Question 6.15: A 1000. g sample of liquid Hg undergoes a reversible transfo...
A 1000. g sample of liquid Hg undergoes a reversible transformation from an initial state P_{i}=1.00 bar , T_{i}=300. K to a final state P_{f}=300. bar, T_{f}=600.K. The density of Hg is 13,534 kg m^{-3},\beta=1.81 × 10^{-4}K^{-1}, C_{p}=27.98 J mol^{-1} K^{-1} and \kappa=3.91 × 10^{-6} bar^{-1}, in the range of the variables in this problem. Recall from Section 3.5 tha C_{P}-C_{V}=\left( TV\beta^{2} /\kappa \right), and that C_{P}\approx C_{V} for liquids and solids.
a. Calculate \Delta U and \Delta H for the transformation. Describe the path assumed for the calculation.
b. Compare the relative contributions to \Delta U and \Delta H from the change in pressure and the change in temperature. Can the contribution from the change in pressure be neglected?
Because U is a state function, \Delta U is independent of the path. We choose the reversible path P_{i}=1.00 bar , T_{i}=300. K\to P_{i}=1.00 bar, T_{f}=600. K\to P_{f}=300. bar, T_{f}=600. K . Note that because we have assumed that all the material constants are independent of P and T in the range of interest, the numerical values calculated will depend slightly on the path chosen. However, if the P and T dependence of the material constant were properly accounted for, all paths would give the same value of \Delta U.
\Delta U =\int_{T_{i}}^{T_{f}}C_{P}dT + \int_{V_{i}}^{V_{f}}\frac{\beta T-\kappa P}{\kappa}dVUsing the definition \kappa=-1/V\left( \partial V/\partial P \right)_{T},\ln V_{f}/V_{i}=-\kappa\left( P_{f}-P_{i} \right) ,and V\left( P \right)=V_{i}e^{-\kappa\left( P-P_{i} \right)}. We can also express P\left( V \right) as P\left( V \right)=P_{i}-\left( 1/\kappa \right)\ln\left( V/V_{i} \right)
\Delta U =\int_{T_{i}}^{T_{f}}C_{P}dT + \int_{V_{i}}^{V_{f}}\left[ \frac{\beta t}{\kappa} -\left( P_{i}-\frac{1}{\kappa}\ln \frac{V}{V_{i}} \right)\right]dV
=\int_{T_{i}}^{T_{f}} C_{P}dT+\left( \frac{\beta T}{\kappa} -P_{i}\right)\int_{V_{i}}^{V_{f}}dV+\frac{1}{\kappa}\int_{V_{i}}^{V_{f}}\ln \frac{V}{V_{i}}dV
Using the standard integral \int_{}^{}\ln\left( \chi/\alpha \right)d\chi=-\chi+\chi \ln\left( \chi/\alpha \right)
\Delta U =\int_{T_{i}}^{T_{f}}C_{P}dT +\left( \frac{\beta T}{\kappa}-P_{i} \right)\left( V_{f}-V_{i} \right)+ \frac{1}{\kappa}\left[ -V+V\ln\frac{V}{V_{i}} \right]^{V_{f}}_{V_{i}}=C_{P}\left( T_{f}-T_{i} \right)+\left( \frac{\beta T}{\kappa}-P_{i} \right)\left( V_{f}-V_{i} \right)+\frac{1}{\kappa}\left( V_{i}-V_{f}+V_{f}\ln \frac{V_{f}}{V_{i}} \right)
V_{i}=\frac{m}{\rho}=\frac{1.000 Kg}{13534 Kg m^{-3}}=7.389 × 10^{-5}m^{3}
V_{f}=V_{i}e^{-\kappa\left( P_{f} -P_{i}\right)}=7.389 × 10^{-5} m^{3}\exp\left( -3.91 × 10^{-6} bar^{-1} × 299 bar\right)
=7.380 ×10^{-5} m^{3}
V_{f}-V_{i}=-9×10^{-8}m^{3}
\Delta U=\frac{1000. g}{200.59 g mol^{-1}}× 27.98 J mol^{-1} K^{-1}×\left( 600. K-300. K \right)
-\left( \frac{1.81 ×10^{-4}K^{-1}×300. K}{3.91 ×10^{-6} bar^{-1}}-1.00 bar \right)×\frac{10^{5} Pa}{1 bar}×9×10^{-8}m^{3}
+\frac{1}{3.91×10^{-6} bar^{-1}}×\frac{10^{5} Pa}{1 bar}×\left( ^{9×10^{-8}m^{3}+7.380×10^{-5}m^{3} }_{× \ln\frac{7.380×10^{-5}m^{3} }{7.389×10^{-5}m^{3} }} \right)
=41.8 × 10^{3} J -100 J+1J\approx 41.7 ×10^{3} J
As shown in Section 3.1,
V\left( P,600. K \right)=\left[ 1+\beta\left( 600.K-300. K \right) \right]V\left( P_{i},300.K \right)e^{-\kappa\left( P-P_{i} \right)}=V e^{-\kappa\left( P-P_{i} \right)}\Delta H=\int_{T_{i}}^{T_{f}} C_{P}dT + \int_{P_{i}}^{P_{f}} VdP=\int_{T_{i}}^{T_{f}}C_{P}dT+ V′ e^{\kappa P_{i}}\int_{P_{i}}^{P_{f}} e^{-\kappa P} dP
=nC_{P,m}\left( T_{f}-T_{i} \right)+\frac{V′_{i}e^{\kappa P_{i}}}{\kappa}\left( e^{-\kappa P_{i}} -e^{-\kappa P_{f}}\right)
=\frac{1000.g}{200.59 g mol^{-1}}×27.98 J mol^{-1} K^{-1}×\left( 600. K-300. K \right)
+7.389 ×10^{-5}m^{3}×\left( 1 + 300. K×1.81 ×10^{-4}K^{-1} \right)
×\frac{\exp \left( 3.91×10^{-6}bar^{-1}×1 bar\right)}{-3.91×10^{-6} bar^{-1} ×\left( 1 bar/10^{5}Pa\right)}
×\left[ \exp \left( – 3.91 × 10^{-6} bar^{-1 } ×1 bar\right)-\exp\left( -3.91×10^{-6} bar^{-1} × 300. bar \right)\right]
\Delta H=41.8 × 10^{3} J +2.29 × 10^{3} J =44.1 ×10^{3} J
The temperature dependent contribution to \Delta U is 98\% of the total change in U, and the corresponding value for \Delta H is \sim 95\%. The contribution from the change in pressure to \Delta U and to \Delta H is small.