U = n*cv*T In terms of a constant called the adiabatic index (which depends on gas flavor), cv can be expressed as: cv = R/(k - 1) Thus U = n*R*T/(k - 1) For diatomic gasses like air's components, k = 7/5. Thus: U = 5/2*n*R*T In the monatomic gas case, k=5/3. Thus, U = 3/2*n*R*T Thus, for the noble gasses, U = 3/2*n*R*T
Yea, exactly. I think the end result is wrong. The amount of work done on the system should be a positive value. All of the math is correct, so the only thing to do is add a negative sign into the original formula.
morninglight25 let me explain it better. If the result were positive (+324) the system should be increasing volume, but it is the opposite. So, because we have -324 J mans the pump is doind negative work, so the system is doing positive, and that matches, becuase, on the graph, rhe volume is decreasing
morninglight25 let me explain it better. If the result were positive (+324) the system should be increasing volume, but it is the opposite. So, because we have -324 J mans the pump is doind negative work, so the system is doing positive, and that matches, becuase, on the graph, rhe volume is decreasing
Thank you finally I understand something in thermo.
Great problem. Thorough explication! Thanks!
why is there a 5/2 in front of the gas equation nRdeltaT?
U = n*cv*T
In terms of a constant called the adiabatic index (which depends on gas flavor), cv can be expressed as:
cv = R/(k - 1)
Thus
U = n*R*T/(k - 1)
For diatomic gasses like air's components, k = 7/5. Thus: U = 5/2*n*R*T
In the monatomic gas case, k=5/3. Thus,
U = 3/2*n*R*T
Thus, for the noble gasses,
U = 3/2*n*R*T
It's great! Thank you very much!
Sir please make a video how to find internal energy of diatomic gases and triatomic
whats the best way to master thermodynamics.am a chemical engineering student
if there is a way please do share with me.
this should be isotropic process
You said that -W was the work done by the pump ON the system. I thought that the negative sign on -W means the work done BY the system?
Yea, exactly. I think the end result is wrong. The amount of work done on the system should be a positive value. All of the math is correct, so the only thing to do is add a negative sign into the original formula.
morninglight25 it is the same thing. the work done by the system on thr pump is negative, thus, making thr pump doing work on the system
morninglight25 let me explain it better. If the result were positive (+324) the system should be increasing volume, but it is the opposite. So, because we have -324 J mans the pump is doind negative work, so the system is doing positive, and that matches, becuase, on the graph, rhe volume is decreasing
morninglight25 let me explain it better. If the result were positive (+324) the system should be increasing volume, but it is the opposite. So, because we have -324 J mans the pump is doind negative work, so the system is doing positive, and that matches, becuase, on the graph, rhe volume is decreasing
This makes no sense at all