Can the Coefficient b in Van der Waals Equation Be Negative

  • Thread starter Thread starter mrawls
  • Start date Start date
  • Tags Tags
    Van der waals
Click For Summary

Homework Help Overview

The discussion revolves around the Van der Waals equation and the determination of the coefficient "b" based on given parameters such as "a", temperature, volume, number of moles, and pressure. The original poster expresses concern about obtaining a negative value for "b" and questions whether this indicates an error in their calculations.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to solve for "b" using the Van der Waals equation and provides their calculations. Other participants inquire about the values used in the calculations and suggest that a mistake in the pressure value may have affected the outcome.

Discussion Status

Participants are actively discussing the calculations and values involved. The original poster acknowledges a mistake in the pressure value, which they believe may have led to the negative result for "b". There is a suggestion that "b" should not be negative, indicating a shared understanding of its physical significance.

Contextual Notes

The original poster mentions challenges with unit conversions due to non-SI units provided by their professor, which may impact their calculations.

mrawls
Messages
4
Reaction score
0
solving for "b" with van der Waals

1. Homework Statement

The problem gives me "a", the temperature, volume, number of moles, and pressure. It asks me to solve for "b".

I have set up the equation to solve for b, and done it. The answer comes out with the right units but it is negative.

2. Homework Equations

Can the value of the coefficient be negative or does that mean I did it wrong?


3. The Attempt at a Solution

The equation:

p = nRT/(V-nb) - a(n/V)^2

Using that equation, I came up with:

b = (Vp + Va(n/V)^2 - nRT) / (np + na(n/V)^2)

I came up with b = -4.58x(10^-4) (m^3) (mol^-1)
Report Post Edit/Delete Message
 
Physics news on Phys.org


The equation looks right. Could you provide us with the values given please?
 


Yes, sorry I forgot to list them before.

a = .580(m^6)(Pa)(mol^-2)
T = 273K
P = 5,000,000 Pa
R = 8.3144 Pa(m^3)/(mol)K

I then solved for a(n/V)^2 and plugged that answer into the equation (I got 4734693.9 Pa for that)

I actually realized a huge mistake I made when posting these values. The given pressure was 5 MPa, but in my equation I plugged in only 1 MPa... I am almost certain that must have been the problem.
 


I think you left out the Volume there.

Did plugging in the correct value for P fix your problem?

b should definitely not be negative as it is a measure of how much volume, on average, is displaced by a given particle.
 


Wow I did forget to list the volume too.. Sorry I had a busy day haha I was not thinking clearly.

Using the correct pressure resulted in the right answer, with the right units!

Sadly my professor likes to give every value in non-SI unrelated units and we have to do all sorts of conversions :(
 

Similar threads

Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 23 ·
Replies
23
Views
12K
Replies
2
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 6 ·
Replies
6
Views
5K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K