Difference between isobar and isochor heat capacity

[Alexis21]

Hello,

I want to show:
[itex] C_p - C_v = -T \big( \frac {\partial V}{\partial p} \big)_{T,n} \big( \frac {\partial p}{\partial T} \big)_{V,n}^2 [/itex]

I started by doing this:
[itex] C_p - C_v= \big( \frac {\partial H}{\partial T} \big)_{p,N} - \big( \frac {\partial U}{\partial T} \big)_{V,n} [/itex]

Applying the definitions of enthalpy and energy:
[itex] dH = TdS + V dp + \mu dn [/itex]
and
[itex]dU = TdS - p dV + \mu dn[/itex]

I can rewrite the equation like this:
[itex]= V \big(\frac {\partial p}{\partial T} \big) + p \big( \frac {\partial V}{\partial T} \big) [/itex]
(while TdS and µdn terms cancel out each other)

Now I do not know how to continue. Can anyone help :)

 

[Chandra Prayaga]

Start with:

T dS = dU + p dV (Combined 1st and 2nd laws) (1)

Write U as a function of T and V:

T dS = [(∂U/∂T)_V dT + (∂U/∂V)_T dV] + p dV

= [(∂U/∂T)_V + p ] dV + (∂U/∂T)_V dT

The last term is C_V dT, so:

T dS = [(∂U/∂T)_V + p ] dV + C_V dT

Now do an isobaric process, and divide by dT to get:

C_P - C_V = [(∂U/∂T)_V + p ] dV (∂V/∂T)_P (2)


which is a standard relation derived, for example, in Sears

Now, starting again with the combined 1st and 2nd law (equ (1)):

dU + p dV = T dS

Divide by dV keeping T constant :

( ∂U/∂V)_T + p = T (∂S/∂V)_T (3)

Substituting this for the term in square brackets in equ (2)

C_P - C_V = T (∂S/∂V)_T (∂V/∂T)_P (4)

Use one of Maxwell's equations (Ref: Sears) to substitute for the first partial derivative on the right of equ (4):

C_P - C_V = T (∂P/∂T)_V (∂V/∂T)_P (5)

Now use the standard relation for partial derivatives (also ref: Sears)

(∂V/∂T)_P (∂T/∂P)_V (∂P/∂V)_T = -1

to substitute for the second partial derivative on the right of equ (5) and get what you need.

Please let me know if you need clarification

 

[Alexis21]

Thank you for your answer :)

I have got a question on that:
What does it exactly mean when you say 'Now do an isobaric process'. I don't see how the C_p drops in.

 

[Chandra Prayaga]

So let us start with the equation before that line:


T dS = [(∂U/∂T)_V + p] dV + C_V dT

In an isobaric process, each of the changes dS, dV and dT will have certain values. So when we divide by that value of dT, we get:

T (∂S/∂T)_p = [(∂U/∂T)_V + p] (∂V/∂T)_p + C_V

The constant p on the partial derivatives signifying the isobaric process.

The left hand side is precisely C_p. Now take the C_V to the left and you get

C_p - C_V = [(∂U/∂T)_V + p] (∂V/∂T)_p

This is equation (2) in what I wrote earlier. Incidentally, there was a typo in the earlier equation (2). There is a dV extra which should be erased.

Let me know if you need any more help

 

[sardar]

Hello,

The difference between isobar and isochor heat capacity is related to the way heat is transferred to a system at constant pressure (isobaric) or constant volume (isochoric). Isobaric heat capacity (Cp) is the amount of heat required to increase the temperature of a substance by 1 degree while keeping the pressure constant. On the other hand, isochoric heat capacity (Cv) is the amount of heat required to increase the temperature of a substance by 1 degree while keeping the volume constant.

To relate these two quantities, we can use the thermodynamic identity:
dU = TdS - pdV + \mu dn

Where U is the internal energy, T is the temperature, S is the entropy, p is the pressure, V is the volume, and \mu is the chemical potential. By differentiating this equation with respect to temperature at constant volume, we get:
\big( \frac {\partial U}{\partial T} \big)_{V,n} = T \big( \frac {\partial S}{\partial T} \big)_{V,n} - p \big( \frac {\partial V}{\partial T} \big)_{V,n} + \mu \big( \frac {\partial n}{\partial T} \big)_{V,n}

Similarly, by differentiating at constant pressure, we get:
\big( \frac {\partial H}{\partial T} \big)_{p,n} = T \big( \frac {\partial S}{\partial T} \big)_{p,n} + V \big( \frac {\partial p}{\partial T} \big)_{p,n} + \mu \big( \frac {\partial n}{\partial T} \big)_{p,n}

Subtracting these two equations, we get:
C_p - C_v = -T \big( \frac {\partial V}{\partial p} \big)_{T,n} \big( \frac {\partial p}{\partial T} \big)_{V,n}^2

This equation shows the relationship between the two heat capacities and how they are related to changes in volume and pressure at constant temperature and number of moles. I hope this helps clarify the difference between isobaric and isochoric heat capacity. Let me know if you have any further questions.