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Finding b in Van der Waals Equation?

  • #1
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.
 

Answers and Replies

  • #2
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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?
 
  • #3
dRic2
Gold Member
635
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##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##
 
  • #4
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.
 
  • #5
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"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?
 
  • #6
Dr. Courtney
Education Advisor
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If you have experimental data, you could do a best fit to your data to find b.
 
  • #7
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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.
 
  • #8
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.
 
  • #9
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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##?
 
  • #10
Dr. Courtney
Education Advisor
Insights Author
Gold Member
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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.
 

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