物理 热力学深化 麦克斯韦关系与热力学导数 麦克斯韦关系与热力学导数
熵是热力学中最核心的物理量，却偏偏无法直接测量。实验室里能精确读出的是温度、压强、体积——这些量的偏导数组合，通过热力学势的数学结构，与熵的偏导数建立起严格的等量关系。这就是麦克斯韦关系存在的意义：把“不可测量的偏导数”与“可测量的物理量”连接起来，使得热力学计算真正落地于实验。
麦克斯韦关系的来源
热力学势都是多变量函数，满足数学上的一个基本性质：对连续可微的二元函数 f ( x , y ) f(x,y) f ( x , y )
∂ 2 f ∂ x ∂ y = ∂ 2 f ∂ y ∂ x \frac{\partial^2 f}{\partial x\,\partial y} = \frac{\partial^2 f}{\partial y\,\partial x} ∂ x ∂ y ∂ 2 f = ∂ y ∂ x ∂ 2 f 将这一性质逐一应用到四个热力学势的全微分，每个势各给出一个麦克斯韦关系。
从内能 U ( S , V ) U(S,V) U ( S , V )
全微分为 d U = T d S − P d V dU = T\,dS - P\,dV d U = T d S − P d V S S S V V V
( ∂ T ∂ V ) S = − ( ∂ P ∂ S ) V \left(\frac{\partial T}{\partial V}\right)_S = -\left(\frac{\partial P}{\partial S}\right)_V ( ∂ V ∂ T ) S = − ( ∂ S ∂ P ) V 全微分为 d H = T d S + V d P dH = T\,dS + V\,dP d H = T d S + V d P
( ∂ T ∂ P ) S = ( ∂ V ∂ S ) P \left(\frac{\partial T}{\partial P}\right)_S = \left(\frac{\partial V}{\partial S}\right)_P ( ∂ P ∂ T ) S = ( ∂ S ∂ V ) P
从亥姆霍兹自由能 F ( T , V ) F(T,V) F ( T , V )
全微分为 d F = − S d T − P d V dF = -S\,dT - P\,dV d F = − S d T − P d V
( ∂ S ∂ V ) T = ( ∂ P ∂ T ) V \left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial P}{\partial T}\right)_V ( ∂ V ∂ S ) T = ( ∂ T ∂ P ) V
从吉布斯自由能 G ( T , P ) G(T,P) G ( T , P )
全微分为 d G = − S d T + V d P dG = -S\,dT + V\,dP d G = − S d T + V d P
( ∂ S ∂ P ) T = − ( ∂ V ∂ T ) P \left(\frac{\partial S}{\partial P}\right)_T = -\left(\frac{\partial V}{\partial T}\right)_P ( ∂ P ∂ S ) T = − ( ∂ T ∂ V ) P
四个麦克斯韦关系中，最常用的是来自 F F F G G G T T T P P P V V V U U U H H H
例题 利用麦克斯韦关系，将 ( ∂ S ∂ V ) T \left(\dfrac{\partial S}{\partial V}\right)_T ( ∂ V ∂ S ) T
由 F F F
( ∂ S ∂ V ) T = ( ∂ P ∂ T ) V \left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial P}{\partial T}\right)_V ( ∂ V ∂ S ) T = ( ∂ T ∂ P ) V 右侧是固定体积时压强对温度的偏导数，称为压强温度系数，可通过实验直接测量。对理想气体 P V = n R T PV = nRT P V = n RT V V V T T T
( ∂ P ∂ T ) V = n R V = P T \left(\frac{\partial P}{\partial T}\right)_V = \frac{nR}{V} = \frac{P}{T} ( ∂ T ∂ P ) V = V n R = T P 于是 ( ∂ S ∂ V ) T = P / T \left(\dfrac{\partial S}{\partial V}\right)_T = P/T ( ∂ V ∂ S ) T = P / T
热力学记忆图
四个麦克斯韦关系涉及多个偏导数，符号正负容易混淆。物理学家玻恩（Born）设计了一种几何图形来辅助记忆，通常称为热力学记忆图或热力学方块。
图形是一个正方形，四个角放置四个热力学势，四条边的中点放置对应的自然变量：
读取麦克斯韦关系的规则：在正方形的任意一条对角线上，对角线两端的自然变量构成偏导数的分子和分母，而共轭变量（位于相邻边上）则决定偏导数的形式。关于箭头与正负号：
方块内部的对角线箭头通常从 P P P V V V − P d V -PdV − P d V S S S T T T − S d T -SdT − S d T
在推导麦克斯韦关系式时，顺着箭头方向通常带有负号。
利用玻恩方块图推导麦克斯韦关系式（Maxwell relations），不仅能够快速写出复杂的偏导数，还能通过一个非常巧妙的“看分子定正负”法则绝对准确地判断符号。
推导过程分为两步：画 U 型路径构造等式，以及看箭头判断正负。
第一步：画 U 型路径（构造等式）
选择方块图四条边上的任意一个热力学势（例如 F F F F F F V V V T T T
左侧路径：S → V S \rightarrow V S → V T T T ( ∂ S ∂ V ) T \left(\frac{\partial S}{\partial V}\right)_T ( ∂ V ∂ S ) T
右侧路径：P → T P \rightarrow T P → T V V V ( ∂ P ∂ T ) V \left(\frac{\partial P}{\partial T}\right)_V ( ∂ T ∂ P ) V
现在等式框架有了：
( ∂ S ∂ V ) T = ± ( ∂ P ∂ T ) V \left(\frac{\partial S}{\partial V}\right)_T = \pm \left(\frac{\partial P}{\partial T}\right)_V ( ∂ V ∂ S ) T = ± ( ∂ T ∂ P ) V 第二步：分子法则（判断正负）
核心口诀：只看等式两边的“分子”变量，看它们身上的箭头状态。还记得我们确定的两条单向箭头吗？(P → V P \rightarrow V P → V S → T S \rightarrow T S → T
同向为正 ( + ) (+) ( + ) ：如果方块图中的箭头对这两个分子变量的指向相同（即：箭头同时指向它们，或者同时背离它们），则等式为正。异向为负 ( − ) (-) ( − ) ：如果方块图中的箭头对这两个分子变量的指向不同（即：一个被箭头指着，另一个箭头背离它），则等式带负号。
实战推演验证
亥姆霍兹自由能 F F F
等式：( ∂ S ∂ V ) T = ± ( ∂ P ∂ T ) V \left(\frac{\partial S}{\partial V}\right)_T = \pm \left(\frac{\partial P}{\partial T}\right)_V ( ∂ V ∂ S ) T = ± ( ∂ T ∂ P ) V
看分子：分子是 S S S P P P
看箭头：在方块中，S S S T T T P P P V V V
结论：状态相同（都背离），取正号。
结果：( ∂ S ∂ V ) T = ( ∂ P ∂ T ) V \left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial P}{\partial T}\right)_V ( ∂ V ∂ S ) T = ( ∂ T ∂ P ) V
内能 U U U
等式：( ∂ T ∂ V ) S = ± ( ∂ P ∂ S ) V \left(\frac{\partial T}{\partial V}\right)_S = \pm \left(\frac{\partial P}{\partial S}\right)_V ( ∂ V ∂ T ) S = ± ( ∂ S ∂ P ) V
看分子：分子是 T T T P P P
看箭头：T T T P P P
结论：状态不同（一指一离），取负号。
结果：( ∂ T ∂ V ) S = − ( ∂ P ∂ S ) V \left(\frac{\partial T}{\partial V}\right)_S = -\left(\frac{\partial P}{\partial S}\right)_V ( ∂ V ∂ T ) S = − ( ∂ S ∂ P ) V
记忆图只是辅助记忆工具，使用时最容易出错的是符号。来自内能 U U U G G G H H H F F F
三个基本可测量量
处理热力学偏导数时，目标是把所有未知的导数化简为以下三个在实验室中可直接测量的量：
定义 κ T \kappa_T κ T ( ∂ V / ∂ P ) T < 0 (\partial V/\partial P)_T < 0 ( ∂ V / ∂ P ) T < 0 κ T > 0 \kappa_T > 0 κ T > 0
例题 验证理想气体的热膨胀系数 α \alpha α κ T \kappa_T κ T
对 1 mol 1\ \text{mol} 1 mol P V = R T PV = RT P V = RT V = R T / P V = RT/P V = RT / P
α = 1 V ( ∂ V ∂ T ) P = 1 V ⋅ R P = R P V = 1 T \alpha = \frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_P = \frac{1}{V} \cdot \frac{R}{P} = \frac{R}{PV} = \frac{1}{T} α = V 1 ( ∂ T ∂ V ) P = V 1 ⋅ P R = P V R = T 1 κ T = − 1 V ( ∂ V ∂ P ) T = − 1 V ⋅ ( − R T P 2 ) = R T P V ⋅ P = 1 P \kappa_T = -\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_T = -\frac{1}{V} \cdot \left(-\frac{RT}{P^2}\right) = \frac{RT}{PV \cdot P} = \frac{1}{P} κ T = − V 1 ( ∂ P ∂ V ) T = − V 1 ⋅ ( − P 2 RT ) = P V ⋅ P RT = P 1 这两个结果将在后续推导 C P − C V C_P - C_V C P − C V
内压力与理想气体的判断
将麦克斯韦关系与热力学基本关系结合，可以化简很多看起来无从入手的偏导数。一个重要例子是等温条件下内能随体积的变化，称为内压力（internal pressure）：
π T = ( ∂ U ∂ V ) T \pi_T = \left(\frac{\partial U}{\partial V}\right)_T π T = ( ∂ V ∂ U ) T 利用 d U = T d S − P d V dU = T\,dS - P\,dV d U = T d S − P d V V V V
( ∂ U ∂ V ) T = T ( ∂ S ∂ V ) T − P \left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial S}{\partial V}\right)_T - P ( ∂ V ∂ U ) T = T ( ∂ V ∂ S ) T − P 用来自 F F F ( ∂ S ∂ V ) T = ( ∂ P ∂ T ) V \left(\dfrac{\partial S}{\partial V}\right)_T = \left(\dfrac{\partial P}{\partial T}\right)_V ( ∂ V ∂ S ) T = ( ∂ T ∂ P ) V ( ∂ P ∂ T ) V \left(\dfrac{\partial P}{\partial T}\right)_V ( ∂ T ∂ P ) V α / κ T \alpha/\kappa_T α / κ T
π T = ( ∂ U ∂ V ) T = T ⋅ α κ T − P \boxed{\pi_T = \left(\frac{\partial U}{\partial V}\right)_T = T\cdot\frac{\alpha}{\kappa_T} - P} π T = ( ∂ V ∂ U ) T = T ⋅ κ T α − P 对理想气体，代入 α = 1 / T \alpha = 1/T α = 1/ T κ T = 1 / P \kappa_T = 1/P κ T = 1/ P
π T = T ⋅ 1 / T 1 / P − P = P − P = 0 \pi_T = T \cdot \frac{1/T}{1/P} - P = P - P = 0 π T = T ⋅ 1/ P 1/ T − P = P − P = 0 理想气体的内压力为零，说明其内能与体积无关，仅取决于温度。实际气体分子间存在相互吸引力，膨胀时需要克服这些力做功，因此 π T > 0 \pi_T > 0 π T > 0
例题 范德瓦耳斯气体的状态方程为 ( P + a V m 2 ) ( V m − b ) = R T \left(P + \dfrac{a}{V_m^2}\right)(V_m - b) = RT ( P + V m 2 a ) ( V m − b ) = RT a a a b b b V m V_m V m
由状态方程解出 P P P
P = R T V m − b − a V m 2 P = \frac{RT}{V_m - b} - \frac{a}{V_m^2} P = V m − b RT − V m 2 a 计算 ( ∂ P ∂ T ) V m = R V m − b \left(\dfrac{\partial P}{\partial T}\right)_{V_m} = \dfrac{R}{V_m - b} ( ∂ T ∂ P ) V m = V m − b R
π T = T ⋅ R V m − b − P = R T V m − b − ( R T V m − b − a V m 2 ) = a V m 2 \pi_T = T \cdot \frac{R}{V_m - b} - P = \frac{RT}{V_m - b} - \left(\frac{RT}{V_m - b} - \frac{a}{V_m^2}\right) = \frac{a}{V_m^2} π T = T ⋅ V m − b R − P = V m − b RT − ( V m − b RT − V m 2 a ) = V m 2 a 范德瓦耳斯气体的内压力为 a / V m 2 a/V_m^2 a / V m 2 a a a a a a
C P − C V C_P - C_V C P − C V 定压热容 C P C_P C P C V C_V C V
将熵看作 T T T V V V
d S = ( ∂ S ∂ T ) V d T + ( ∂ S ∂ V ) T d V dS = \left(\frac{\partial S}{\partial T}\right)_V dT + \left(\frac{\partial S}{\partial V}\right)_T dV d S = ( ∂ T ∂ S ) V d T + ( ∂ V ∂ S ) T d V 在恒压条件下（d P = 0 dP = 0 d P = 0 T T T ( ∂ S ∂ V ) T = ( ∂ P ∂ T ) V \left(\dfrac{\partial S}{\partial V}\right)_T = \left(\dfrac{\partial P}{\partial T}\right)_V ( ∂ V ∂ S ) T = ( ∂ T ∂ P ) V
( ∂ S ∂ T ) P = ( ∂ S ∂ T ) V + ( ∂ P ∂ T ) V ( ∂ V ∂ T ) P \left(\frac{\partial S}{\partial T}\right)_P = \left(\frac{\partial S}{\partial T}\right)_V + \left(\frac{\partial P}{\partial T}\right)_V \left(\frac{\partial V}{\partial T}\right)_P ( ∂ T ∂ S ) P = ( ∂ T ∂ S ) V + ( ∂ T ∂ P ) V ( ∂ T ∂ V ) P 两边乘以 T T T C P C_P C P C V C_V C V
C P − C V = T ( ∂ P ∂ T ) V ( ∂ V ∂ T ) P C_P - C_V = T\left(\frac{\partial P}{\partial T}\right)_V \left(\frac{\partial V}{\partial T}\right)_P C P − C V = T ( ∂ T ∂ P ) V ( ∂ T ∂ V ) P 利用链式关系 ( ∂ P ∂ T ) V = α κ T \left(\dfrac{\partial P}{\partial T}\right)_V = \dfrac{\alpha}{\kappa_T} ( ∂ T ∂ P ) V = κ T α ( ∂ V ∂ T ) P = V α \left(\dfrac{\partial V}{\partial T}\right)_P = V\alpha ( ∂ T ∂ V ) P = V α
C P − C V = T V α 2 κ T \boxed{C_P - C_V = \frac{TV\alpha^2}{\kappa_T}} C P − C V = κ T T V α 2
由于 T > 0 T > 0 T > 0 V > 0 V > 0 V > 0 α 2 ≥ 0 \alpha^2 \geq 0 α 2 ≥ 0 κ T > 0 \kappa_T > 0 κ T > 0 C P ≥ C V C_P \geq C_V C P ≥ C V α = 0 \alpha = 0 α = 0 4 ∘ C 4\ ^\circ\text{C} 4 ∘ C C P = C V C_P = C_V C P = C V
例题 验证理想气体满足 C P − C V = n R C_P - C_V = nR C P − C V = n R C P − C V C_P - C_V C P − C V
对 n mol n\ \text{mol} n mol α = 1 / T \alpha = 1/T α = 1/ T κ T = 1 / P \kappa_T = 1/P κ T = 1/ P V = n R T / P V = nRT/P V = n RT / P
C P − C V = T ⋅ ( n R T / P ) ⋅ ( 1 / T ) 2 1 / P = T ⋅ n R / P ⋅ P / T 2 ⋅ T 1 = n R C_P - C_V = \frac{T \cdot (nRT/P) \cdot (1/T)^2}{1/P} = \frac{T \cdot nR/P \cdot P/T^2 \cdot T}{1} = nR C P − C V = 1/ P T ⋅ ( n RT / P ) ⋅ ( 1/ T ) 2 = 1 T ⋅ n R / P ⋅ P / T 2 ⋅ T = n R 每摩尔理想气体 C P − C V = R = 8.314 J ⋅ mol − 1 ⋅ K − 1 C_P - C_V = R = 8.314\ \text{J} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} C P − C V = R = 8.314 J ⋅ mol − 1 ⋅ K − 1
对于固态铜（室温 T = 298 K T = 298\ \text{K} T = 298 K
C P − C V = 298 × 7.11 × 10 − 6 × ( 5.0 × 10 − 5 ) 2 7.3 × 10 − 12 C_P - C_V = \frac{298 \times 7.11 \times 10^{-6} \times (5.0 \times 10^{-5})^2}{7.3 \times 10^{-12}} C P − C V = 7.3 × 1 0 − 12 298 × 7.11 × 1 0 − 6 × ( 5.0 × 1 0 − 5 ) 2 = 298 × 7.11 × 10 − 6 × 2.5 × 10 − 9 7.3 × 10 − 12 = 5.30 × 10 − 12 7.3 × 10 − 12 ≈ 0.73 J ⋅ mol − 1 ⋅ K − 1 = \frac{298 \times 7.11 \times 10^{-6} \times 2.5 \times 10^{-9}}{7.3 \times 10^{-12}} = \frac{5.30 \times 10^{-12}}{7.3 \times 10^{-12}} \approx 0.73\ \text{J} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} = 7.3 × 1 0 − 12 298 × 7.11 × 1 0 − 6 × 2.5 × 1 0 − 9 = 7.3 × 1 0 − 12 5.30 × 1 0 − 12 ≈ 0.73 J ⋅ mol − 1 ⋅ K − 1 铜的 C P − C V ≈ 0.73 J ⋅ mol − 1 ⋅ K − 1 C_P - C_V \approx 0.73\ \text{J} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} C P − C V ≈ 0.73 J ⋅ mol − 1 ⋅ K − 1 R R R 8.8 % 8.8\% 8.8% C P ≈ C V C_P \approx C_V C P ≈ C V
绝热压缩系数与声速
等温压缩系数描述等温条件下压强对体积的影响，而声波在介质中传播时，压缩过程发生得极快，来不及与外界交换热量，更接近绝热过程。因此描述声波传播时需要用绝热压缩系数：
κ S = − 1 V ( ∂ V ∂ P ) S \kappa_S = -\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_S κ S = − V 1 ( ∂ P ∂ V ) S 利用热力学偏导数的链式化简（过程与推导 C P − C V C_P - C_V C P − C V
κ T κ S = C P C V = γ \frac{\kappa_T}{\kappa_S} = \frac{C_P}{C_V} = \gamma κ S κ T = C V C P = γ 其中 γ = C P / C V \gamma = C_P / C_V γ = C P / C V C P > C V C_P > C_V C P > C V κ T > κ S \kappa_T > \kappa_S κ T > κ S
牛顿最初用等温压缩系数计算声速，结果偏低约 18 % 18\% 18% κ S \kappa_S κ S κ T \kappa_T κ T
v s = 1 ρ κ S = γ P ρ = γ R T M v_s = \sqrt{\frac{1}{\rho\,\kappa_S}} = \sqrt{\frac{\gamma P}{\rho}} = \sqrt{\frac{\gamma RT}{M}} v s = ρ κ S 1 = ρ γ P = M γ RT 其中 ρ \rho ρ M M M
例题 计算空气在 20 ∘ C 20\ ^\circ\text{C} 20 ∘ C T = 293 K T = 293\ \text{K} T = 293 K M = 29 × 10 − 3 kg ⋅ mol − 1 M = 29 \times 10^{-3}\ \text{kg} \cdot \text{mol}^{-1} M = 29 × 1 0 − 3 kg ⋅ mol − 1 γ = 1.40 \gamma = 1.40 γ = 1.40 R = 8.314 J ⋅ mol − 1 ⋅ K − 1 R = 8.314\ \text{J} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} R = 8.314 J ⋅ mol − 1 ⋅ K − 1
v s = γ R T M = 1.40 × 8.314 × 293 29 × 10 − 3 v_s = \sqrt{\frac{\gamma RT}{M}} = \sqrt{\frac{1.40 \times 8.314 \times 293}{29 \times 10^{-3}}} v s = M γ RT = 29 × 1 0 − 3 1.40 × 8.314 × 293 = 3414.8 0.029 = 117 751 ≈ 343 m ⋅ s − 1 = \sqrt{\frac{3414.8}{0.029}} = \sqrt{117\,751} \approx 343\ \text{m} \cdot \text{s}^{-1} = 0.029 3414.8 = 117 751 ≈ 343 m ⋅ s − 1 实验测量值约为 343 m ⋅ s − 1 343\ \text{m} \cdot \text{s}^{-1} 343 m ⋅ s − 1 γ \gamma γ 291 m ⋅ s − 1 291\ \text{m} \cdot \text{s}^{-1} 291 m ⋅ s − 1 15 % 15\% 15%
练习题
选择题
1. 下列哪个麦克斯韦关系来自亥姆霍兹自由能 F ( T , V ) F(T, V) F ( T , V )
A. ( ∂ T ∂ V ) S = − ( ∂ P ∂ S ) V \left(\dfrac{\partial T}{\partial V}\right)_S = -\left(\dfrac{\partial P}{\partial S}\right)_V ( ∂ V ∂ T ) S = − ( ∂ S ∂ P ) V
B. ( ∂ S ∂ V ) T = ( ∂ P ∂ T ) V \left(\dfrac{\partial S}{\partial V}\right)_T = \left(\dfrac{\partial P}{\partial T}\right)_V ( ∂ V ∂ S ) T = ( ∂ T ∂ P ) V
C. ( ∂ S ∂ P ) T = − ( ∂ V ∂ T ) P \left(\dfrac{\partial S}{\partial P}\right)_T = -\left(\dfrac{\partial V}{\partial T}\right)_P ( ∂ P ∂ S ) T = − ( ∂ T ∂ V ) P
D. ( ∂ T ∂ P ) S = ( ∂ V ∂ S ) P \left(\dfrac{\partial T}{\partial P}\right)_S = \left(\dfrac{\partial V}{\partial S}\right)_P ( ∂ P ∂ T ) S = ( ∂ S ∂ V ) P
答案：B
亥姆霍兹自由能 F F F d F = − S d T − P d V dF = -S\,dT - P\,dV d F = − S d T − P d V
∂ 2 F ∂ V ∂ T = − ( ∂ S ∂ V ) T , ∂ 2 F ∂ T ∂ V = − ( ∂ P ∂ T ) V \frac{\partial^2 F}{\partial V\,\partial T} = -\left(\frac{\partial S}{\partial V}\right)_T, \qquad \frac{\partial^2 F}{\partial T\,\partial V} = -\left(\frac{\partial P}{\partial T}\right)_V ∂ V ∂ T ∂ 2 F = − ( ∂ V ∂ S ) T , ∂ T ∂ V ∂ 2 F = − ( ∂ T ∂ P ) V 两者相等，负号约去，得 ( ∂ S ∂ V ) T = ( ∂ P ∂ T ) V \left(\dfrac{\partial S}{\partial V}\right)_T = \left(\dfrac{\partial P}{\partial T}\right)_V ( ∂ V ∂ S ) T = ( ∂ T ∂ P ) V U U U G G G H H H
2. 对任意纯物质，下列关于 C P C_P C P C V C_V C V
A. C P < C V C_P < C_V C P < C V
B. C P = C V C_P = C_V C P = C V
C. C P ≥ C V C_P \geq C_V C P ≥ C V α = 0 \alpha = 0 α = 0
D. 固体中 C P < C V C_P < C_V C P < C V C P > C V C_P > C_V C P > C V
答案：C
由公式 C P − C V = T V α 2 / κ T C_P - C_V = TV\alpha^2/\kappa_T C P − C V = T V α 2 / κ T T > 0 T > 0 T > 0 V > 0 V > 0 V > 0 α 2 ≥ 0 \alpha^2 \geq 0 α 2 ≥ 0 κ T > 0 \kappa_T > 0 κ T > 0 C P ≥ C V C_P \geq C_V C P ≥ C V α = 0 \alpha = 0 α = 0 4 ∘ C 4\ ^\circ\text{C} 4 ∘ C C P ≥ C V C_P \geq C_V C P ≥ C V
3. 声波在介质中传播时，使用绝热压缩系数 κ S \kappa_S κ S κ T \kappa_T κ T
A. 声波频率很高，κ S \kappa_S κ S
B. 声波压缩过程极快，来不及与外界交换热量，近似为绝热过程
C. 等温压缩系数 κ T \kappa_T κ T
D. 绝热压缩系数与等温压缩系数数值相同，可以互换
答案：B
声波传播时，压缩和膨胀发生在极短时间内（频率 ∼ 10 2 \sim 10^2 ∼ 1 0 2 10 4 Hz 10^4\ \text{Hz} 1 0 4 Hz 0.1 ms 0.1\ \text{ms} 0.1 ms κ S \kappa_S κ S κ T \kappa_T κ T κ T / κ S = γ > 1 \kappa_T / \kappa_S = \gamma > 1 κ T / κ S = γ > 1
4. 对于 1 mol 1\ \text{mol} 1 mol C V = 3 2 R C_V = \frac{3}{2}R C V = 2 3 R γ = 5 / 3 \gamma = 5/3 γ = 5/3
A. κ S = κ T \kappa_S = \kappa_T κ S = κ T
B. κ S = 3 5 κ T \kappa_S = \dfrac{3}{5}\kappa_T κ S = 5 3 κ T
C. κ S = 3 5 κ T \kappa_S = \dfrac{3}{5}\kappa_T κ S = 5 3 κ T
D. κ S > κ T \kappa_S > \kappa_T κ S > κ T
答案：C
由关系 κ T / κ S = γ = 5 / 3 \kappa_T / \kappa_S = \gamma = 5/3 κ T / κ S = γ = 5/3 κ S = ( 3 / 5 ) κ T < κ T \kappa_S = (3/5)\kappa_T < \kappa_T κ S = ( 3/5 ) κ T < κ T κ S < κ T \kappa_S < \kappa_T κ S < κ T
计算题
5. 某有机液体在温度 T = 400 K T = 400\ \text{K} T = 400 K V m = 3.0 × 10 − 5 m 3 ⋅ mol − 1 V_m = 3.0 \times 10^{-5}\ \text{m}^3 \cdot \text{mol}^{-1} V m = 3.0 × 1 0 − 5 m 3 ⋅ mol − 1 α = 4.0 × 10 − 4 K − 1 \alpha = 4.0 \times 10^{-4}\ \text{K}^{-1} α = 4.0 × 1 0 − 4 K − 1 κ T = 6.0 × 10 − 10 Pa − 1 \kappa_T = 6.0 \times 10^{-10}\ \text{Pa}^{-1} κ T = 6.0 × 1 0 − 10 Pa − 1 C P = 80.0 J ⋅ mol − 1 ⋅ K − 1 C_P = 80.0\ \text{J} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} C P = 80.0 J ⋅ mol − 1 ⋅ K − 1
（1）计算该液体的 C P − C V C_P - C_V C P − C V
（2）计算定容摩尔热容 C V C_V C V
（3）计算热容比 γ = C P / C V \gamma = C_P / C_V γ = C P / C V κ S \kappa_S κ S
（4）与固体铜（C P − C V ≈ 0.73 J ⋅ mol − 1 ⋅ K − 1 C_P - C_V \approx 0.73\ \text{J} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} C P − C V ≈ 0.73 J ⋅ mol − 1 ⋅ K − 1 C P C_P C P C V C_V C V
（1）计算 C P − C V C_P - C_V C P − C V
C P − C V = T V m α 2 κ T = 400 × 3.0 × 10 − 5 × ( 4.0 × 10 − 4 ) 2 6.0 × 10 − 10 C_P - C_V = \frac{TV_m\alpha^2}{\kappa_T} = \frac{400 \times 3.0 \times 10^{-5} \times (4.0 \times 10^{-4})^2}{6.0 \times 10^{-10}} C P − C V = κ T T V m α 2 = 6.0 × 1 0 − 10 400 × 3.0 × 1 0 − 5 × ( 4.0 × 1 0 − 4 ) 2 分子：
400 × 3.0 × 10 − 5 × 16.0 × 10 − 8 = 1.92 × 10 − 9 400 \times 3.0 \times 10^{-5} \times 16.0 \times 10^{-8} = 1.92 \times 10^{-9} 400 × 3.0 × 1 0 − 5 × 16.0 × 1 0 − 8 = 1.92 × 1 0 − 9 C P − C V = 1.92 × 10 − 9 6.0 × 10 − 10 = 3.2 J ⋅ mol − 1 ⋅ K − 1 C_P - C_V = \frac{1.92 \times 10^{-9}}{6.0 \times 10^{-10}} = 3.2\ \text{J} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} C P − C V = 6.0 × 1 0 − 10 1.92 × 1 0 − 9 = 3.2 J ⋅ mol − 1 ⋅ K − 1 （2）计算 C V C_V C V
C V = C P − 3.2 = 80.0 − 3.2 = 76.8 J ⋅ mol − 1 ⋅ K − 1 C_V = C_P - 3.2 = 80.0 - 3.2 = 76.8\ \text{J} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} C V = C P − 3.2 = 80.0 − 3.2 = 76.8 J ⋅ mol − 1 ⋅ K − 1 （3）计算 γ \gamma γ κ S \kappa_S κ S
γ = C P C V = 80.0 76.8 ≈ 1.042 \gamma = \frac{C_P}{C_V} = \frac{80.0}{76.8} \approx 1.042 γ = C V C P = 76.8 80.0 ≈ 1.042 κ S = κ T γ = 6.0 × 10 − 10 1.042 ≈ 5.76 × 10 − 10 Pa − 1 \kappa_S = \frac{\kappa_T}{\gamma} = \frac{6.0 \times 10^{-10}}{1.042} \approx 5.76 \times 10^{-10}\ \text{Pa}^{-1} κ S = γ κ T = 1.042 6.0 × 1 0 − 10 ≈ 5.76 × 1 0 − 10 Pa − 1 液体中 γ \gamma γ 1 1 1
（4）与固体铜的比较
液体的热膨胀系数 α \alpha α κ T \kappa_T κ T C P − C V C_P - C_V C P − C V R ≈ 8.3 J ⋅ mol − 1 ⋅ K − 1 R \approx 8.3\ \text{J} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} R ≈ 8.3 J ⋅ mol − 1 ⋅ K − 1
6. 利用麦克斯韦关系计算一个重要结果：证明在等温过程中，理想气体的内能不随体积变化，而范德瓦耳斯气体的内能随体积增大而增大。
已知范德瓦耳斯气体状态方程（1 mol 1\ \text{mol} 1 mol
( P + a V m 2 ) ( V m − b ) = R T \left(P + \frac{a}{V_m^2}\right)(V_m - b) = RT ( P + V m 2 a ) ( V m − b ) = RT 其中 a = 0.364 J ⋅ m 3 ⋅ mol − 2 a = 0.364\ \text{J} \cdot \text{m}^3 \cdot \text{mol}^{-2} a = 0.364 J ⋅ m 3 ⋅ mol − 2 N 2 \text{N}_2 N 2 b = 3.87 × 10 − 5 m 3 ⋅ mol − 1 b = 3.87 \times 10^{-5}\ \text{m}^3 \cdot \text{mol}^{-1} b = 3.87 × 1 0 − 5 m 3 ⋅ mol − 1
（1）写出内压力公式 π T = ( ∂ U ∂ V m ) T \pi_T = \left(\dfrac{\partial U}{\partial V_m}\right)_T π T = ( ∂ V m ∂ U ) T
（2）对理想气体，证明 π T = 0 \pi_T = 0 π T = 0
（3）对范德瓦耳斯气体，求 π T \pi_T π T V m = 1.0 × 10 − 3 m 3 ⋅ mol − 1 V_m = 1.0 \times 10^{-3}\ \text{m}^3 \cdot \text{mol}^{-1} V m = 1.0 × 1 0 − 3 m 3 ⋅ mol − 1 N 2 \text{N}_2 N 2
（1）内压力的一般表达式
由热力学基本关系 d U = T d S − P d V m dU = T\,dS - P\,dV_m d U = T d S − P d V m V m V_m V m F F F ( ∂ S ∂ V m ) T = ( ∂ P ∂ T ) V m \left(\dfrac{\partial S}{\partial V_m}\right)_T = \left(\dfrac{\partial P}{\partial T}\right)_{V_m} ( ∂ V m ∂ S ) T = ( ∂ T ∂ P ) V m
π T = ( ∂ U ∂ V m ) T = T ( ∂ P ∂ T ) V m − P \pi_T = \left(\frac{\partial U}{\partial V_m}\right)_T = T\left(\frac{\partial P}{\partial T}\right)_{V_m} - P π T = ( ∂ V m ∂ U ) T = T ( ∂ T ∂ P ) V m − P （2）理想气体的内压力
对理想气体 P = R T / V m P = RT/V_m P = RT / V m V m V_m V m T T T
( ∂ P ∂ T ) V m = R V m \left(\frac{\partial P}{\partial T}\right)_{V_m} = \frac{R}{V_m} ( ∂ T ∂ P ) V m = V m R 代入：
π T = T ⋅ R V m − R T V m = R T V m − R T V m = 0 \pi_T = T \cdot \frac{R}{V_m} - \frac{RT}{V_m} = \frac{RT}{V_m} - \frac{RT}{V_m} = 0 π T = T ⋅ V m R − V m RT = V m RT − V m RT = 0 理想气体内压力为零，内能与体积无关。
（3）范德瓦耳斯气体的内压力
由状态方程展开：
P = R T V m − b − a V m 2 P = \frac{RT}{V_m - b} - \frac{a}{V_m^2} P = V m − b RT − V m 2 a 固定 V m V_m V m T T T
( ∂ P ∂ T ) V m = R V m − b \left(\frac{\partial P}{\partial T}\right)_{V_m} = \frac{R}{V_m - b} ( ∂ T ∂ P ) V m = V m − b R 代入内压力公式：
π T = T ⋅ R V m − b − ( R T V m − b − a V m 2 ) = R T V m − b − R T V m − b + a V m 2 = a V m 2 \pi_T = T \cdot \frac{R}{V_m - b} - \left(\frac{RT}{V_m - b} - \frac{a}{V_m^2}\right) = \frac{RT}{V_m-b} - \frac{RT}{V_m-b} + \frac{a}{V_m^2} = \frac{a}{V_m^2} π T = T ⋅ V m − b R − ( V m − b RT − V m 2 a ) = V m − b RT − V m − b RT + V m 2 a = V m 2 a 范德瓦耳斯气体的内压力为 π T = a / V m 2 > 0 \pi_T = a/V_m^2 > 0 π T = a / V m 2 > 0
在 V m = 1.0 × 10 − 3 m 3 ⋅ mol − 1 V_m = 1.0 \times 10^{-3}\ \text{m}^3 \cdot \text{mol}^{-1} V m = 1.0 × 1 0 − 3 m 3 ⋅ mol − 1 N 2 \text{N}_2 N 2
π T = 0.364 ( 1.0 × 10 − 3 ) 2 = 0.364 1.0 × 10 − 6 = 3.64 × 10 5 Pa ≈ 3.6 atm \pi_T = \frac{0.364}{(1.0 \times 10^{-3})^2} = \frac{0.364}{1.0 \times 10^{-6}} = 3.64 \times 10^5\ \text{Pa} \approx 3.6\ \text{atm} π T = ( 1.0 × 1 0 − 3 ) 2 0.364 = 1.0 × 1 0 − 6 0.364 = 3.64 × 1 0 5 Pa ≈ 3.6 atm 这意味着，N 2 \text{N}_2 N 2 1 m 3 1\ \text{m}^3 1 m 3 3.64 × 10 5 J 3.64 \times 10^5\ \text{J} 3.64 × 1 0 5 J
