Temperature and Pressure from Counting Microstates
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You must be comfortable with the definition of the microcanonical ensemble and the fundamental postulate that all accessible microstates are equally probable. Familiarity with basic multivariable calculus, specifically partial derivatives and the chain rule, is required to follow the statistical definitions of thermodynamic variables.
Why this matters
This lecture bridges the gap between abstract state counting and the tangible thermodynamic quantities of temperature and pressure.
In the previous lecture, we established that entropy is a measure of the number of microstates, \( S = k \ln \Omega \). However, this definition is abstract. To connect statistical mechanics to the laboratory, we must show that the derivative of this entropy with respect to energy yields the familiar thermodynamic temperature, and that the derivative with respect to volume yields pressure. By deriving these quantities from the counting of microstates, we demonstrate that the Second Law of Thermodynamics is not an independent axiom but a consequence of probability and the sheer scale of macroscopic systems.
The Statistical Definition of Temperature
Imagine two large systems, System A and System B, placed in thermal contact. They can exchange energy, but the total energy \( U_{\text{total}} \) is fixed. The number of ways to distribute this energy between the two systems is the product of their individual multiplicities, \( \Omega_{\text{total}} = \Omega_A(U_A) \Omega_B(U_B) \). Nature does not "choose" the most probable state; rather, the system spends the vast majority of its time in the energy partition that maximises this product. Because the logarithm is a monotonic function, maximizing the product is equivalent to maximizing the sum of the logarithms. This leads us to define a quantity, temperature, as the slope of the entropy curve with respect to energy. If this slope is the same for both systems, the energy distribution is stable. This statistical slope is what we measure with a thermometer.
Deriving Temperature from Entropy
We begin with the Boltzmann entropy formula, \( S = k \ln \Omega \). For an isolated system where volume \( V \) and particle number \( N \) are fixed, entropy is a function of internal energy \( U \) alone. The statistical definition of temperature \( T \) is derived by requiring that the entropy of a composite system is additive and that equilibrium is reached when the total entropy is maximized.
Consider a small change in energy \( dU \). The change in entropy is given by the total differential:
From thermodynamics, we know that for a reversible process, \( dS = \frac{\delta Q}{T} \). If we identify the heat added \( \delta Q \) with the change in internal energy \( dU \) (for a system doing no work), we can equate the coefficients. This yields the fundamental statistical definition of temperature:
Substituting the Boltzmann relation \( S = k \ln \Omega \), we obtain:
This equation tells us that temperature is a measure of how rapidly the number of accessible microstates increases as we add energy to the system. A low temperature corresponds to a steep increase in \( \ln \Omega \) with \( U \) (a little energy opens up many states), while a high temperature corresponds to a shallow increase.
Pressure as a Statistical Force
While temperature relates to energy exchange, pressure relates to volume exchange. Consider a system where the volume \( V \) can change. The total differential of entropy now includes a term for volume:
From the first law of thermodynamics, \( dU = \delta Q - P dV \). Rearranging for heat, \( \delta Q = dU + P dV \). Substituting this into the thermodynamic definition \( dS = \frac{\delta Q}{T} \), we get:
Comparing the coefficient of \( dV \) in this thermodynamic expression with the statistical expression above, we identify the statistical definition of pressure:
Or, equivalently:
Using the Boltzmann entropy again, this can be written as:
Physically, this means that pressure is proportional to the rate at which the number of microstates increases when the volume is expanded. If expanding the volume allows many more configurations (high \( \partial \ln \Omega / \partial V \)), the system exerts a high pressure to resist compression or drive expansion.
Consistency with the Ideal Gas Law
To verify these definitions, we apply them to the ideal gas, where the multiplicity \( \Omega \) is known from the previous lecture. For an ideal gas of \( N \) particles, the number of microstates scales with volume as \( \Omega \propto V^N \). Therefore, \( \ln \Omega = N \ln V + \text{const} \).
Calculating the derivative with respect to volume:
Substituting this into the statistical pressure formula:
Rearranging gives \( PV = N k T \), which is the ideal gas law. This confirms that the statistical definitions of temperature and pressure are consistent with classical thermodynamics and provide a microscopic justification for the macroscopic equation of state.
The derivation
Boltzmann Entropy and Statistical Temperature
For an isolated composite of two macroscopic systems exchanging energy at fixed total U = U1 + U2, the equilibrium energy partition is the one that maximises the total number of accessible microstates Ω = Ω1Ω2. Requiring this maximum forces the entropy to take the form S = k ln Ω (additive, monotonic in Ω) and identifies the statistical temperature through 1/T = (∂S/∂U)V,N, reproducing the thermodynamic entropy of the Clausius relation dS = δQ/T.
A common misconception is that temperature is an intrinsic property of a single particle. In statistical mechanics, temperature is a property of the ensemble. For a single particle, the concept of temperature is ill-defined because \( \Omega \) is not a smooth function of \( U \). Temperature only emerges as a meaningful macroscopic variable when \( N \) is large enough that \( \ln \Omega \) is a smooth, concave function of \( U \). C
Worked example
Problem. A system of \(N\) particles has multiplicity \(\Omega(U,V) = C\,V^N U^{3N/2}\), with \(C\) a constant. Find its temperature and pressure, and the resulting equation of state.
- Entropy: \(S = k\ln\Omega = k\ln C + Nk\ln V + \tfrac32 Nk\ln U\).
- Temperature: \(\dfrac1T = \left(\dfrac{\partial S}{\partial U}\right)_{V,N} = \dfrac{3Nk}{2U}\), so \(U = \tfrac32 NkT\).
- Pressure: \(\dfrac PT = \left(\dfrac{\partial S}{\partial V}\right)_{U,N} = \dfrac{Nk}{V}\), so \(PV = NkT\).
- Combining the two: \(P = \dfrac{2U}{3V}\): the pressure is two-thirds of the kinetic energy density.
- Number check: \(N = 1.00 \times 10^{23}\), \(V = 1.00 \times 10^{-3}\ \mathrm{m^3}\) and \(U = 600\ \mathrm{J}\) give \(T = 290\ \mathrm{K}\) and \(P = 2 \times 600/(3 \times 10^{-3}) = 4.0 \times 10^{5}\ \mathrm{Pa}\), and indeed \(NkT/V = 1.00 \times 10^{23} \times 1.38 \times 10^{-23} \times 290/10^{-3} = 4.0 \times 10^{5}\ \mathrm{Pa}\).
Units check. \(U/V\) is \(\mathrm{J\,m^{-3}} = \mathrm{Pa}\); \(U/(Nk)\) is in K.
Three quick tests
- If \( \frac{\partial \ln \Omega}{\partial U} \) is zero, what is the temperature?
Answer
Since \( 1/T \propto \partial \ln \Omega / \partial U \), if the derivative is zero, \( 1/T = 0 \), implying \( T \to \infty \). - Why must \( \frac{\partial^2 S}{\partial U^2} \) be negative for stability?
Answer
A negative second derivative ensures that entropy is a concave function of energy, meaning the equilibrium state is a maximum, not a minimum or saddle point. - How does the statistical definition of pressure relate to the force exerted by gas molecules?
Answer
Statistically, pressure is the tendency of the system to increase its volume to access more microstates. Microscopically, this corresponds to the momentum transfer of molecules colliding with the container walls.
Problem set for PU-203 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes checked line by line by the editor; every number in the worked example was recomputed.