# Partial derivative of thermodynamics properties

I know that in finding the partial derivative of certain thermodynamics property such as $$H=H(P,V)$$, we can hold the other variable as constant. But what will happen if the relation have more than two variables? For example, if a certain thermodynamics property of pure substance is given by $$\Gamma$$ where, $$\Gamma=S-\frac{U}{T}-\frac{PV}{T}$$

If I need to to find the value of $$\left(\frac{\partial\Gamma}{\partial P}\right)_T$$, when I perform a partial derivative can I also consider other thermodynamics properties (S and U) as constant and the equation becomes $$\left(\frac{\partial\Gamma}{\partial P}\right)_T=0-0-\frac{V}{T}$$

• The relationship your wrote does not have consistent units. Jan 9, 2021 at 14:42
• Sorry, I am using MathJax on this site for the first time, it was my typing error. Jan 9, 2021 at 14:49
• Shouldn't that be a + sign in front of the PV? Jan 9, 2021 at 15:22
• @ChetMiller why? If I factorize -1/T from the term U and PV it will give me (U+PV) so I think the negative sign should be correct. Jan 9, 2021 at 15:31
• Oops. My mistake. Never mind. Jan 9, 2021 at 15:39

When you evaluate that partial derivative, you do not need to consider other properties, U and S, as constant. You have $$d(T\ \Gamma)=TdS+SdT-dU+VdP+PdV$$But, $$dU=TdS-PdV$$Therefore, $$d(T\ \Gamma)=\Gamma dT + Td\Gamma=SdT-VdP$$So, $$\left(\frac{\partial \Gamma}{\partial P}\right)_T=-\frac{V}{T}$$
My conclusion is the same as in Chet Miller's answer. Still, in this case, I would have better start from the guideline of identifying the two independent variables an expression like $$\Gamma$$ is a function of. It makes explicit a step that has implicitly been used in Chet Miller's second equation.
In quantity like $$\Gamma=S-\frac{U}{T}-\frac{PV}{T}$$ we have the entropy, which is a fundamental thermodynamic function if it is seen as a function of $$U$$ and $$V$$, minus the product of $$U$$ (the first variable) times $$\frac1T=\left.\frac{\partial{S}}{\partial{U}}\right|_V$$, minus the product of $$V$$ (second variable of $$S$$) times $$\frac{P}{T}=\left.\frac{\partial{S}}{\partial{V}}\right|_V$$. Therefore, we have an expression which can be used to perform a double Legendre transform of $$S(U,V)$$ into another fundamental equation $$\Gamma(1/T,P/T)$$.
It is a property of the Legendre transform that the derivative of $$\Gamma$$ with respect to one of its variables (in this case $$P/T$$) gives the minus the conjugate variable $$V$$, as a function of the same variables $$\Gamma$$ depends on. Then, $$\left.\frac{\partial \Gamma}{\partial{(P/T)}}\right|_T=T\left.\frac{\partial \Gamma}{\partial{P}}\right|_T= -V$$ Where now $$V$$ should be intended as a function of $$1/T$$,$$P/T)$$ or equivalently $$P$$ and $$T$$.