7.10 Calculate \( \Delta H \) and \( \Delta U \) for the transformation of one mole of an ideal gas from \( 27^{\circ} \mathrm{C} \) and \( 1 \mathrm{~atm} \) to \( 327{ }^{\circ} \mathrm{C} \) and \( 17 \mathrm{~atm} . \bar{C}_{p}=20.9+0.042 \mathrm{~T} \mathrm{~J} / \mathrm{K} \mathrm{mol} \).
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The final temperature is 327°C, which is 327 + 273.15 = 600.15 K. The initial temperature is 27°C, which is 27 + 273.15 = 300.15 K. So, the change in temperature, ΔT, is 600.15 - 300.15 = 300 K. Show more…
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