Example Problem with Complete Solution

3B-1 : ΔU and ΔH for Isothermal Expansion of Superheated Water Vapor
2 pts
Superheated water vapor at 300°C expands isothermally in a piston-and-cylinder device from 10 atm to 5 atm.  Calculate the change in the molar enthalpy and molar internal energy in units of KJ / mol.
Read : Because the water vapor is superheated, it has 2 degrees of freedom.  In this case both the T and P must be specified to completely determine the state.  Because the state is completely determined, we can use the given T and P values to look up properties like U and H in the Superheated Tables in the Steam Tables.
Given: T1 =
300
 oC P1 =
10
 atm
T2 =
300
 oC P2 =
5
 atm
Find: ΔU
???
 kJ/mol
ΔH
???
 kJ/mol
Assumptions: None.
Solution : Use the NIST Webbook to determine the properties of superheated water vapor at the initial and final pressures.  As always, use the ASHRAE convention.  A portion of the thermodynamic table used in this problem is provided below.
Temp.
 (°C)
Pressure (atm)
Internal Energy (kJ/mol)
Enthalpy (kJ/mol)
Phase
    
300
4
50.533
55.252
vapor
    
300
5
50.499
55.206
vapor
    
300
6
50.465
55.159
vapor
    
300
7
50.43
55.112
vapor
    
300
8
50.395
55.065
vapor
    
300
9
50.359
55.018
vapor
    
300
10
50.324
54.97
vapor
    
300
11
50.288
54.921
vapor
    
The internal energy and enthalpy at the given pressures are:
P = 10 atm P = 5 atm
U1 =
50.324
 KJ/mol U2 =
50.499
 KJ/mol
H1 =
54.97
 KJ/mol H2 =
55.206
 KJ/mol
Remember that the change in any property is defined as the final state minus the initial state.
Answers : ΔU = U2 - U1 =
0.175
 KJ/mol
ΔH = H2 - H1 =
0.236
 KJ/mol