Question

A rigid 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.

Answer 1

`T_2=562.2 C`

ΔU =195 KJ

W=0 KJ

Step-by-step explanation:

Given that

`P_1=0.5 MPa`

`P_2=1.2 MPa`

`T_1=75 C`

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

`(P_1)/(T_1)=(P_2)/(T_2)`

`(0.5)/(273+75)=(1.2)/(T_2)`

`T_2=562.2 C`

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.