Question 8.3.2: Cooling of an Ideal Gas Assuming ideal-gas behavior, calcula......

Neglecting kinetic energy changes, the energy balance for the open system of Part 1 is Q = ΔH, and that for the closed system of Part 2 is Q = ΔU. (Show it.) The problem is therefore to evaluate ΔH and ΔU for the two specified processes.
1. From Table B.2, Appendix B, the heat capacity of N_{2} at a constant pressure of 1 atm is

where T is in °C. Since we are assuming ideal-gas behavior, the enthalpy change for the gas is independent of any pressure change that may occur, and hence, from Equation 8.3-10a,

If you were using APEx to perform the preceding calculation, you would simply enter the formula of Equation 8.3-12a

=Enthalpy(“nitrogen”, 20,100) [or =Enthalpy (“nitrogen”, 20,100,“C”, “g”)]
in a spreadsheet cell, and the value 2.332 (in kJ/mol) would be returned. Finally,

\dot{Q} = \Delta \dot{H} = \dot{n}\Delta \hat{H} \\= 100 \begin{array}{c|c}mol& 2.332  kJ \\ \hline min & mol\end{array} = \boxed{233  kJ/min}

2. To evaluate ΔU, we need the number of moles n, which may be calculated using the ideal-gas equation of state, and Δ\hat{U}, which we calculate from Equation 8.3-12b

\Delta \hat{U} ≈ \int_{90°C}^{30°C} (C_{p})_{N_{2}}(T)  dT –  R(30°C  – 90°C)

Substituting the formula from Table B.2 for the heat capacity and integrating as in Part (1) or entering the APEx formula =Enthalpy(“nitrogen”,90,30) for the integral, and substituting 8.314 × 10^{-3} kJ/(mol · K) for R, yields Δ\hat{U} = – 1.250 kJ/mol.

Calculate n:
At the initial condition (the only point at which we know P, V, and T)

n= PV/RT\\ =\frac{(3.00  bar)(5.00  L)}{[0.08314  L\cdot bar /(mol\cdot K)](363  K)} = 0.497  mol

Calculate Q:

Q=\Delta U=n\Delta \hat{U} \\ =(0.497  mol) (-  1.250  kJ/mol) = \boxed{- 0.621  kJ}