Calculation of Helmholtz energy

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Homework Statement



The 4 fundamental equations of thermodynamics are:

dE = TdS - PdV
dH = TdS + VdP
dG = VdP - SdT
dA = - PdV - SdT


Supose a gas obeys the equation of state

P = [itex]\frac{nRT}{V}[/itex] - [itex]\frac{an^{2}}{V^{2}}[/itex]

Use one of the fundamental equations to find the change in Helmholtz energy (A) when one mole of gas expands isothermally from 20 L to 40 L at 300 K. Let a = 0.1 Pa m6 mol-2. (1 L = 10-3 m3).

Homework Equations



Well the four fundamental equations should be a given. In particular, the fourth one for Helmholtz energy dA.

The Attempt at a Solution



Well I tried integrating the fourth fundamental equation

[itex]\int[/itex]dA = -[itex]\int[/itex]PdV -[itex]\int[/itex]SdT

And since the process is isothermal, the last term is zero, and the Helmholtz energy is just the product of the pressure P and the change in volume ΔV.

But how would I obtain the pressure P? My first guess would be to plug in the known values into the equation of state:

P = [itex]\frac{nRT}{V}[/itex] - [itex]\frac{an^{2}}{V^{2}}[/itex]

I'm letting R = 8.314 [itex]\frac{Pa m^{3}}{K mol}[/itex] since that would lead to a dimensionally correct answer in Pa. My problem is what to plug in for volume considering I have two values.
 

Answers and Replies

  • #2
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But how would I obtain the pressure P? My first guess would be to plug in the known values into the equation of state:

P = [itex]\frac{nRT}{V}[/itex] - [itex]\frac{an^{2}}{V^{2}}[/itex]
Yes. But just plug the expression in the integral, substitute the values in the end.

My problem is what to plug in for volume considering I have two values.

Well, what do you think? :rolleyes:

You are integrating with respect to volume. Do you know about definite integrals?
 
  • #3
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Yep! This was actually a very DOH! moment for me.

So I forgot that I can just plug in the equation of state and write P in terms of V.

This leads to the differential equation:
-[itex]\int[/itex]dA = [itex]\int[/itex][itex]\frac{nRT}{V}[/itex]-[itex]\frac{an^{2}}{v^{2}}[/itex]dV,

from 20L to 40L

This comes out as
nRT ln([itex]\frac{V_{2}}{V_{1}}[/itex])+[itex]\frac{an^{2}}{V_{2}-V_{1}}[/itex]

And plugging in, I come up with A = -1733.85 Pa m[itex]^{3}[/itex] = -1733.85 J

Does this look good?
 
  • #4
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This comes out as
nRT ln([itex]\frac{V_{2}}{V_{1}}[/itex])+[itex]\frac{an^{2}}{V_{2}-V_{1}}[/itex]

I suggest you to check this again. The second part doesn't look correct. What is ##\displaystyle \int \frac{1}{x^2}dx##?
 
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  • #5
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I suggest you to check this again. The second part doesn't look correct. What is ##\displaystyle \int \frac{1}{x^2}dx##?

-[itex]\int[/itex][itex]\frac{1}{x}[/itex]

I understand - From 20L to 40L, the later term should have the difference [itex]\frac{1}{V_{2}}[/itex] - [itex]\frac{1}{V_{1}}[/itex] multiplied by an[itex]^{2}[/itex]
 
  • #6
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-[itex]\int[/itex][itex]\frac{1}{x}[/itex]
You mean -1/x?
I understand - From 20L to 40L, the later term should have the difference [itex]\frac{1}{V_{2}}[/itex] - [itex]\frac{1}{V_{1}}[/itex] multiplied by an[itex]^{2}[/itex]
Check again, you missed the minus sign.

When you solve ##\displaystyle \int_a^b \frac{1}{x^2} dx##, you get
$$-\left(\frac{1}{b}-\frac{1}{a}\right)$$
Do you see how?
 

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