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

Answer:

$T_2=562.2 C$

ΔU =195 KJ

W=0 KJ

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

$\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}$

$\dfrac{0.5}{273+75}=\dfrac{1.2}{T_2}$

$T_2=562.2 C$

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.