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A gas is allowed to expand in an insulated container against a constant external pressure of $2.5 \mathrm{~atm}$ from $2.5 \mathrm{~L}$ to $4.5 \mathrm{~L}$, the change in internal energy of the gas in joules is
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Verified Answer
The correct answer is:
$-506.5 \mathrm{~J}$
$\mathrm{q}=0$ due to insulated container From first law of thermodynamics,
$$
\begin{aligned}
& \Delta \mathrm{U}=\mathrm{q}+\mathrm{w} \\
& \mathrm{W}=-\operatorname{Pext}\left[\mathrm{V}_2-\mathrm{V}_1\right] \\
& =-2.5[4.5-2.5] \mathrm{atm}-\mathrm{L} \\
& =-2.5 \times 2 \mathrm{~atm}-\mathrm{L} \\
& =-5 \times 101.3 \mathrm{~J} \\
& =-506.5 \mathrm{~J} \\
& \Delta \mathrm{U}=\mathrm{q}+\mathrm{w} \\
& =0-506.5 \\
& =-506.5 \mathrm{~J}
\end{aligned}
$$
$$
\begin{aligned}
& \Delta \mathrm{U}=\mathrm{q}+\mathrm{w} \\
& \mathrm{W}=-\operatorname{Pext}\left[\mathrm{V}_2-\mathrm{V}_1\right] \\
& =-2.5[4.5-2.5] \mathrm{atm}-\mathrm{L} \\
& =-2.5 \times 2 \mathrm{~atm}-\mathrm{L} \\
& =-5 \times 101.3 \mathrm{~J} \\
& =-506.5 \mathrm{~J} \\
& \Delta \mathrm{U}=\mathrm{q}+\mathrm{w} \\
& =0-506.5 \\
& =-506.5 \mathrm{~J}
\end{aligned}
$$
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