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

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 $$P$$?

• Are you asking why $C_p$ is not a function of volume as well as pressure, or why $C_p$ is not a function of volume instead of pressure? Jun 5, 2020 at 19:03
• i am asking why it is not a function of volume as well as pressure? why is volume excluded? Jun 5, 2020 at 19:29

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.

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.

• $C_p$ and $C_v$ are constants not functions of P and T or P and V ( atleast at room temperatures). At higer temperatures they are function of just temperatures ( debye model) Jun 6, 2020 at 4:03
• @Shashaank Specific heat capacities are the derivatives of thermodynamic potentials, and are therefore generally functions of the thermodynamic variables. The fact that they may be approximately constant in certain models of certain systems is irrelevant. Jun 6, 2020 at 4:07
• how is c = dQ/dT a derivative.... It is just a ratio not a derivative... Isn't it... because Q is not an exact differential. It's should be just a ratio, albeit the notation we use is misleading Jun 6, 2020 at 4:10
• @Shashaank it is a ratio of two infinitesimal quantities, which is what a derivative is. Strictly speaking it is not a partial derivative of U, but in every meaningful sense it has the characteristic of a derivative. Jun 6, 2020 at 4:14
• @Shashaank If you prefer not to think of it as a derivative, then $C_v$ is the equal to $T$ divided by the partial derivative of $T$ with respect to the entropy $S$. In either case, those are both functions of $V$ (in principle). Jun 6, 2020 at 4:22

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 $$C_v$$.

• so are you saying $C_P$ is a function of P,T and V? Jun 5, 2020 at 19:51
• @DhatriDongre no it's a function of either of the 2 variables Jun 6, 2020 at 4:03

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.

• firstly, i was asking for heat capacity and not specific capacity,though i am not so sure it would make a difference. secondly, i havent covered enthalpy concept yet, so i am afraid i dont understand much of it Jun 5, 2020 at 19:39
• @DhatriDongre Firstly, heat capacity is simply specific capacity multiplied by mass. It makes no difference with respect to the derivation. Secondly, the specific heat at constant pressure is defined in terms of the enthalpy. Sorry if you haven't covered enthalpy yet, but that's the reason it doesn't depend on volume and that's what you asked about. Maybe you should come back after you have learned more. Jun 5, 2020 at 19:42
• But what about this equation,$C_p=C_v +\left [\left(\frac{\partial U}{\partial V}\right)_T +P \right]V\beta$? Jun 5, 2020 at 19:44
• @DhatriDongre What equation? Jun 5, 2020 at 19:45
• But according to your definition $C_p$ and $C_v$ are just constant ( like they should be at low temperatures of course) because dh/dT is not a differentiation but just a ratio... Right, they are constants not function of P and T Jun 6, 2020 at 3:59