# Partial derivatives with chain rule

Consider three state functions $$p,v,t$$ which satisfy one relation: $$pv = t$$, so only two are independent. A four state function, call it $$\Gamma$$ , is then given by the formula: $$\Gamma = vp^2$$.

$$(\text{a})$$ Express $$\Gamma$$ purely in terms of $$p$$ and $$t$$ (i.e., find $$\Gamma(p,t)$$) then compute the following partial derivatives and show that they are not equal: $$\frac{\partial \Gamma}{\partial p}\Bigg|_{v} \quad \text{and} \quad \frac{\partial \Gamma}{\partial p}\Bigg |_{t}$$

### Attempt at solution

$$\Gamma(p,t) = \Gamma(p,v(p,t)) = \left(\frac{t}{p}\right)p^2 = tp.$$ Then, using the multivariable chain rule \begin{align} \frac{\partial \Gamma}{\partial p}\Bigg|_{v} &= \frac{\partial \Gamma}{\partial p} + \frac{\partial \Gamma}{\partial v} \frac{\partial v}{\partial t} = t + \frac{1}{p}\\ \frac{\partial \Gamma}{\partial p}\Bigg|_{t} &=\frac{\partial \Gamma}{\partial p} + \frac{\partial \Gamma}{\partial v} \frac{\partial v}{\partial p} = t +\left(-\frac{t}{p^2}\right), \end{align} both of which are evidently not equal.

I'm really unsure about my solution, I'm almost positive that I have at least one mistake somewhere. Some help would be appreciated.

I think they give you the hint for a reason. If I just take the partial derivative of derivative of $$\Gamma(p,v)$$ with respect to $$p$$, it means that $$v$$ is a constant. If I take the partial derivative of $$\Gamma(p,t)$$ with respect to $$p$$, it means that $$t$$ is constant.$$\frac{\partial \Gamma}{\partial p}\Bigg|_{v}=\frac{\partial}{\partial p}\Gamma(p,v)=2pv=2t$$ On the other hand, $$\frac{\partial \Gamma}{\partial p}\Bigg|_{t}=\frac{\partial}{\partial p}\Gamma(p,t)=t$$ Notice that they are different.
This $$pv=t$$ equation seems like the equation for the ideal gas, with temperature in units of energy. If you look at your equations, $$t$$ and $$1/p$$ have different dimensions, so they cannot be added together.