Arigid container container contain 3kg of air. the initial pressure = 0.5mpa and final pressure = 1.2mpa, the initial temp. =75 degree celcius. heat absorb =195kj. assume cv =0.714kj/kg, find final temperature, change in u from 1-2 and work done from 1-2.

ΔU =195 KJ

W=0 KJ

Explanation:

Given that

Mass of air m=3 kg

heat absorb by air =195 KJ

Cv=0.741 KJ/kgK

If we assume that air is as ideal gas so

Given that container is rigid it means that volume of system is constant so W=0

From first law of thermodynamics

Q=ΔU + W

⇒Q= ΔU                  (W=0)

So ΔU =195 KJ

And work done will be zero because because it is a constant volume process.

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