Partial derivatives with chain rule

Question

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.

Answer 1

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.