Thermodynamics: air expansion in a cylinder

[sara lopez]

Homework Statement

Consider air expansion inside a cylinder. Assume that the volume and initial pressure is 1 ft^3 and 1500 PSI ABS respectively. If the expansion process is reversible and the path is given by P.V^1.4=constant. Calculate the total work done by the gas to reach the final volume of 8 ft^3, express the result in BTU if used:
a) The Van der Waals equation

2. Homework Equations
(P + a^2 / v^2) * (v - b) = RT[/B]
(V−b)T^R/CV=constant.

The Attempt at a Solution

T1(v1 − b)^ R/cV = T2(v2 − b)^ R/cV
W = −∆E = cV (T2 − T1) − a(1/v2 − 1/v1)
how do I replace T2 and T1?
I have two ideas: T1= (P + a^2 / v^2) * (v - b) /R
T2 = (P + a^2 / v^2) * (v - b) /R
or
T1= (P + a^2 / v^2) * (v - b) /R
T2= T1 ( (v1 − b)^ R/cV / (v2 − b)^ R/cV )

 

[Chestermiller]

If you know the P-V relationship for the process, why do you need the van der Waals equation to calculate the work?

 

[sara lopez]

I think that the process is for ideal gas, and i already did that, but for van der waals it couldn't be the process, how can I find the work without that process?

 

[Chestermiller]

I think that the process is for ideal gas, and i already did that, but for van der waals it couldn't be the process, how can I find the work without that process?

You don't know anything about the heat added or removed, so you can't do a van der waals gas. If you assume that the process is adiabatic and reversible, then you can. But why do you think that the problem statement is referring to an ideal gas, particularly if the initial pressure is 100 atm?

 

[sara lopez]

yes, I have to assume that the process is adiabatic and reversible, and because the first point was find the work with an ideal gas

 

[Chestermiller]

yes, I have to assume that the process is adiabatic and reversible, and because the first point was find the work with an ideal gas

Did you leave that out of the problem statement? What else did you leave out?

 

[sara lopez]

the problem only says to find the work with an ideal gas and with the van der waals equation, but I already did with an ideal gas, i don`t how to do it with the van der waals equation

 

[Chestermiller]

the problem only says to find the work with an ideal gas and with the van der waals equation, but I already did with an ideal gas, i don`t how to do it with the van der waals equation

The starting point for the analysis is the general equation for dU as a function of dT and dV:
$$dU=C_vdT+\left[T\left(\frac{\partial P}{\partial T}\right)_V-P\right]dV$$
If you substitute the equation for a van der waals gas into the second term in this equation, what do you get for dU?

 

[sara lopez]

U = nCv (T2- T1) + a (1/ v1 - 1/v2)

 

[sara lopez]

U= nCv (T2-T1) + a (1/v1 - 1/v2)

 

[Chestermiller]

I was asking for dU in terms of dT and dV. When you integrated, how did you know that Cv was not a function of V? From your equation for dU in terms of dT and dV, can you tell if Cv is a function of V? How?

 

[Chestermiller]

A trick on this problem is to recognize that , even though you don't know the number of moles n or the initial temperature T1, you know the product nT1. This will help you get the amount of work by determining delta U.

 

[sara lopez]

So how do I know if Cv is a function of V?

 

[Chestermiller]

So how do I know if Cv is a function of V?

You found that, for a van der waals gas, $$du=c_vdT+\frac{a}{v^2}dv$$where I have used lower case symbols to represent the properties per mole. From this equation,$$\left(\frac{\partial u}{\partial T}\right)_v=c_v$$and $$\left(\frac{\partial u}{\partial v}\right)_T=\frac{a}{v^2}$$So that means that $$\frac{\partial^2 u}{\partial T \partial v}=\left(\frac{\partial c_v}{\partial v}\right)_T=\left(\frac{\partial (a/v^2)}{\partial T}\right)_v$$But, the right hand side of this equation is equal to zero. Therefore, ##c_v## is not a function of v for a van der Waals gas; it is only a function of T. Therefore, it must be equal to the limiting value at infinite v, which denotes the ideal gas value. Therefore ##c_v## for a van der Waals gas is the value for the same gas in the ideal gas limit.

Sara, I think I may have been incorrect in what I said in post #12. For a van der Waals gas, it may not be possible to get the total work done W, but only the work per mole w (unless they told you the number of moles or the initial temperature). So let's work it out per mole first (anyway).

You found that $$w=-\Delta u=-\left[c_v(T_2-T_1)+a\left(\frac{1}{v_1}-\frac{1}{v_2}\right)\right]$$
You should also be able to show that $$\frac{T_2}{T_1}=\left(\frac{(v_1-b)}{(v_2-b)}\right)^{\frac{R}{c_v}}$$
Therefore, $$w=-\left[c_vT_1\left(\left(\frac{(v_1-b)}{(v_2-b)}\right)^{\frac{R}{c_v}}-1\right)+a\left(\frac{1}{v_1}-\frac{1}{v_2}\right)\right]$$
This is the work per mole. But, I don't know how to get the number of moles without knowing T1.

Chet

 

[sara lopez]

To replace T in the equation like
T=(P + a^2 / v^2) * (v - b) / R
And put everything in terms of P and V

 

[Chestermiller]

To replace T in the equation like
T=(P + a^2 / v^2) * (v - b) / R
And put everything in terms of P and V

The v's in this equation are the volume per mole, not the actual volume. So, vn = V. But you don't know n.

Also, in your equation, there should be an a, not an a².

 

[sara lopez]

Yes, it was my mistake

 

[sara lopez]

But I can take n=1

 

[Chestermiller]

But I can take n=1

Did they tell you that there is 1 mole?

 

[sara lopez]

No, but I can assume that

 

[Chestermiller]

No, but I can assume that

Really. Who says? What initial temperature do you get if you assume that?

 

[sara lopez]

I don`t have temperature, What else can I do?

 

[Chestermiller]

I don`t have temperature, What else can I do?

If you know the initial pressure, the initial volume, and the number of moles, you can use the van der Waals equation to calculate the initial temperature. What value do you get?

 

[sara lopez]

i don`t have the number of moles, that's why I was thinking of use n=1

 

[Chestermiller]

i don`t have the number of moles, that's why I was thinking of use n=1

What do you get for the starting temperature when you assume n = 1?

 

[sara lopez]

T1 = https://www.physicsforums.com/file://localhost/Users/Carolina/Library/Group%20Containers/UBF8T346G9.Office/msoclip1/01/clip_image001.png T1 = - 175,3 K

 

[sara lopez]

and what're the units for this work?

 

[Chestermiller]

T1 = https://www.physicsforums.com/file://localhost/Users/Carolina/Library/Group%20Containers/UBF8T346G9.Office/msoclip1/01/clip_image001.png T1 = - 175,3 K

Does this answer seem reasonable to you? What does the ideal gas law predict with n = 1?

 

[Chestermiller]

and what're the units for this work?

What are your thoughts on this?

 

[sara lopez]

bar*m^3 and then pass them to BTU

 

[sara lopez]

bar*m^3 and then pass them to BTU

because i have for a =1,358 bar (m^3/kmol)^2 and b= 0,0364 m^3/kmol

 

[Chestermiller]

because i have for a =1,358 bar (m^3/kmol)^2 and b= 0,0364 m^3/kmol

Why would that give a negative temperature?

 

[sara lopez]

Why would that give a negative temperature?

P2 is -48,354 Bar

 

[Chestermiller]

because the volume molar is less than b

0.0364 m^3/kmol = 0.0000364 m^3/mol

 

[sara lopez]

0.0364 m^3/kmol = 0.0000364 m^3/mol

sorry, now the temperature 2 = 15304,4 K, isn't high?