Simultaneous infinitesimal and significant change in state variables

Question

While studying thermodynamics, I came across the equation

$dV = (𝛿V/𝛿P)dP+ (𝛿V/ 𝛿T)dT $

where $V$ is the volume of a container, $P$ in the pressure inside the container and $T$ is the temperature of the gas inside the container.

This equation is apparently valid for an infinitesimal change in pressure and temperature. What if there is a simultaneous infinitesimal change in temperature but significant change in pressure or vice versa? Would it still be possible to use this equation?

Answer 1

The equation you have stated gives the infinitesimal change in $V$ due to infinitesimal changes in $P$ and $T$. Suppose the initial state of the system is $(T_1,P_1)$. Then the precise way of writing your equation is: $$dV\Bigr|_{(T_1,P_1)}=\frac{\partial V}{\partial P}\Biggr|_{(T_1,P_1)}dP+\frac{\partial V}{\partial T}\Biggr|_{(T_1,P_1)}dT$$ in which the partial derivatives are evaluated at the initial state. If the change in $P$ or $T$ is finite, then the change in $V$ will also be finite. So above equation will not directly apply when one of the variables undergoes a finite change; instead the equation must be integrated over.

Suppose the system changes from $(T_1,P_1)$ to $(T_2,P_2)$. To obtain the change in $V$ above equation must be integrated over a path in the $T$-$P$ plane: $$V_2-V_1=\int_{P_1}^{P_2}\frac{\partial V}{\partial P}\Biggr|_TdP+\int_{T_1}^{T_2}\frac{\partial V}{\partial T}\Biggr|_PdT$$ Since $V$ is a state function, every path of integration connecting the two states gives the same answer. The form of the equation suggests that integration becomes simple if the path from state 1 to state 2 consists of constant-$P$ and constant-$T$ curves. We then have two special choices:

• Along $T_1=$ constant, go from $(T_1,P_1)$ to $(T_1,P_2)$, and then along $P_2=$ constant, go from $(T_1,P_2)$ to $(T_2,P_2)$: $$V_2-V_1=\int_{P_1}^{P_2}\frac{\partial V}{\partial P}\Biggr|_{T_1}dP+\int_{T_1}^{T_2}\frac{\partial V}{\partial T}\Biggr|_{P_2}dT$$
• Along $P_1=$ constant, go from $(T_1,P_1)$ to $(T_2,P_1)$, and then along $T_2=$ constant, go from $(T_2,P_1)$ to $(T_2,P_2)$: $$V_2-V_1=\int_{T_1}^{T_2}\frac{\partial V}{\partial T}\Biggr|_{P_1}dT+\int_{P_1}^{P_2}\frac{\partial V}{\partial P}\Biggr|_{T_2}dP$$

Suppose the change in $T$ is infinitesimal, i.e. $T_2=T_1+dT$. Using the first path in the list above we get: \begin{align} V_2-V_1 &=\int_{P_1}^{P_2}\frac{\partial V}{\partial P}\Biggr|_{T_1}dP+\int_{T_1}^{T_2}\frac{\partial V}{\partial T}\Biggr|_{P_2}dT\\ &=\int_{P_1}^{P_2}\frac{\partial V}{\partial P}\Biggr|_{T_1}dP+\int_{T_1}^{T_1+dT}\frac{\partial V}{\partial T}\Biggr|_{P_2}dT\\ &=\int_{P_1}^{P_2}\frac{\partial V}{\partial P}\Biggr|_{T_1}dP+\frac{\partial V}{\partial T}\Biggr|_{(T_1,P_2)}dT \end{align} in which the last equation is arrived at by writing a Taylor expansion of $\frac{\partial V}{\partial T}\Biggr|_{P_2}$ about $T_1$ and retaining only terms of linear order in $dT$.