Thermodynamics: air expansion in a cylinder

[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?

 

[Chestermiller]

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

1 mole is obviously too few. Try 100 moles.

 

[sara lopez]

1 mole is obviously too few. Try 100 moles.

n= 100, T2= 15771,3

 

[sara lopez]

Thanks for the help :D

 

[Chestermiller]

n= 100, T2= 15771,3

How is that possible?

 

[Chestermiller]

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)

$$T_1=\frac{(28.3-3.64)(102+(13440)/(28.3^2))}{(100)(0.0821)}=358 K$$

 

[sara lopez]

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