-1) \mathrm{m}^3 \\
& =-900 \mathrm{Nm}=-900 \mathrm{~J}
\end{aligned}
$$
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