A non flow system undergoes a reversible process according to law p=4.5/v+2 wher

Question

A non flow system undergoes a reversible process according to law p=4.5/v+2 where pressure p is in bar , specific volume v in m3/kg. During this process volume changes from 0.12 m3/kg. if the specific gas constant is 278j/kg-k, find p1 and T1 (in si units)
p2 and T2, expansion work done during the process and whether the work is done on the system or BY the system, if the specific heats cp=1000j/kgk and cv=772j/kgk what is the change in internal energy and enthalpy of the gas going from process 1 to process 2

Explanation / Answer

p = 4.5/v + 2

Putting v1 = 0.12 m^3/kg, we get p1 = 4.5/0.12 +2 = 39.5 bar

Ideal gas quation: p1*v1 = R*T1 or (39.5*10^5)*0.12 = 278*T1

so, T1 = 1705 K

The final volume v2 is missing in the question. please recofirm the question. thanks.

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