[itex](\frac{\partial U}{\partial P})_T[/itex] derivation

In summary, the conversation discussed using Maxwell relations to simplify the calculation of partial derivatives in thermodynamics. The person asking the question outlined their approach and asked for confirmation. The expert summarized the process and pointed out the use of the "we know that" equation, which is derived from the Maxwell relations.
  • #1
wololo
27
0

Homework Statement


Capture.PNG


Homework Equations


Maxwell relations

The Attempt at a Solution


Here is how I proceeded. Am I allowed to go from line 1 to 2? It almost seems too simple.
[tex]
dU=TdS-PdV \\
(\frac{\partial U}{\partial P})_T=T(\frac{\partial S}{\partial P})_T-P(\frac{\partial V}{\partial P})_T \\
\text{We know that}-(\frac{\partial S}{\partial P})_T=(\frac{\partial V}{\partial T})_P \\
(\frac{\partial U}{\partial P})_T=-T(\frac{\partial V}{\partial T})_P-P(\frac{\partial V}{\partial P})_T \\
(\frac{\partial U}{\partial P})_T=-[T(\frac{\partial V}{\partial T})_P+P(\frac{\partial V}{\partial P})_T]
[/tex]
 
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  • #2
This seems fine, as long as you are allowed to use the "we know that" equation, which comes from the Maxwell relations.
 

Related to [itex](\frac{\partial U}{\partial P})_T[/itex] derivation

1. What is the definition of [itex](\frac{\partial U}{\partial P})_T[/itex]?

The partial derivative of internal energy [itex]U[/itex] with respect to pressure [itex]P[/itex] at constant temperature [itex]T[/itex] is defined as [itex](\frac{\partial U}{\partial P})_T[/itex]. This quantity represents the change in internal energy for a small change in pressure at a constant temperature.

2. How is [itex](\frac{\partial U}{\partial P})_T[/itex] calculated?

[itex](\frac{\partial U}{\partial P})_T[/itex] can be calculated using the Maxwell relations, which relate partial derivatives of thermodynamic quantities. In this case, [itex](\frac{\partial U}{\partial P})_T = T(\frac{\partial V}{\partial T})_P - P(\frac{\partial V}{\partial P})_T[/itex], where [itex]V[/itex] is the volume of the system.

3. What is the physical significance of [itex](\frac{\partial U}{\partial P})_T[/itex]?

The partial derivative [itex](\frac{\partial U}{\partial P})_T[/itex] represents the change in internal energy with respect to pressure at a constant temperature. It can be thought of as a measure of the sensitivity of internal energy to changes in pressure at a fixed temperature.

4. What are some applications of [itex](\frac{\partial U}{\partial P})_T[/itex]?

[itex](\frac{\partial U}{\partial P})_T[/itex] is an important quantity in thermodynamics and is used in various applications, such as in the analysis of phase transitions and in the study of thermodynamic properties of gases. It also plays a role in the calculation of work done in isothermal processes.

5. How does [itex](\frac{\partial U}{\partial P})_T[/itex] relate to other thermodynamic quantities?

[itex](\frac{\partial U}{\partial P})_T[/itex] is related to other thermodynamic quantities through the Maxwell relations. In particular, it is related to the isothermal compressibility [itex]\kappa_T[/itex] by [itex](\frac{\partial U}{\partial P})_T = -\frac{1}{\kappa_T}[/itex]. It is also related to the heat capacity at constant pressure [itex]C_P[/itex] by [itex](\frac{\partial U}{\partial P})_T = T(\frac{\partial V}{\partial T})_P = T(\frac{\partial S}{\partial P})_T = T(C_P - C_V)[/itex].

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