Question 3.9: Calculate the change in enthalpy when 124 g of liquid methan...
Calculate the change in enthalpy when 124 g of liquid methanol initially at 1.00 bar and 298 K undergoes a change of state to 2.50 bar and 425 K. The density of liquid methanol under these conditions is 0.791 g cm^{-3}, and C_{P,m} for liquid methanol is 81.1 J K^{-1} mol^{-1}.
Because H is a state function, any path between the initial and final states will give the same \Delta H. We choose the path methanol \left( l,1.00 bar, 298 K \right)\to methanol \left( l,1.00 bar, 425 K \right)\to methanol \left( l,2.50 bar, 425 K \right). The first step is isothermal, and the second step is isobaric. The total change in H is
\Delta H=n \int_{T_{i}}^{T_{f}}C_{P,m}dT + \int_{P_{i}}^{P_{f}} VdP\approx C_{P,m}\left( T_{f}-T_{i} \right) + V\left( P_{f}-P_{i} \right)=81.1 J K^{-1} mol^{-1}×\frac{124 g}{32.04 g mol^{-1}}× \left( 425 K -298 K \right)
+\frac{124 g}{0.791 g cm^{-3}}× \frac{10^{-6}m^{3}}{cm^{3}} × \left( 2.50 bar – 1,00 bar \right) × \frac{10^{5} Pa}{bar}
=39.9 × 10^{3} J +23.5 J \approx 39.0 KJ
Note that the contribution to \Delta H from the change in T is far greater than that from the change in P.
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