Thermodynamics: air expansion in a cylinder

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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.
  • #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
 
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  • #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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