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(a) Determine the specific enthalpy (kJ/mol) of n-pentane vapor at 200C and 2.0 atm
Chapter 8, Problem 8.44(choose chapter or problem)
(a) Determine the specific enthalpy (kJ/mol) of n-pentane vapor at \(200^{\circ} \mathrm{C}\) and 2.0 atm relative to n-pentane liquid at \(20^{\circ} \mathrm{C}\) and 1.0 atm, assuming ideal-gas behavior for the vapor. Show clearly the process path you construct for this calculation and give the enthalpy changes for each step. State where you used the ideal-gas assumption.
(b) What is the enthalpy of n-pentane liquid at \(20^{\circ} \mathrm{C}\) and 1.0 atm relative to n-pentane vapor at \(200^{\circ} \mathrm{C}\) and 2.0 atm? (This part should not take much time to complete.)
(c) Beginning with the value of \(\hat{H}\) calculated in Part (a) and still assuming ideal-gas behavior, determine the specific internal energy of the vapor at \(200^{\circ} \mathrm{C}\) and 2.0 atm. Once again, state where you used the ideal-gas assumption.
Questions & Answers
QUESTION:
(a) Determine the specific enthalpy (kJ/mol) of n-pentane vapor at \(200^{\circ} \mathrm{C}\) and 2.0 atm relative to n-pentane liquid at \(20^{\circ} \mathrm{C}\) and 1.0 atm, assuming ideal-gas behavior for the vapor. Show clearly the process path you construct for this calculation and give the enthalpy changes for each step. State where you used the ideal-gas assumption.
(b) What is the enthalpy of n-pentane liquid at \(20^{\circ} \mathrm{C}\) and 1.0 atm relative to n-pentane vapor at \(200^{\circ} \mathrm{C}\) and 2.0 atm? (This part should not take much time to complete.)
(c) Beginning with the value of \(\hat{H}\) calculated in Part (a) and still assuming ideal-gas behavior, determine the specific internal energy of the vapor at \(200^{\circ} \mathrm{C}\) and 2.0 atm. Once again, state where you used the ideal-gas assumption.
ANSWER:Step 1 of 3
(a)
The cycle occurs as the n-pentane changes its state from \(20^{\circ} \mathrm{C}\) (liquid) to \(200^{\circ} \mathrm{C}\) (vapor) is,
\(\begin{array}{l}\text{n-pentane } \left(20^{\circ} \mathrm{C}\right. \text{, liquid } ) \rightarrow \text{ n-pentane } \left(36.1^{\circ} \mathrm{C}\right. \text{, liquid } )\\ \mathrm{n} \text{-pentane } \left(200^{\circ} \mathrm{C}\right. \text{, vapor } ) \leftarrow \mathrm{n} \text{-pentane } \left(36.1^{\circ} \mathrm{C}\right. \text{, vapor } )\end{array}\)
Consider the first consider, i.e.
\(\mathrm{n} \text {-pentane }\left(20^{\circ} \mathrm{C} \text {, liquid }\right) \rightarrow \mathrm{n} \text {-pentane }\left(36.1^{\circ} \mathrm{C} \text {, liquid }\right)\)
First, n-pentane changes from \(20^{\circ} \mathrm{C}\) (liquid) to \(36.1^{\circ} \mathrm{C}\) (boiling point of n-pentane in the liquid state).
\(\begin{aligned}\Delta H_1&=\int_{20^{\circ}\mathrm{C}}^{361^{\circ}\mathrm{C}}C_pdT\\ &=C_p\left(36.1\mathrm{C}-20^{\circ}\mathrm{C}\right)\\ &=(168\mathrm{\ J}/\mathrm{mol}\cdot\mathrm{K})(16.1\mathrm{\ K})\\ &=2704.8\mathrm{\ J}/\mathrm{mol}\\ &=2.705\mathrm{\ kJ}/\mathrm{mol}\end{aligned}\)
Second, n-pentane changes from \(36.1^{\circ} \mathrm{C}\) (boiling point of n-pentane in the liquid state) to \(36.1^{\circ} \mathrm{C}\) (boiling point of n-pentane in the vapor state). In this case, enthalpy is equal to the latent heat of vaporization of n-pentane at 36.1 degrees Celsius.
\(\Delta H_2=25.79\mathrm{\ KJ}/\mathrm{mol}\)
Third, n-pentane changes from \(36.1^{\circ} \mathrm{C}\) (boiling point of n-pentane in the vapor state) to n-pentane 200 degree Celsius i.e.,
\(\begin{array}{l}\text{n-pentane } \left(200^{\circ} \mathrm{C}\right. \text{, vapor } ) \leftarrow \text{ n-pentane } \left(36.1^{\circ} \mathrm{C}\right. \text{, vapor } )\\ \begin{aligned}\Delta H & =\int_{36.1 \mathrm{C}}^{200^{\circ} \mathrm{C}} C_{p} d T \\ & =C_{p}\left(200^{\circ} \mathrm{C}-36.1^{\circ} \mathrm{C}\right) \\ & =(120 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K})(163.9 \mathrm{~K}) \\ & =19668 \mathrm{~J} / \mathrm{mol} \\ & =19.67 \mathrm{~kJ} / \mathrm{mol}\end{aligned}\end{array}\)
The total specific enthalpy of n-pentane vapor is,
\(\begin{aligned}\Delta H&=\Delta H_1+\Delta H_2+\Delta H_3\\ &=2.705\mathrm{\ KJ}/\mathrm{mol}+25.79\ \mathrm{KJ}/\mathrm{mol}+19.67\ \mathrm{KJ}/\mathrm{mol}\\ &=48.165\mathrm{\ KJ}/\mathrm{mol}\\ &\approx48.1\mathrm{\ KJ}/\mathrm{mol}\end{aligned}\)