Thermodynamics: air expansion in a cylinder

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SUMMARY

The discussion focuses on calculating the total work done by air during expansion in a cylinder from an initial volume of 1 ft³ and pressure of 1500 PSI ABS to a final volume of 8 ft³, using both the ideal gas law and the Van der Waals equation. The participants clarify the need to assume adiabatic and reversible conditions for accurate calculations. Key equations include the Van der Waals equation, \( (P + \frac{a^2}{v^2})(v - b) = RT \), and the work done per mole, \( w = -\Delta u = -\left[c_v(T_2 - T_1) + a\left(\frac{1}{v_1} - \frac{1}{v_2}\right)\right] \). The final temperature calculations yield values of 352 K and 358 K for the ideal gas and Van der Waals gas, respectively.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically adiabatic and reversible processes.
  • Familiarity with the Van der Waals equation for real gases.
  • Knowledge of the ideal gas law and its applications.
  • Ability to perform calculations involving specific heat capacities, \( c_v \), and work done in thermodynamic processes.
NEXT STEPS
  • Study the derivation and applications of the Van der Waals equation in real gas scenarios.
  • Learn about adiabatic processes and their implications in thermodynamics.
  • Explore the concept of work done in thermodynamic systems, particularly for non-ideal gases.
  • Investigate the relationship between pressure, volume, and temperature in gas laws through practical examples.
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Students and professionals in mechanical engineering, chemical engineering, and physics, particularly those focusing on thermodynamics and gas behavior in various conditions.

  • #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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