# Finding b in Van der Waals Equation?

• DaynaClarke
In summary: U## for this gas? Does it depend on b?In summary, the Van der Waals equation of state for a gas is p(V-b)=nRT, where b is the excluded volume for two particles of the gas. In an isobaric expansion from a volume of 1m3 to 2m3, with 1 mole of the gas, the change in internal energy is ΔU=nCv(T2-T1), where T1 and T2 are the initial and final temperatures, respectively. Using the algebraic expressions for T1 and T2, it can be seen that b does not affect the temperature change, and therefore does not need to be identified. The gas in this problem is non-
DaynaClarke
Member has been warned not to remove the template.
I've got a question that requires me to use the Van der Waals equation in the form:
p(V-b)=nRT
The process is isobaric, the volume changes from 1m3 to 2m3, and there is 1 mole of the unidentified gas.
Ultimately, I need to find initial and final values of T. So I rearranged the formula:
p(V-b)/nR=T

So I need b. I think b is the volume per mole, so that's how I worked it out - as V/n. However, this gave me b as 1m3/mol and 2m3/mol. That gives me initial and final temperatures of 0K, which I'm guessing is wrong.

In short, if someone could point me in the right direction as to what b actually is, I would very much appreciate it. Thank you.

DaynaClarke said:
I've got a question that requires me to use the Van der Waals equation in the form:
p(V-b)=nRT
The process is isobaric, the volume changes from 1m3 to 2m3, and there is 1 mole of the unidentified gas.
Ultimately, I need to find initial and final values of T. So I rearranged the formula:
p(V-b)/nR=T

So I need b. I think b is the volume per mole, so that's how I worked it out - as V/n. However, this gave me b as 1m3/mol and 2m3/mol. That gives me initial and final temperatures of 0K, which I'm guessing is wrong.

In short, if someone could point me in the right direction as to what b actually is, I would very much appreciate it. Thank you.
What is the exact statement of the problem?

##b## should be the excluded volume for two particles of the gas (https://en.wikipedia.org/wiki/Van_der_Waals_equation) so it depends on the kind of gas you are working with. And it makes sense because because VdW introduced his formula to take in account for different kind of molecules (Pv=RT handles all gases the same way).
If you don't know anything about the gas I don't know how you can find a correct value for ##b##

Chestermiller said:
What is the exact statement of the problem?

"The equation of state of n moles of a gas is p(V-b)=nRT. The molar heat capacities of the gas at constant pressure and constant volume, cp and cv, satisfy cp - cv = R, where R is the gas constant, and γ = cp/cv = 5/3.

Find the change in internal energy of one mole of the gas in an isobaric expansion at a pressure of 1 bar from a volume of 1 m3 to 2 m3."

I have values for cp and cv, and I know that ΔU= Q - W = n⋅cp⋅dT - pdV. However, due to being unable to identify values for b at initial and final volumes, I don't know how to find initial and final temperatures.

DaynaClarke said:
"The equation of state of n moles of a gas is p(V-b)=nRT. The molar heat capacities of the gas at constant pressure and constant volume, cp and cv, satisfy cp - cv = R, where R is the gas constant, and γ = cp/cv = 5/3.

Find the change in internal energy of one mole of the gas in an isobaric expansion at a pressure of 1 bar from a volume of 1 m3 to 2 m3."

I have values for cp and cv, and I know that ΔU= Q - W = n⋅cp⋅dT - pdV.
This equation is incorrect. For this equation of state, $$\Delta U=nC_v(T_2-T_1)$$
Do you know how to show that this in the change in internal energy for this equation of state?
However, due to being unable to identify values for b at initial and final volumes, I don't know how to find initial and final temperatures.
Algebraically, in terms of p, ##V_1## and b, what is the initial temperature?
Algebraically, in terms of p, ##V_2##, and b, what is the final temperature?
From these algebraic expressions, what is ##(T_2-T_1)##? Does it involve b?

If you have experimental data, you could do a best fit to your data to find b.

Dr. Courtney said:
If you have experimental data, you could do a best fit to your data to find b.
That's not necessary. To solve this problem, the value of b is not needed.

Chestermiller said:
This equation is incorrect. For this equation of state, $$\Delta U=nC_v(T_2-T_1)$$
Do you know how to show that this in the change in internal energy for this equation of state?

Is it because the gas is ideal, and the internal energy of an ideal gas is purely kinetic, so it only considers the heat? I'll admit I'm not sure why it's Cv and not Cp because volume isn't constant, but pressure is.

Algebraically, in terms of p, ##V_1## and b, what is the initial temperature?
$$\frac{p(V_1-b)}{nR} = T_1$$
Algebraically, in terms of p, ##V_2##, and b, what is the final temperature?
$$\frac{p(V_2-b)}{nR} = T_2$$
From these algebraic expressions, what is ##(T_2-T_1)##? Does it involve b?
$$T_2-T_1 = \frac{p(V_1-V_2)}{nR}$$
So it doesn't contain b at all, it cancels out.

DaynaClarke said:
Is it because the gas is ideal, and the internal energy of an ideal gas is purely kinetic, so it only considers the heat? I'll admit I'm not sure why it's Cv and not Cp because volume isn't constant, but pressure is.
No. This gas is not ideal. Are you familiar with the following equation for a general non-ideal gas:
$$dU=nC_vdT-\left[p-T\left(\frac{\partial p}{\partial T}\right)_V\right]dV$$
If not, then your teacher did you a disservice by assigning you this problem. In any event, for the particular equation of state in the present problem, what do you get for the term in brackets?

Regarding Cv and Cp, you are aware of the following definitions, correct?:
$$nC_v=\left(\frac{\partial U}{\partial T}\right)_V$$and $$nC_p=\left(\frac{\partial H}{\partial T}\right)_p$$So, Cp is always associated with H and Cv is always associated with U. And, for this gas, if you did the math above correctly, you know that U is independent of V, and depends only on T.

$$\frac{p(V_1-b)}{nR} = T_1$$

$$\frac{p(V_2-b)}{nR} = T_2$$

$$T_2-T_1 = \frac{p(V_1-V_2)}{nR}$$
So it doesn't contain b at all, it cancels out.
These results are correct. So, what do you get if you substitute this result for the temperature change into the equation for ##\Delta U##?

DaynaClarke said:
Is it because the gas is ideal, and the internal energy of an ideal gas is purely kinetic, so it only considers the heat? I'll admit I'm not sure why it's Cv and not Cp because volume isn't constant, but pressure is.$$\frac{p(V_1-b)}{nR} = T_1$$

$$\frac{p(V_2-b)}{nR} = T_2$$

$$T_2-T_1 = \frac{p(V_1-V_2)}{nR}$$
So it doesn't contain b at all, it cancels out.

Great job. One thing I often tell students is that if you are not given a value that appears in an equation, and you are not asked for it, and you don't have the information needed to find it, it probably cancels out.

Chestermiller

## 1. What is the Van der Waals Equation?

The Van der Waals Equation is a mathematical formula that describes the behavior of real gases. It accounts for the intermolecular forces and molecular volume that are neglected in the ideal gas law.

## 2. What is the significance of finding b in the Van der Waals Equation?

The value of b in the Van der Waals Equation represents the excluded volume of the molecules, or the space that the molecules take up in a given volume. It is used to correct for the volume of the molecules in real gases, making the equation more accurate.

## 3. How is b calculated in the Van der Waals Equation?

The value of b is calculated by taking the total volume of the molecules and dividing it by the number of moles of gas. It is often experimentally determined and can vary depending on the type of gas.

## 4. Why is it important to use the Van der Waals Equation instead of the ideal gas law?

The ideal gas law assumes that gas molecules have no volume and do not interact with each other. However, in real gases, molecules do have volume and interact with each other. By using the Van der Waals Equation, we can account for these factors and obtain more accurate results.

## 5. How does the value of b affect the behavior of a gas?

The value of b has a significant impact on the behavior of a gas. A larger b value means that the molecules take up more space and have stronger intermolecular forces, causing the gas to deviate more from ideal behavior. A smaller b value indicates that the molecules are more compact and have weaker intermolecular forces, resulting in a gas that behaves more like an ideal gas.

### Similar threads

• Introductory Physics Homework Help
Replies
2
Views
949
• Introductory Physics Homework Help
Replies
6
Views
2K
• Introductory Physics Homework Help
Replies
31
Views
769
• Introductory Physics Homework Help
Replies
6
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
2K
• Introductory Physics Homework Help
Replies
10
Views
20K
• Thermodynamics
Replies
4
Views
538
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
8
Views
2K
• Classical Physics
Replies
23
Views
2K