Great video, thanks! So does this sound right? From A to B, the system (gas in the cylinder) does 535 J of work on the surroundings, since W is positive (using the handbook's sign convention for work as positive when done by the system). From C to A the surroundings does 371 J of work on the system. Net work is 535 - 371 = 164 J out of the system, and overall Q = 164 J into the system since dU = Q - W = 0 for the cyclic process (no kinetic or potential effects)
Well done! That's spot on and honestly, I should've include a question asking for the net work. I'm glad you went decided to take it a step further and find a value for that. Very insightful.
We would need to convert L to cubic meters then atm to pascals separately then we would get joules. We are still using page.3 for convert these. 1 L = 0.001 cubic meters (m^3) 1 atm = 101328 pascals = 101326 N/m^2 Multiply the two: 1L - atm = 0.001 m^3 - 101326 N/m^2 = 101.328 N - m = 101.328 Joules
For others, the reason you can add the "n" variable for moles into the equation is you can derive that from the gas constant equation and the molecular weight equation in the FE handbook. It's written as Rbar (bar on top of R symbol) in the handbook. That confused me at first, so hope it helps others!
This question is for you and Farouq.. R = R (bar)/ M, M = m/n, where is m in the problem and the answer? we should use it as well. so the correct equation for W = n/m*R*T* Ln(v2/v1)
@@samMarvel It looks isentropic, but you can see that T_A and T_C are the same so that means it's isothermal. Being isentropic doesn't normally have the same temperature for the two states (look at the Rankine cycle Ts diagram). Additionally like @logantaylor8565 said, there's on mention of entropies so that is another hint to avoid the eqs for isentropic work.
I initially got the incorrect answer because I used the formula in the FE manual for boundary work with constant temperature, which reads W = R*T*ln(P1/P2), but this gives a different answer than what you did? Its stated in the section under Closed Thermodynamic Systems. Am I misunderstanding something or is the manual wrong?
Great video, thanks! So does this sound right? From A to B, the system (gas in the cylinder) does 535 J of work on the surroundings, since W is positive (using the handbook's sign convention for work as positive when done by the system). From C to A the surroundings does 371 J of work on the system. Net work is 535 - 371 = 164 J out of the system, and overall Q = 164 J into the system since dU = Q - W = 0 for the cyclic process (no kinetic or potential effects)
Well done! That's spot on and honestly, I should've include a question asking for the net work. I'm glad you went decided to take it a step further and find a value for that. Very insightful.
Where in the handbook is the L.atm conversion to Joules? I don't see it on pg 3, only for the universal gas constants
We would need to convert L to cubic meters then atm to pascals separately then we would get joules.
We are still using page.3 for convert these.
1 L = 0.001 cubic meters (m^3)
1 atm = 101328 pascals = 101326 N/m^2
Multiply the two:
1L - atm = 0.001 m^3 - 101326 N/m^2 = 101.328 N - m = 101.328 Joules
For others, the reason you can add the "n" variable for moles into the equation is you can derive that from the gas constant equation and the molecular weight equation in the FE handbook. It's written as Rbar (bar on top of R symbol) in the handbook. That confused me at first, so hope it helps others!
This question is for you and Farouq..
R = R (bar)/ M, M = m/n, where is m in the problem and the answer? we should use it as well. so the correct equation for W = n/m*R*T* Ln(v2/v1)
@directhubfeexam
How did you know that the process C to A was isothermal and not isentropic?
I'm wondering this as well. I just assumed the entropy stayed the same because they didn't mention anything about it.
@@logantaylor8565 yeah, even from the thumbnail, the process looks isentropic.
@@samMarvel It looks isentropic, but you can see that T_A and T_C are the same so that means it's isothermal. Being isentropic doesn't normally have the same temperature for the two states (look at the Rankine cycle Ts diagram). Additionally like @logantaylor8565 said, there's on mention of entropies so that is another hint to avoid the eqs for isentropic work.
I initially got the incorrect answer because I used the formula in the FE manual for boundary work with constant temperature, which reads W = R*T*ln(P1/P2), but this gives a different answer than what you did? Its stated in the section under Closed Thermodynamic Systems. Am I misunderstanding something or is the manual wrong?
Pressure is not constant on process No. 3