Proving Internal Pressure of Ideal & VdW Gases

In summary, the conversation discusses the problem of proving the internal pressure of an ideal gas to be 0 and for a Van der Waals gas to be ((n^2)a)/(v^2). The conversation mentions the given thermal equations of state for both gases and suggests using the identity \left(\frac{\partial U}{\partial V}\right)_T=T\left(\frac{\partial p}{\partial T}\right)_V-p to derive the caloric equation of state. The conversation ends with the individual coming up with a solution for the problem.
  • #1
Jennifer Lyn
5
0
Like in the other problem I posted- This is the other question that I missed and just can't find a solution for.

Homework Statement


Prove the internal pressure is 0 for an ideal gas and ((n^2)a)/(v^2) for a Van der Waals gas.

Homework Equations


1. VdQ Eqn: p= (nRT)/(v-b) - ((n^2)a)/(v^2)
2. (partial S/partial V) for constant T = (partial p/partial T) for contant V.
3. dU = TdS - pdV
4. pi sub t (internal pressure) = (partial U/partial V) for constant T


The Attempt at a Solution



a) Ideal Gas
0 = (partial U/partial V) for const T
int 0 dv = int du
0 = int (TdS - pdV)
int p dv = int T ds
int (nRT/v) dv = int (Pv/nR) dS
nRT x int(1/V) dv = pv/nR x int 1 dS
... and I get kind of lost here, though I know that what I've already done is wrong.. :(

b) VdW gas
I actual have to get going to school, but I'll come back and type up what I've done (incorrectly :( for this part afterwards).
 
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  • #2
Hello Jennifer,

I suppose you are given the so called thermal equation of state [itex]p=p(T,V,n)[/itex] for both

the ideal gas

[tex]p=\frac{nRT}{V}[/tex]

and the Van der Waals gas

[tex]p=\frac{nRT}{V-nb}-\frac{n^2a}{V^2}[/tex]

The inner pressure [itex]\left(\frac{\partial U}{\partial V}\right)_T[/itex] can be calculated after finding the so called caloric equation of state [itex]U=U(T,V,n)[/itex] for both cases.

Another straightforward method would be to use the following identity which shows that the caloric and thermal equations of state are not independent of each other:

[tex]\left(\frac{\partial U}{\partial V}\right)_T=T\left(\frac{\partial p}{\partial T}\right)_V-p[/tex]

Do you know how to derive this identity?
 
  • #3
I think so..
[tex]\Pi[/tex]t = [tex]\partial[/tex]u/[tex]\partial[/tex]v for constant t
= ( 1/[tex]\partial[/tex]v[tex]\times[/tex](Tds - pdv) )
= T [tex]\times[/tex] ([tex]\partial[/tex]p/[tex]\partial[/tex]t) - p

I think that's right. I still don't know how to get from that Maxwell relation to the ideal gas and Van der Waals eqn, though.
 
Last edited:
  • #4
Ok, I think I figured it out, from my previous post (sorry- I am still getting used to using the tools for math on this board) I replace the vanderwaals eqn into P in my partial p and then just solve from there.
 
  • #5
Thanks everyone!
 

1. What is the difference between an ideal gas and a van der Waals gas?

An ideal gas is a theoretical gas that follows the gas laws exactly, meaning that its particles have no volume and do not interact with each other. A van der Waals gas takes into account the volume of the gas particles and the attractive forces between them, making it a more realistic model for real gases.

2. How do you determine the internal pressure of an ideal gas?

The internal pressure of an ideal gas can be calculated using the ideal gas law, which states that pressure is equal to the number of moles of gas multiplied by the gas constant and temperature, divided by the volume of the gas.

3. What factors affect the internal pressure of a van der Waals gas?

The internal pressure of a van der Waals gas is affected by the volume of the gas particles, the strength of the intermolecular forces, and the temperature of the gas. These factors can cause deviations from ideal gas behavior.

4. How do you experimentally prove the internal pressure of a gas?

To prove the internal pressure of a gas, one can perform experiments to measure the pressure, volume, and temperature of the gas at different conditions. These values can then be used to calculate the internal pressure and compare it to theoretical predictions.

5. What are some real-world applications of understanding the internal pressure of gases?

Understanding the internal pressure of gases is important in many fields, including chemistry, physics, and engineering. It is used in the design and operation of gas storage tanks, the compression and expansion of gases in engines, and in the production and purification of industrial gases.

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