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Reaction: $\ce{2Al + 3Cl2 → 2AlCl3}$

Answer: We must have either temperature, or the values of P and ΔV in order to calculate ΔE from ΔΗ.

Since ΔH is given, we can use the formula ΔE=ΔH-PΔV to calculate change in internal energy, and only pressure and ΔV are required to calculate ΔΕ. So my question is, why do we need temperature in this question?

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Because:

1) your question assumes that the volumes of the solids ($\ce{Al}$ and $\ce{AlCl3}$) are negligible compared to the volume of the $\ce{Cl2}$ gas. This means that $\Delta V \approx V(\ce{Cl2})$.

2) your question assumes that chlorine gas behaves ideally. This means that $V(\ce{Cl2}) = \frac{nRT}{p}$.

In the end, it means you only need two of the three variables ($p$, $T$, $V$) in order to obtain all the information you want.

I would however be very careful about using molar quantities, in particular $V_m$ (since $p$ and $T$ are intensive properties), if the question is asking you to find $\Delta H$ and $\Delta E$ in terms of $\text{kJ mol}^{-1}$.

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