Difference between isobar and isochor heat capacity

In summary, enthalpy and energy are combined to get U. U is a function of T and V. The last term in U is CV dT. The equation for isobaric process is CP - CV = T (∂S/∂V)T (∂V/∂T)P.
  • #1
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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 :)
 
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  • #2
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 CV dT, so:

T dS = [(∂U/∂T)V + p ] dV + CV dT

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

CP - CV = [(∂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)

CP - CV = 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):

CP - CV = 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
 
  • #3
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.
 
  • #4
So let us start with the equation before that line:


T dS = [(∂U/∂T)V + p] dV + CV 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 + CV

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

The left hand side is precisely Cp. Now take the CV to the left and you get

Cp - CV = [(∂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
 
  • #5


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.
 

What is the difference between isobar and isochor heat capacity?

Isobar and isochor heat capacity are two different ways of measuring the heat capacity of a substance. Isobar heat capacity measures the heat required to raise the temperature of a substance at a constant pressure, while isochor heat capacity measures the heat required to raise the temperature of a substance at a constant volume.

How are isobar and isochor heat capacity related?

Isobar and isochor heat capacity are related by the equation Cp = CV + R, where Cp is the isobar heat capacity, CV is the isochor heat capacity, and R is the gas constant. This means that isobar heat capacity is always greater than isochor heat capacity for the same substance.

Why is it important to know the difference between isobar and isochor heat capacity?

It is important to know the difference between isobar and isochor heat capacity because it helps us understand how different substances respond to changes in temperature and pressure. It also allows us to accurately calculate the heat capacity of a substance in different conditions, which is essential in various fields such as thermodynamics, chemistry, and engineering.

Can isobar and isochor heat capacity be measured experimentally?

Yes, isobar and isochor heat capacity can be measured experimentally using different methods such as calorimetry and differential scanning calorimetry. These experiments involve measuring the amount of heat absorbed or released by a substance at a constant pressure or volume, respectively.

How do isobar and isochor heat capacity affect the behavior of gases?

Isobar and isochor heat capacity play a crucial role in determining the behavior of gases. For example, gases with a high isochor heat capacity are more resistant to changes in temperature and will maintain a more stable volume under pressure. On the other hand, gases with a high isobar heat capacity are more likely to expand when heated at a constant pressure.

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