An ideal gas expands by $1.5 \mathrm{~L}$ against a constant external pressure of $2 \mathrm{~atm}$ at $298…
An ideal gas expands by $1.5 \mathrm{~L}$ against a constant external pressure of $2 \mathrm{~atm}$ at $298 \mathrm{~K}$. Calculate the work done?
- $\quad-75 \mathrm{~J}$
- $\quad-303.9 \mathrm{~J}$
- $13.3 \mathrm{~J}$
- $-30 \mathrm{~J}$
Solution
$\begin{aligned}
\mathrm{W} & =-\mathrm{P}_{\text {ext }} \Delta \mathrm{V} \\
& =-2 \mathrm{~atm} \times(1.5 \mathrm{~L}) \\
& =-3 \mathrm{~atm} \mathrm{~L} \times 1.01325=-3.0398 \mathrm{dm}^3 \mathrm{bar}
\end{aligned}$
Now, $1 \mathrm{dm}^3$ bar $=100 \mathrm{~J}$
$\text { Hence, } \begin{aligned}
-3.0398 \mathrm{dm}^3 \text { bar } \times \frac{100 \mathrm{~J}}{1 \mathrm{dm}^3 \mathrm{bar}} & =-303.98 \mathrm{~J} \\
& \cong-303.9 \mathrm{~J}
\end{aligned}$
Asked in: MHT CET 2023 (09 May Shift 2)
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