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

In summary: When you solve ##\displaystyle \int_a^b \frac{1}{x^2} dx##, you get $$-\left(\frac{1}{b}-\frac{1}{a}\right)$$I see!
  • #1
Youngster
38
0

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.
 
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  • #2
Youngster said:
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
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
Youngster said:
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##?
 
Last edited:
  • #5
Pranav-Arora said:
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
Youngster said:
-[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?
 

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

1. What is Helmholtz energy and why is it important?

Helmholtz energy, also known as free energy, is a thermodynamic property that measures the amount of energy available to do work in a system. It is important because it helps us understand and predict the behavior of a system under different conditions.

2. How is Helmholtz energy calculated?

Helmholtz energy can be calculated using the formula A = U - TS, where A is Helmholtz energy, U is internal energy, T is temperature, and S is entropy. This formula is based on the first law of thermodynamics, which states that energy cannot be created or destroyed, only transferred or converted.

3. What is the significance of Helmholtz energy in phase transitions?

In phase transitions, the Helmholtz energy is used to determine the stability of a system. When the Helmholtz energy is at a minimum, the system is in its most stable state. This helps us understand why certain substances undergo phase transitions at certain temperatures and pressures.

4. Can Helmholtz energy be negative?

Yes, Helmholtz energy can be negative. This usually occurs when the internal energy of a system is greater than the product of its temperature and entropy. Negative Helmholtz energy indicates that the system has more available energy to do work than it would at a higher energy state.

5. How is Helmholtz energy related to other thermodynamic properties?

Helmholtz energy is related to other thermodynamic properties through various equations, such as the Maxwell relations. It is also related to the Gibbs free energy, which is another thermodynamic property that takes into account both the enthalpy and entropy of a system. Additionally, the Helmholtz energy can be used to calculate other important thermodynamic properties, such as the heat capacity and chemical potential.

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