为何$dQ$与$dL$不是恰当微分而$dU$是？能否从解析层面验证？

Great question—this cuts right to the core of state functions vs. path functions, which is foundational to thermodynamics. Let’s break this down clearly, with both conceptual and analytical explanations:

First: What defines an exact differential? #

An exact (or "proper") differential corresponds to a state function: a quantity that depends only on the system’s current state (e.g., pressure $P$, volume $V$, temperature $T$), not the path taken to reach that state. For a differential $df = M(x,y)dx + N(x,y)dy$ to be exact, it must satisfy Clairaut’s theorem:
$$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$$
Integrating an exact differential between two states gives a result that only depends on the initial and final states, never the path.

Why $dU$ is an exact differential #

Internal energy $U$ is a pure state function. For any system, $U$ is uniquely determined by its state parameters—for an ideal gas, $U = nC_VT$, where $n$ is moles, $C_V$ is molar constant-volume heat capacity, and $T$ is temperature.

No matter what process you use to go from state $A(P_1,V_1,T_1)$ to state $B(P_2,V_2,T_2)$, the change $\Delta U = U(B) - U(A)$ will always be identical. This directly means $dU$ is an exact differential: its integral doesn’t depend on the path taken.

Analytical verification for $dU$ (ideal gas example) #

Let’s use $P$ and $V$ as state variables. For an ideal gas, $T = \frac{PV}{nR}$, so $U = \frac{C_V}{R}PV$. Taking the differential:
$$dU = \frac{C_V}{R}P dV + \frac{C_V}{R}V dP$$
Here, $M = \frac{C_V}{R}P$ and $N = \frac{C_V}{R}V$. Compute the partial derivatives:
$$\frac{\partial M}{\partial V} = \frac{C_V}{R}, \quad \frac{\partial N}{\partial P} = \frac{C_V}{R}$$
They’re equal, so $dU$ satisfies the exact differential condition perfectly.

Why $dQ$ and $dL$ are NOT exact differentials #

Heat $Q$ and work $L$ are path functions: their values depend entirely on the specific process (path) used to transition between states, not just the initial and final states.

For example, take an ideal gas expanding from $V_1$ to $V_2$:

• In an isothermal expansion, work done $L = nRT\ln\left(\frac{V_2}{V_1}\right)$, and heat $Q = L$ (since $\Delta U=0$ for ideal gas isothermal processes).
• In an isobaric expansion, work done $L = P(V_2 - V_1)$, and heat $Q = nC_P\Delta T$ (where $C_P = C_V + R$).

The values of $Q$ and $L$ are completely different for these two paths, even though the initial and final states are identical. This means $\int dQ$ and $\int dL$ depend on the path—so their differentials can’t be exact.

Analytical verification for $dQ$ (ideal gas example) #

From the first law of thermodynamics: $dQ = dU + PdV$. Substitute our earlier expression for $dU$:
$$dQ = \frac{C_V}{R}P dV + \frac{C_V}{R}V dP + P dV = \frac{C_P}{R}P dV + \frac{C_V}{R}V dP$$
Here, $M = \frac{C_P}{R}P$ and $N = \frac{C_V}{R}V$. Compute partial derivatives:
$$\frac{\partial M}{\partial V} = 0, \quad \frac{\partial N}{\partial P} = \frac{C_V}{R}$$
Since $C_V \neq 0$, these are not equal—so $dQ$ fails the exact differential test.

Analytical verification for $dL$ (volume work example) #

For volume work, $dL = P dV$. Treating $P$ and $V$ as state variables, we can write this as:
$$dL = P dV + 0 \cdot dP$$
Here, $M = P$ and $N = 0$. Compute partial derivatives:
$$\frac{\partial M}{\partial P} = 1, \quad \frac{\partial N}{\partial V} = 0$$
These are not equal, so $dL$ is not an exact differential.

A quick side note: Even though $dQ$ isn’t exact, multiplying it by $1/T$ (an integrating factor) gives $dS = \frac{dQ_{\text{rev}}}{T}$, which is an exact differential—this is the definition of entropy, another key state function in thermodynamics.

内容的提问来源于stack exchange，提问作者Landau