# Thermodynamic Relations: Step-by-Step Solution We are to prove the following thermodynamic identities (for a one-component system): 1. \( V \left(\frac{\partial P}{\partial T}\right)_u = S \) 2. \( V \left(\frac{\partial P}{\partial u}\right)_T = N \) and then **express the pressure \(P\) of an ideal classical gas in terms of the variables \(u\) (energy density) and \(T\)**, and **verify the relations**. --- ## 1. **Deriving the Formulae Thermodynamically** ### **A. Fundamental Thermodynamic Relations** - Internal energy differential for a simple system: \[ dU = T dS - P dV + \mu dN \] For a fixed number of particles (\(dN=\)): \[ dU = T dS - P dV \] - **Energy density**: \( u = U/V \) - **Entropy density**: \( s = S/V \) ### **B. Expressing \(P = P(u,T)\)** Let us consider \(P\) as a function of \(u\) and \(T\): \[ P = P(u, T) \] --- ### **C. The First Identity: \( V \left(\frac{\partial P}{\partial T}\right)_u = S \)** #### **Step 1: Express \(P\) as a function of \(U, V, S\)** From the fundamental relation: \[ dU = T dS - P dV \] Write \(U = U(S, V)\), so: \[ dU = \left(\frac{\partial U}{\partial S}\right)_V dS + \left(\frac{\partial U}{\partial V}\right)_S dV \] Comparing with above: \[ \left(\frac{\partial U}{\partial S}\right)_V = T, \quad \left(\frac{\partial U}{\partial V}\right)_S = -P \] #### **Step 2: Legendre Transform to \(u\) and \(T\)** Since \(u=U/V\), let's consider: \[ dU = V du + u dV \] Combining with the previous expression: \[ V du + u dV = T dS - P dV \] \[ V du = T dS - (P + u) dV \] If we keep \(V\) constant (\(dV=\)), this simplifies to: \[ V du = T dS \implies dS = \frac{V}{T} du \] But we want to find \(\left(\frac{\partial P}{\partial T}\right)_u\). #### **Step 3: Use Maxwell Relation** Recall the thermodynamic identity: \[ dS = \frac{1}{T} dU + \frac{P}{T} dV \] So, \(S = S(U, V)\), and therefore: \[ dS = \left(\frac{\partial S}{\partial U}\right)_V dU + \left(\frac{\partial S}{\partial V}\right)_U dV \] But from the fundamental relation above: \[ \left(\frac{\partial S}{\partial U}\right)_V = \frac{1}{T}, \quad \left(\frac{\partial S}{\partial V}\right)_U = \frac{P}{T} \] Now, \(P = P(U, V)\), or, per unit volume, \(P = P(u, V)\). But since \(u = U/V\), if \(U\) and \(V\) are independent, \(u\) and \(V\) are independent. Now, let's consider \(P\) as a function of \(u\) and \(T\). #### **Step 4: Jacobian Manipulation** Let us consider the general thermodynamic relationships for the variables \((u, T)\): We can write: \[ S = S(u, V) \] But for a fixed number of particles, we can focus on \(u\) and \(T\). The Maxwell relation most directly useful here is: \[ \left(\frac{\partial S}{\partial V}\right)_U = \left(\frac{\partial P}{\partial T}\right)_V \] But let's try to express everything per unit volume: \[ s = S/V \] \[ u = U/V \] So, \[ ds = \frac{1}{V} dS - \frac{S}{V^2} dV \] \[ du = \frac{1}{V} dU - \frac{U}{V^2} dV \] But let's use the following trick: Consider the total differential of pressure as a function of \(u\) and \(T\): \[ dP = \left(\frac{\partial P}{\partial u}\right)_T du + \left(\frac{\partial P}{\partial T}\right)_u dT \] Now, consider the Gibbs relation for \(N\) particles: \[ dU = T dS - P dV \] Divide by \(V\): \[ du = T ds - P \frac{dV}{V} \] At constant \(u\) (\(du=\)): \[ = T ds - P \frac{dV}{V} \implies ds = \frac{P}{T} \frac{dV}{V} \] But if we keep \(u\) and \(T\) constant, and vary \(V\), then \(ds = \), so \(P = \)? That can't be correct. Alternatively, let's recall for the grand potential: \[ \Omega = U - T S - \mu N \] but perhaps that's not helpful here. #### **Step 5: Using Known Thermodynamic Identities** From thermodynamic identities, for a system with \(N\) particles, we have: \[ dP = s dT + n d\mu \] where \(s = S/V\), \(n = N/V\). If we write \(P\) as a function of \(T\) and \(u\), then: \[ \left(\frac{\partial P}{\partial T}\right)_u = \left(\frac{\partial P}{\partial T}\right)_n + \left(\frac{\partial P}{\partial n}\right)_T \left(\frac{\partial n}{\partial T}\right)_u \] But this becomes complicated. Alternatively, for an ideal gas, let's directly verify the formulae. --- ## 2. **Express \(P\) of an Ideal Gas in Terms of \(u\) and \(T\)** For an ideal classical gas: - \( P V = N k_B T \implies P = n k_B T \), where \( n = N/V \) - The internal energy: \( U = \frac{f}{2} N k_B T \), where \( f \) is the degrees of freedom (for monatomic, \( f=3 \)) - So, \( u = U/V = \frac{f}{2} n k_B T \) Therefore, \[ n = \frac{2u}{f k_B T} \] So, \[ P = n k_B T = \frac{2u}{f} \] So, for a monatomic ideal gas (\(f=3\)): \[ P = \frac{2}{3} u \] Importantly, **\(P\) does not depend explicitly on \(T\) at fixed \(u\)**. --- ## 3. **Verification of the Formulae** ### **A. \( V \left(\frac{\partial P}{\partial T}\right)_u = S \)** For the ideal gas: \[ P = \frac{2}{f} u \] So, \[ \left(\frac{\partial P}{\partial T}\right)_u = \] Thus, \[ V \left(\frac{\partial P}{\partial T}\right)_u = \] But the entropy \(S\) is not zero. This suggests that for an ideal gas, at constant \(u\), \(P\) does not change with \(T\). So, **the identity is not satisfied for an ideal gas**. In general, this identity applies to systems where \(u\) and \(T\) are independent variables, and \(P\) depends on both. ### **B. \( V \left(\frac{\partial P}{\partial u}\right)_T = N \)** From above, \[ P = \frac{2}{f} u \] So, \[ \left(\frac{\partial P}{\partial u}\right)_T = \frac{2}{f} \] Thus, \[ V \left(\frac{\partial P}{\partial u}\right)_T = V \cdot \frac{2}{f} \] But, recall \(u = \frac{f}{2} n k_B T\), so \[ n = \frac{2u}{f k_B T} \implies N = n V = \frac{2u V}{f k_B T} \] But at constant \(T\), \(u\) is proportional to \(N\). Alternatively, for the ideal gas, this formula gives a constant, not \(N\). --- ## 4. **Summary (Final Answers)** ### **Step-by-step summary:** 1. **Thermodynamic identities** link entropy, energy, pressure, and temperature. 2. For the ideal gas, - \( P = \frac{2}{f} u \) - \( u = \frac{f}{2} n k_B T \) 3. **Verification:** - \( V \left(\frac{\partial P}{\partial T}\right)_u = \neq S \) - \( V \left(\frac{\partial P}{\partial u}\right)_T = V \frac{2}{f} \neq N \) ### **Final Answers** #### **Ideal Gas Pressure in Terms of \(u\) and \(T\):** \[ P = \frac{2}{f} u \] #### **Verification of Formulae:** For an ideal classical gas, - \( V \left(\frac{\partial P}{\partial T}\right)_u = \) - \( V \left(\frac{\partial P}{\partial u}\right)_T = V \frac{2}{f} \) These do **not** generally equal \(S\) and \(N\) for an ideal gas. The formulae are more general, coming from thermodynamic identities, but for the ideal gas, pressure depends only on \(u\) (not \(T\)) when \(u\) is fixed. --- ## **References** - Callen, H.B., "Thermodynamics and an Introduction to Thermostatistics" (2nd ed.), Sections 3.4–3.6 - Pathria & Beale, "Statistical Mechanics" (3rd ed.), Section 1.1 --- ## **Summary Table** | Formula | Ideal Gas Result | General Meaning | |-----------------------------------------|------------------------------------|----------------------| | \( V \left(\frac{\partial P}{\partial T}\right)_u \) | | Not \(S\) for ideal gas | | \( V \left(\frac{\partial P}{\partial u}\right)_T \) | \(V \frac{2}{f}\) | Not \(N\) for ideal gas | | \( P(u, T) \) | \( \frac{2}{f} u \) | |