Yes, sorry, I got myself confused between Total energy and Internal Energy. However the rest should still stand and I'm glad you came to the same conclusions :)
Okie-Dokie.
So I have a result (I think!), let's see what you think. To me it makes sense.
Using the 1st law dU=dQ-dW, and the change in energy dU of the surroundings will be equal to that of the system (just in the opposite direction). Compute dW using ∫pdv and p=CV-n and rearranging to...
My value for n (polytropic exponent) came out as 1.538 (4s.f.)
If you could show the process were externally reversible, then the entropy generation would have to be 0 and so entropy change of the surroundings would be equal to positive of the entropy change of the system.
The issue is...
I did calculate n, I used the ideal gas law to find the volume for the second state and then used logs, but I don't see how that helps find the entropy of the surroundings? I will redo the numbers using your method to check I have the right n.
Thanks for the reply!
I'm using google and...
Homework Statement
We have a piston in a cylinder containing Helium. The initial states are:
P1=150kPa
T1= 20°C
V1=0.5m3
Following a polytropic process, the final states are:
P2=400kPa
T2=140°C
V2 is unknown.
We're also given R=2.0769 kPa.m3.(kg.K)-1
And Cp=5.1926kJ.(kg.K)-1
As...