Why is heat capacity at const pressure, $C_p$ not a function of volume?

Question

I am reading heat and thermodynamics by Zemansky and while defining heat capacities at constant pressure and volume, it is said that heat capacity at constant pressure is a function of $P$ and $T$. Why not V? and likewise for heat capacity at constant volume is a function of $V$ and $T$, why not $P$?

Answer 1

The equation of state for a thermodynamical system with a fixed number of particles takes the form $$f(p,V,T)=0$$ for some function $f$. For example, in the case of the ideal gas, one has that

$$f(p,V,T) = pV-nRT = 0$$

As a result, $p,V,$ and $T$ are not all independent of one another, in the sense that you are not free to specify all three independently.

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A function like $C_p$ which is defined for a particular thermodynamic system cannot be considered a function of $p,V,$ and $T$ because those three variables cannot be chosen independently. It would be like saying $f(x,2x) = 2x^2$ is a function of both $x$ and $2x$, which doesn't make sense.

Instead, we work in a framework where we either consider $P$ to be a function of $V$ (and other variables) or vice-versa. We can go back and forth between these viewpoints via Legendre transformation. More concretely, when we talk about the internal energy $U$ or the Helmholtz energy $A$, then we are considering $p$ to be a function of $V$. When we talk about the enthalpy $H$ or the Gibbs energy $G$, then we are considering $V$ to be a function of $p$.

But what about this equation $C_p = C_v + \left[\left(\frac{\partial U}{\partial V}\right)_T + P\right] V\beta$

Written out with all of the arguments for the functions, this equation becomes

$$C_p(P,T) = C_v\big(V(P,T) , T\big) + \left[\left(\frac{\partial U}{\partial V}\right)_T(P,T) + P\right]V(P,T) \beta$$

In words, give some $P$ and $T$,

1. First use $P,T$ to find $V(P,T)$. Plug this into the first slot of the function $C_v$, and plug $T$ into the second slot.
2. Next, calculate $\left(\frac{\partial U}{\partial V}\right)_T$ and evaluate it at $P$ and $T$. Add $P$ to the result.
3. Multiply the output of step 2 by $V(P,T)$ and $\beta$
4. Add the result of step 3 to the result of step 1

That is the recipe which defines the function you wrote, and which gives the specific heat at constant pressure as a function of $P$ and $T$ only.

Answer 2

Specific heats of solid are generally constant upto good ranges of temperature. At higher temperatures specific heats do become a function of temperature ( according to Einstein and Debye Model) However say if $C_p$ is a function of Pand T then the equation of state can help to calculate $C_p$ as a function of T and V. Similarly for $C_v$.

Answer 3

The specific heat at constant pressure is not a function of volume due to the fact that it is defined in terms of the change in enthalpy with temperature at constant pressure.

The specific heat at constant pressure is defined by

$$c_{p}=\biggl(\frac{\delta h}{\delta T}\biggr)_P$$

Specific enthalpy is

$$dh=du+d(pV)=q-pdv+pdv+vdp$$

For a constant pressure process $dp=0$;

$$dh=q$$

$$q=c_{p}dT$$

$$dh=c_{p}dT$$

$$c_{p}=\frac{dh}{dT}$$

Similarly, in the case of the specific heat at constant volume it is defined by

$$c_{v}=\biggl(\frac{\delta u}{\delta T}\biggr)_V$$

Specific internal energy closed system is

$$du=q-w=q-pdv$$

For a constant volume process $dv=0$;

$$du=q=c_{v}dT$$

$$c_{v}=\frac{du}{dT}$$

Hope this helps.