sara lopez
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because i have for a =1,358 bar (m^3/kmol)^2 and b= 0,0364 m^3/kmolsara lopez said:bar*m^3 and then pass them to BTU
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.
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because i have for a =1,358 bar (m^3/kmol)^2 and b= 0,0364 m^3/kmolsara lopez said:bar*m^3 and then pass them to BTU
Why would that give a negative temperature?sara lopez said:because i have for a =1,358 bar (m^3/kmol)^2 and b= 0,0364 m^3/kmol
P2 is -48,354 BarChestermiller said:Why would that give a negative temperature?
0.0364 m^3/kmol = 0.0000364 m^3/molsara lopez said:because the volume molar is less than b
sorry, now the temperature 2 = 15304,4 K, isn't high?Chestermiller said:0.0364 m^3/kmol = 0.0000364 m^3/mol
1 mole is obviously too few. Try 100 moles.sara lopez said:sorry, now the temperature 2 = 15304,4 K, isn't high?
n= 100, T2= 15771,3Chestermiller said:1 mole is obviously too few. Try 100 moles.
How is that possible?sara lopez said:n= 100, T2= 15771,3
yes, you are right, thanks for the correctionChestermiller 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)
$$T_1=\frac{(28.3-3.64)(102+(13440)/(28.3^2))}{(100)(0.0821)}=358 K$$