In calculus, a derivative measures the rate at which a function changes when one of its variables changes. For a function of a single variable, \(f(x)\), the derivative is defined by
\[ \frac{df}{dx} = \lim_{\Delta x \rightarrow 0} \frac{ f(x+\Delta x)-f(x) } { \Delta x } \]
Geometrically, this derivative is the slope of the tangent line to the graph of the function.
Many thermodynamic quantities, however, depend on more than one variable. For example, the pressure of an ideal gas depends on both temperature and volume:
\[ p = p(V,T) \]
A function of two variables is not represented by a curve but by a surface. Unlike a curve, a surface can have different slopes depending on the direction in which it is examined.
Imagine standing on a hillside. If you walk east-west, you may encounter a steep slope. If you walk north-south, the slope may be much gentler.
A thermodynamic surface behaves in the same way. Consider a function
\[ f(x,y) \]
The slope measured while holding \(y\) constant may be completely different from the slope measured while holding \(x\) constant.
Partial derivatives allow us to measure these directional slopes by varying one variable while keeping all others fixed.
The partial derivative with respect to \(x\) is
\[ \left( \frac{\partial f} {\partial x} \right)_y \]
and represents the slope obtained by moving only in the \(x\)-direction.
Similarly,
\[ \left( \frac{\partial f} {\partial y} \right)_x \]
represents the slope obtained by moving only in the \(y\)-direction.
The partial derivative with respect to \(x\) is defined by
\[ \left( \frac{\partial f} {\partial x} \right)_y = \lim_{\Delta x \rightarrow 0} \frac{ f(x+\Delta x,y)-f(x,y) } { \Delta x } \]
Notice that the variable \(y\) remains fixed throughout the calculation.
Likewise,
\[ \left( \frac{\partial f} {\partial y} \right)_x = \lim_{\Delta y \rightarrow 0} \frac{ f(x,y+\Delta y)-f(x,y) } { \Delta y } \]
where \(x\) remains fixed.
Thermodynamic properties often depend on several variables simultaneously. For example, pressure depends on both volume and temperature:
\[ p = \frac{RT}{V} \]
The quantity
\[ \left( \frac{\partial p} {\partial V} \right)_T \]
tells us how pressure changes when the volume changes while temperature is held constant.
Likewise,
\[ \left( \frac{\partial p} {\partial T} \right)_V \]
tells us how pressure changes when the temperature changes while volume is held constant.
Big picture: Partial derivatives allow us to describe how a thermodynamic quantity changes when one variable changes while all others remain fixed. They are the mathematical language used throughout thermodynamics.
A partial derivative describes how a function changes when only one variable changes and all other variables are held constant. In a real thermodynamic process, however, several variables often change simultaneously.
Consider a function that depends on two variables:
\[ p = p(V,T) \]
Since pressure depends on both volume and temperature, a small change in pressure can result from a change in volume, a change in temperature, or both.
The total differential of pressure is therefore
\[ dp = \left( \frac{\partial p} {\partial V} \right)_T dV + \left( \frac{\partial p} {\partial T} \right)_V dT \]
This expression separates the change in pressure into two independent contributions:
Each term can be interpreted as a slope multiplied by a displacement.
The first term tells us how much the pressure changes because the volume changes:
\[ \left( \frac{\partial p} {\partial V} \right)_T dV \]
The second term tells us how much the pressure changes because the temperature changes:
\[ \left( \frac{\partial p} {\partial T} \right)_V dT \]
To determine the total change in pressure between two states, we integrate the differential expression:
\[ \Delta p = \int \left( \frac{\partial p} {\partial V} \right)_T dV + \int \left( \frac{\partial p} {\partial T} \right)_V dT \]
This equation can be interpreted as the sum of two consecutive changes:
Since pressure is a state function, the total change in pressure depends only on the initial and final states and not on the order in which these changes are carried out.
This observation forms the foundation for many thermodynamic derivations and calculations.
For one mole of an ideal gas, calculate the change in pressure when the gas changes from \(V_1=1.00\ \mathrm{L\,mol^{-1}}\) and \(T_1=200\ \mathrm{K}\) to \(V_2=3.00\ \mathrm{L\,mol^{-1}}\) and \(T_2=400\ \mathrm{K}\).
For one mole of an ideal gas,
\[ p(V,T)=\frac{RT}{V} \]
The total differential is
\[ dp = \left( \frac{\partial p}{\partial V} \right)_T dV + \left( \frac{\partial p}{\partial T} \right)_V dT \]
First calculate the partial derivatives:
\[ \left( \frac{\partial p}{\partial V} \right)_T = -\frac{RT}{V^2} \]
\[ \left( \frac{\partial p}{\partial T} \right)_V = \frac{R}{V} \]
Therefore,
\[ dp = -\frac{RT}{V^2}dV + \frac{R}{V}dT \]
Since pressure is a state function, we may choose any convenient pathway between the initial and final states. Use two steps:
For step 1, \(dT=0\), so
\[ \Delta p_1 = \int_{V_1}^{V_2} -\frac{RT_1}{V^2}dV \]
\[ \Delta p_1 = -RT_1 \int_{V_1}^{V_2} V^{-2}dV \]
\[ \Delta p_1 = RT_1 \left[ \frac{1}{V} \right]_{V_1}^{V_2} \]
\[ \Delta p_1 = RT_1 \left( \frac{1}{V_2} - \frac{1}{V_1} \right) \]
For step 2, \(dV=0\), so
\[ \Delta p_2 = \int_{T_1}^{T_2} \frac{R}{V_2}dT \]
\[ \Delta p_2 = \frac{R}{V_2} (T_2-T_1) \]
Using \(R=0.08206\ \mathrm{L\,atm\,mol^{-1}\,K^{-1}}\):
\[ \Delta p_1 = \left(0.08206 \frac{atm\ L}{mol\ K} \right)(200\ K) \left( \frac{1}{3.00\ L} - \frac{1}{1.00\ L} \right) = -10.94\ \mathrm{atm} \]
\[ \Delta p_2 = \frac{0.08206 \frac{atm\ L}{mol\ K}}{3.00\ L} (400\ K - 200\ K) = +5.47\ \mathrm{atm} \]
Therefore,
\[ \Delta p = \Delta p_1+\Delta p_2 = -10.94\ \mathrm{atm} + 5.47\ \mathrm{atm} \]
\[ \Delta p = -5.47\ \mathrm{atm} \]
Check by calculating the initial and final pressures directly:
\[ p_1 = \frac{RT_1}{V_1} = \frac{\left(0.08206 \frac{atm\ L}{mol\ K} \right)(200\ K)}{1.00\ L} = 16.41\ \mathrm{atm} \]
\[ p_2 = \frac{RT_2}{V_2} = \frac{\left(0.08206 \frac{atm\ L}{mol\ K} \right)(400\ K)}{3.00\ L} = 10.94\ \mathrm{atm} \]
\[ \Delta p = p_2-p_1 = 10.94\ \mathrm{atm} - 16.41\ \mathrm{atm} = -5.47\ \mathrm{atm} \]
Physical interpretation: Increasing the volume lowers the pressure, while increasing the temperature raises the pressure. In this example, the pressure decrease caused by expansion is larger than the pressure increase caused by heating, so the overall pressure decreases.
Big picture: Partial derivatives provide the mathematical language used throughout thermodynamics. They allow changes in a thermodynamic quantity to be separated into contributions from individual variables, making it possible to construct total differentials and calculate finite changes in state functions such as pressure, internal energy, and enthalpy.