What is a partial derivative?
A partial derivative, like a total derivative, is a slope. It gives a magnitude as to how quickly a function changes value when one of the dependent variables changes. Mathematically, a partial derivative is defined for a function \(f(x_1,x_2, \dots x_n)\) by
\[\left( \dfrac{ \partial f}{\partial x_i} \right)_{x_j \neq i} = \lim_{\Delta _i \rightarrow 0} \left( \dfrac{f(x_1+ \Delta x_1 , x_2 + \Delta x_2, \dots, x_i +\Delta x_i, \dots x_n+\Delta x_n) - f(x_1,x_2, \dots x_i, \dots x_n) }{\Delta x_i} \right) \nonumber \]
Because it measures how much a function changes for a change in a given dependent variable, infinitesimal changes in the in the function can be described by
\[ df = \sum_i \left( \dfrac{\partial f}{\partial x_i} \right)_{x_j \neq i} \nonumber \]
So that each contribution to the total change in the function \(f\) can be considered separately.
For simplicity, consider an ideal gas. The pressure can be calculated for the gas using the ideal gas law. In this expression, pressure is a function of temperature and molar volume.
\[ p(V,T) = \dfrac{RT}{V} \nonumber \]
The partial derivatives of p can be expressed in terms of \(T\) and \(V\) as well.
\[ \left( \dfrac{\partial p}{ \partial V} \right)_{T} = - \dfrac{RT}{V^2} \label{max1} \]
and
\[ \left( \dfrac{\partial p}{ \partial T} \right)_{V} = \dfrac{R}{V} \label{max2} \]
So that the change in pressure can be expressed
\[ dp = \left( \dfrac{\partial p}{ \partial V} \right)_{T} dV + \left( \dfrac{\partial p}{ \partial T} \right)_{V} dT \label{eq3} \]
or by substituting Equations \ref{max1} and \ref{max2}
\[ dp = \left( - \dfrac{RT}{V^2} \right ) dV + \left( \dfrac{R}{V} \right) dT \nonumber \]
Macroscopic changes can be expressed by integrating the individual pieces of Equation \ref{eq3} over appropriate intervals.
\[ \Delta p = \int_{V_1}^{V_2} \left( \dfrac{\partial p}{ \partial V} \right)_{T} dV + \int_{T_1}^{T_2} \left( \dfrac{\partial p}{ \partial T} \right)_{V} dT \nonumber \]
This can be thought of as two consecutive changes. The first is an isothermal (constant temperature) expansion from \(V_1\) to \(V_2\) at \(T_1\) and the second is an isochoric (constant volume) temperature change from \(T_1\) to \(T_2\) at \(V_2\). For example, suppose one needs to calculate the change in pressure for an ideal gas expanding from 1.0 L/mol at 200 K to 3.0 L/mol at 400 K. The set up might look as follows.
\[ \Delta p = \underbrace{ \int_{V_1}^{V_2} \left( - \dfrac{RT}{V^2} \right ) dV}_{\text{isothermal expansion}} + \underbrace{ \int_{T_1}^{T_2}\left( \dfrac{R}{V} \right) dT}_{\text{isochoric heating}} \nonumber \]
or
\[ \begin{align*} \Delta p &= \int_{1.0 \,L/mol}^{3.0 \,L/mol} \left( - \dfrac{R( 400\,K)}{V^2} \right ) dV + \int_{200 \,K}^{400,\ K }\left( \dfrac{R}{1.0 \, L/mol} \right) dT \\[4pt] &= \left[ \dfrac{R(200\,K)}{V} \right]_{ 1.0\, L/mol}^{3.0\, L/mol} + \left[ \dfrac{RT}{3.0 \, L/mol} \right]_{ 200\,K}^{400\,K} \\[4pt] &= R \left[ \left( \dfrac{200\,K}{3.0\, L/mol} - \dfrac{200\,K}{1.0\, L/mol}\right) + \left( \dfrac{400\,K}{3.0\, L/mol} - \dfrac{200\,K}{3.0\, L/mol}\right) \right] \\[4pt] &= -5.47 \, atm \end{align*} \]
Alternatively, one could calculate the change as an isochoric temperature change from \(T_1\) to \(T_2\) at \(V_1\) followed by an isothermal expansion from \(V_1\) to \(V_2\) at \(T_2\):
\[ \Delta p = \int_{T_1}^{T_2}\left( \dfrac{R}{V} \right) dT + \int_{V_1}^{V_2} \left( - \dfrac{RT}{V^2} \right ) dV \nonumber \]
or
\[ \begin{align*} \Delta p &= \int_{200 \,K}^{400,\ K }\left( \dfrac{R}{1.0 \, L/mol} \right) dT + \int_{1.0 \,L/mol}^{3.0 \,L/mol} \left( - \dfrac{R( 400\,K)}{V^2} \right ) dV \\[4pt] &= \left[ \dfrac{RT}{1.0 \, L/mol} \right]_{ 200\,K}^{400\,K} + \left[ \dfrac{R(400\,K)}{V} \right]_{ 1.0\, L/mol}^{3.0\, L/mol} \\[4pt] &= R \left[ \left( \dfrac{400\,K}{1.0\, L/mol} - \dfrac{200\,K}{1.0\, L/mol}\right) + \left( \dfrac{400\,K}{3.0\, L/mol} - \dfrac{400\,K}{1.0\, L/mol}\right) \right] \\[4pt] &= -5.47 \, atm \end{align*} \]
This results demonstrates an important property of pressure in that pressure is a state variable, and so the calculation of changes in pressure do not depend on the pathway!