Question

A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oC. The air undergoes a process to a state where the pressure is 2.5 bar, during which the pressure–volume relationship is pV = constant. Assume ideal gas behavior for the air.​​Determine the work and heat transfer, in kJ.

Answer 1

Answer:

Work transfer is - 97.02 KJ. It means that work is given to the system.

Heat transfer = - 97.02 KJ . It means that heat is rejected from the system.

Explanation:

Given that

m= 3 kg

P₁=2 bar

T=T₁=T₂=30 °C

T=303 K

P₂=2.5 bar

PV=  Constant

This is the isothermal process .

We know that work for isothermal process  given as

$W=P_1V_1\ln \dfrac{P_1}{P_2}$

$W=mRT\ln \dfrac{P_1}{P_2}$

For air

R= 0.287 KJ/Kg.K

Now by putting the values

$W=mRT\ln \dfrac{P_1}{P_2}$

$W=5\times 0.287\times 303\ln \dfrac{2}{2.5}$

W= - 97.02 KJ

So the work transfer is - 97.02 KJ. It means that work is given to the system.

We know that for ideal gas internal energy is the only function of temperature.The change in internal energy ΔU

ΔU = m Cv ΔT

Here ΔT= 0

So

ΔU =0

From first law of thermodynamics

Q= ΔU +W

ΔU = 0

Q= W

Q= - 97.02 KJ

Heat transfer = - 97.02 KJ . It means that heat is rejected from the system.