Thermodynamics: air expansion in a cylinder

AI Thread Summary
The discussion focuses on calculating the work done by air expanding in a cylinder from an initial volume of 1 ft³ to a final volume of 8 ft³ under a reversible process defined by P.V^1.4=constant. Participants debate the necessity of using the Van der Waals equation versus the ideal gas law for this calculation, considering the initial conditions of pressure and volume. The conversation highlights the challenges of determining initial temperature and the number of moles, with some suggesting assumptions to simplify the problem. Ultimately, the calculations for temperature using both the ideal gas law and the Van der Waals equation are presented, leading to discussions about the feasibility of the results. The thread concludes with a consensus on the importance of accurately applying thermodynamic principles to derive meaningful results.
sara lopez
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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 )
 
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If you know the P-V relationship for the process, why do you need the van der Waals equation to calculate the work?
 
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?
 
sara lopez said:
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?
 
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
 
sara lopez said:
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?
 
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
 
sara lopez said:
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?
 
U = nCv (T2- T1) + a (1/ v1 - 1/v2)
 
  • #10
U= nCv (T2-T1) + a (1/v1 - 1/v2)
 
  • #11
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?
 
  • #12
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.
 
  • #13
So how do I know if Cv is a function of V?
 
  • #14
sara lopez said:
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
 
  • #15
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
 
  • #16
sara lopez said:
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.
 
  • #17
Yes, it was my mistake
 
  • #18
But I can take n=1
 
  • #19
sara lopez said:
But I can take n=1
Did they tell you that there is 1 mole?
 
  • #20
No, but I can assume that
 
  • #21
sara lopez said:
No, but I can assume that
Really. Who says? What initial temperature do you get if you assume that?
 
  • #22
I don`t have temperature, What else can I do?
 
  • #23
sara lopez said:
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?
 
  • #24
i don`t have the number of moles, that's why I was thinking of use n=1
 
  • #25
sara lopez said:
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?
 
  • #26
T1 = https://www.physicsforums.com/file://localhost/Users/Carolina/Library/Group%20Containers/UBF8T346G9.Office/msoclip1/01/clip_image001.png T1 = - 175,3 K
 
Last edited by a moderator:
  • #27
and what're the units for this work?
 
  • #28
sara lopez said:
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?
 
Last edited by a moderator:
  • #29
sara lopez said:
and what're the units for this work?
What are your thoughts on this?
 
  • #30
bar*m^3 and then pass them to BTU
 
  • #31
sara lopez said:
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
 
  • #32
sara lopez said:
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?
 
  • #33
Chestermiller said:
Why would that give a negative temperature?
P2 is -48,354 Bar
 
  • #34
sara lopez said:
because the volume molar is less than b
0.0364 m^3/kmol = 0.0000364 m^3/mol
 
  • #35
Chestermiller said:
0.0364 m^3/kmol = 0.0000364 m^3/mol
sorry, now the temperature 2 = 15304,4 K, isn't high?
 
Last edited:
  • #36
sara lopez said:
sorry, now the temperature 2 = 15304,4 K, isn't high?
1 mole is obviously too few. Try 100 moles.
 
Last edited:
  • #37
Chestermiller said:
1 mole is obviously too few. Try 100 moles.
n= 100, T2= 15771,3
 
  • #38
Thanks for the help :D
 
  • #39
sara lopez said:
n= 100, T2= 15771,3
How is that possible?
 
  • #40
1 ft^3 = 28.3 liters
1500 psi = 102.0 atm.
n = 100 moles
IDEAL GAS LAW
$$T_1=\frac{PV}{nR}=\frac{(28.3)(102.0)}{(100)(0.0821)}=352 K$$

VAN DER WAAL EQUATION

a=1.344 atm (liters/mole)^2

b=0.0364 (liters/mole)

$$T_1=\frac{(28.3-3.64)(102+(13440)/(28.3^2))}{(100)(0.0821)}=358 K$$
 
  • #41
Chestermiller said:
1 ft^3 = 28.3 liters
1500 psi = 102.0 atm.
n = 100 moles
IDEAL GAS LAW
$$T_1=\frac{PV}{nR}=\frac{(28.3)(102.0)}{(100)(0.0821)}=352 K$$

VAN DER WAAL EQUATION

a=1.344 atm (liters/mole)^2

b=0.0364 (liters/mole)

$$T_1=\frac{(28.3-3.64)(102+(13440)/(28.3^2))}{(100)(0.0821)}=358 K$$
yes, you are right, thanks for the correction
 

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