Thermodynamics: air expansion in a cylinder

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Discussion Overview

The discussion revolves around the calculation of work done by air during expansion in a cylinder, specifically under the assumption of a reversible process described by the equation P.V1.4 = constant. Participants explore the implications of using the Van der Waals equation versus ideal gas assumptions in this context.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents a homework problem involving air expansion in a cylinder, specifying initial conditions and asking for the work done using the Van der Waals equation.
  • Another participant questions the necessity of the Van der Waals equation if the P-V relationship is already known.
  • Some participants express confusion about how to calculate work using the Van der Waals equation, suggesting that the process might be more suited for ideal gas calculations.
  • There is a discussion about the assumptions of adiabatic and reversible processes, with some participants asserting that these assumptions are necessary for the calculations.
  • One participant suggests that the initial temperature can be derived from the initial pressure and volume using the Van der Waals equation, while others express uncertainty about the number of moles and its implications.
  • Participants explore the relationship between internal energy changes and work done, with references to specific equations and conditions for Van der Waals gases.
  • There are multiple attempts to clarify the role of specific heat capacity (Cv) and its dependence on volume, with some participants questioning whether it is a function of volume for Van der Waals gases.
  • Discussions also touch on the implications of assuming a certain number of moles and the resulting calculations for temperature and work.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best approach to calculate the work done during the expansion, with ongoing debate about the applicability of the Van der Waals equation versus ideal gas assumptions. Uncertainty remains regarding the assumptions made about the number of moles and the initial temperature.

Contextual Notes

Participants note limitations in their assumptions, particularly regarding the number of moles and the initial temperature, which affect the calculations. There is also a lack of clarity about the conditions under which the Van der Waals equation can be applied compared to ideal gas behavior.

  • #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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