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If standard molar enthalpy change and standard molar internal energy change measured in bomb calorimeter are equal, which one of the following statements is correct?
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Verified Answer
The correct answer is:
$\Delta \mathrm{n}=0$, at constant pressure
$\Delta \mathrm{H}^{\circ}=\Delta \mathrm{U}^{\circ}$ (given)
$$
\begin{aligned}
& \Delta \mathrm{H}^{\circ}=\Delta \mathrm{U}^{\circ}+\mathrm{P} \Delta \mathrm{V} \\
& =\Delta \mathrm{U}^{\circ}+\Delta \mathrm{nRT} \\
& =\Delta \mathrm{U}^{\circ}
\end{aligned}
$$
$\Rightarrow \Delta \mathrm{nRT}=0$ which means $\Delta \mathrm{n}=0$ with constant pressure.
$$
\begin{aligned}
& \Delta \mathrm{H}^{\circ}=\Delta \mathrm{U}^{\circ}+\mathrm{P} \Delta \mathrm{V} \\
& =\Delta \mathrm{U}^{\circ}+\Delta \mathrm{nRT} \\
& =\Delta \mathrm{U}^{\circ}
\end{aligned}
$$
$\Rightarrow \Delta \mathrm{nRT}=0$ which means $\Delta \mathrm{n}=0$ with constant pressure.
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