# Calculation of Helmholtz energy

Youngster

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

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).

## 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

$\int$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.

## Answers and Replies

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

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

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?

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

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

-$\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$^{2}$

Saitama
-$\int$$\frac{1}{x}$
You mean -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$^{2}$
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?