How to Find the Change in Helmholtz Energy for an Isothermal Gas Expansion?

[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

\intdA = -\intPdV -\intSdT

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.

 

[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:
-\intdA = \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 $\displaystyle \int \frac{1}{x^2}dx$?

 

[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?