Question

Temperature and pressure may be defined as $$T=\left(\frac{\partial u}{\partial s}\right)_{v} \text { and } P=-\left(\frac{\partial u}{\partial v}\right).$$ Using these definitions, prove that for a simple compressible substance $$\left(\frac{\partial s}{\partial v}\right)_{u}=\frac{P}{T}.$$

Short answer

Based on the step by step solution provided, answer the following question: Q: Prove the relationship, considering the definitions of temperature and pressure given by $T=\left(\frac{\partial u}{\partial s}\right)_{v}$ and $P=-\left(\frac{\partial u}{\partial v}\right)$: $$\left(\frac{\partial s}{\partial v}\right)_u = \frac{P}{T}$$ A: By applying the chain rule and partial derivatives, we differentiate the expressions for T and P and obtained the following relationship: $$\frac{\partial T}{\partial v}=-\frac{\partial P}{\partial s}$$ Using this result and differentiating the expression for P with respect to T and v, we get: $$\frac{\partial P}{\partial T}\frac{\partial P}{\partial s}=-\frac{\partial P}{\partial P}\frac{\partial P}{\partial v}$$ Isolating the partial derivative of s with respect to v, we get: $$\left(\frac{\partial s}{\partial v}\right)_u =-\frac{\partial P/\partial T}{\partial P/\partial P}$$ Plugging in the expressions for T and P, we arrive at the desired result: $$\left(\frac{\partial s}{\partial v}\right)_u = \frac{P}{T}$$

Step-by-step solution

Differentiate the expression for T with respect to v

Differentiating the expression for T with respect to v, while keeping u constant, we have $$\frac{\partial T}{\partial v}=\frac{\partial}{\partial v}\left(\frac{\partial u}{\partial s}\right)_{v}.$$ Now, applying the chain rule, we get $$\frac{\partial T}{\partial v}=\left(\frac{\partial^2 u}{\partial s\partial v}\right)_u.$$

Differentiate the expression for P with respect to s

Differentiating the expression for P with respect to s, while keeping u constant, we have $$\frac{\partial P}{\partial s}=\frac{\partial}{\partial s}\left(-\frac{\partial u}{\partial v}\right).$$ Applying the chain rule, we get $$\frac{\partial P}{\partial s}=-\left(\frac{\partial^2 u}{\partial v\partial s}\right)_u.$$

Recognize the similarity in both expressions

Notice that from Step 1 and Step 2, we have $$\frac{\partial T}{\partial v}=\left(\frac{\partial^2 u}{\partial s\partial v}\right)_u \quad \text{and} \quad \frac{\partial P}{\partial s}=-\left(\frac{\partial^2 u}{\partial v\partial s}\right)_u.$$ Since partial derivatives commute, it is true that $$\left(\frac{\partial^2 u}{\partial s\partial v}\right)_u =\left(\frac{\partial^2 u}{\partial v\partial s}\right)_u.$$ Thus, we can rewrite the expressions above as $$\frac{\partial T}{\partial v}=-\frac{\partial P}{\partial s}.$$

Differentiate the expression for P with respect to P and T

We can rewrite the expression for P as $$\frac{\partial P}{\partial T}\frac{\partial T}{\partial v}+\frac{\partial P}{\partial P}\frac{\partial P}{\partial v}=0,$$ since P is constant with respect to v. Using the result from Step 3, we have $$\frac{\partial P}{\partial T}(-\partial P/\partial s)+\frac{\partial P}{\partial P}\frac{\partial P}{\partial v}=0.$$ Rearranging the terms, we get $$\frac{\partial P}{\partial T}\frac{\partial P}{\partial s}=-\frac{\partial P}{\partial P}\frac{\partial P}{\partial v}.$$

Solve for the partial derivative of s with respect to v

From Step 4, we can isolate the partial derivative of s with respect to v, $$\left(\frac{\partial s}{\partial v}\right)_u =-\frac{\partial P/\partial T}{\partial P/\partial P}.$$ Now, we can simply plug in the expressions we derived for T and P to prove the given equation. Using the definitions of T and P, we can rewrite the equation as $$\left(\frac{\partial s}{\partial v}\right)_u = \frac{\left(\frac{\partial u}{\partial s}\right)_{v}}{-\left(\frac{\partial u}{\partial v}\right)}.$$ Now, multiplying both sides by the negative sign, we obtain the desired result: $$\left(\frac{\partial s}{\partial v}\right)_u = \frac{P}{T}.$$