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14.6: Computing Property Changes in Closed Systems

Now we recall the fundamental thermodynamic relationships for closed systems, which we derived above:

dU = T dS – P dV(14.4)

dH = T dS + VdP(14.7)

dG = V dP – S dT(14.9)

dA = – P dV  – S dT(14.12)

where U, H, G, and A are state functions. There is something remarkable about the above expressions: they allow for the direct calculation of the change in a state function as a function of changes in other two. An important lesson to be learned here: when dealing with thermodynamic state properties, we are most interested in changes in the value of state properties rather than their actual values.

As we have said, the above relationships allow us to visualize that the changes of each thermodynamic property are dependent on the change of two others in a closed system.

U = U ( S , V )(14.21a)

H = H ( S , P )(14.21b)

G = G ( P, T )(14.21c)

A = A ( V , T )(14.21d)

Therefore, recalling what we did in equation (14.14), we express the total differential of each of these properties as:

Contact your instructor if you are unable to see or interpret this graphic.(14.22a) Contact your instructor if you are unable to see or interpret this graphic.(14.22b)Contact your instructor if you are unable to see or interpret this graphic.(14.22c) Contact your instructor if you are unable to see or interpret this graphic.(14.22d)

Comparing these equations term to term, the following realizations can be made:

Contact your instructor if you are unable to see or interpret this graphic.Contact your instructor if you are unable to see or interpret this graphic.(14.23a) Contact your instructor if you are unable to see or interpret this graphic.Contact your instructor if you are unable to see or interpret this graphic.(14.23b) Contact your instructor if you are unable to see or interpret this graphic.  ;  Contact your instructor if you are unable to see or interpret this graphic.(14.23c) Contact your instructor if you are unable to see or interpret this graphic.  ;  Contact your instructor if you are unable to see or interpret this graphic.(14.23d)

Additionally, we can go one step further. U, H, G, and A are state functions, and as such, their total differentials (equations 14.4, 14.7, 14.9, and 14.12) must be exact. Recall that for a total differential df=Mdx+Ndy to be an exact differential, it must satisfy the equation:

Contact your instructor if you are unable to see or interpret this graphic.(14.24)

Equation (14.24) is the exactness criteria for a function of two independent variables. It was previously stated in (14.20) for a state function of three independent variables. Its application yields:

Contact your instructor if you are unable to see or interpret this graphic.(14.25a)
Contact your instructor if you are unable to see or interpret this graphic.(14.25b)
Contact your instructor if you are unable to see or interpret this graphic.(14.25c)
Contact your instructor if you are unable to see or interpret this graphic.(14.25d)

Equations (14.25) are known as Maxwell’s relationships. Maxwell’s relationships are very useful for manipulating thermodynamic equations. For instance, it is always desirable for practical purposes to express thermodynamic properties such as enthalpy (H) and entropy (S) as functions of measurable properties such as pressure (P) and temperature (T):

H = H (P, T)(14.26a)

S = S (P, T)(14.26a)

Starting with the total differential of H and S as a function of P and T, we can prove that the relationships between the parameters H and S and the parameters P and T are given by the expressions:

Contact your instructor if you are unable to see or interpret this graphic.(14.27a)
Contact your instructor if you are unable to see or interpret this graphic.(14.27b)

Additionally, since these expressions for dH and dS must also be exact differentials, you could prove that the heat capacity at constant pressure (CP) for an ideal gas (PV = RT) does not depend on pressure. The thermodynamic definition of CP and Cv are:

Contact your instructor if you are unable to see or interpret this graphic.  Heat capacity at constant pressure(14.28a) Contact your instructor if you are unable to see or interpret this graphic.  Heat capacity at constant volume(14.28b)

The heat capacity at constant volume (CV) for an ideal gas does not depend on pressure, either. You could prove this by proving first that Cp=CV+R (R=universal gas constant) for ideal gases, using above tools.

Contributors

  • Prof. Michael Adewumi (The Pennsylvania State University). Some or all of the content of this module was taken from Penn State's College of Earth and Mineral Sciences' OER Initiative.