Calculation of Helmholtz energy

1. Sep 23, 2013

Youngster

1. The problem statement, all variables and given/known data

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 = $\frac{nRT}{V}$ - $\frac{an^{2}}{V^{2}}$

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).

2. Relevant equations

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

3. The attempt at a solution

Well I tried integrating the fourth fundamental equation

$\int$dA = -$\int$PdV -$\int$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 = $\frac{nRT}{V}$ - $\frac{an^{2}}{V^{2}}$

I'm letting R = 8.314 $\frac{Pa m^{3}}{K mol}$ 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.

2. Sep 23, 2013

Pranav-Arora

Yes. But just plug the expression in the integral, substitute the values in the end.

Well, what do you think?

You are integrating with respect to volume. Do you know about definite integrals?

3. Sep 23, 2013

Youngster

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:
-$\int$dA = $\int$$\frac{nRT}{V}$-$\frac{an^{2}}{v^{2}}$dV,

from 20L to 40L

This comes out as
nRT ln($\frac{V_{2}}{V_{1}}$)+$\frac{an^{2}}{V_{2}-V_{1}}$

And plugging in, I come up with A = -1733.85 Pa m$^{3}$ = -1733.85 J

Does this look good?

4. Sep 24, 2013

Pranav-Arora

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

Last edited: Sep 24, 2013
5. Sep 25, 2013

Youngster

-$\int$$\frac{1}{x}$

I understand - From 20L to 40L, the later term should have the difference $\frac{1}{V_{2}}$ - $\frac{1}{V_{1}}$ multiplied by an$^{2}$

6. Sep 25, 2013

Pranav-Arora

You mean -1/x?
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?