Question

An ideal gas with γ=1.4 occupies 5.0 L at 300 K and 120 kPa pressure and is heated at constant volume until its pressure has doubled. It's then compressed adiabatically until its volume is one-fourth its original value, then cooled at constant volume to 300 K , and finally allowed to expand isothermally to its original state.

Find the net work done on the gas. W= ___J

Answer 1

Constant volume process:(1-2)

P₁/P₂ = T₁/T₂

120/240 = 300/T₂

T₂= 600 k

Work done during constant volume process is zero.

Adiabatic Compression(2-3):

PV = constant.

P₂V₂ = P₃V₃

P₃= P₂(V₂/V₃)

= 240(5/1.25)^1.4 = 1671.46 kpa

From ideal gas equation, we will find T₃.

P₂/P₃ = T₂/T₃.

240/1671.46= 600/T₃

T₃ = 4177.65 k.

Work done during this process can be given as:

W_2-3 = P₂V₂[(1/-1){(V₂/V₃)^-1 -1}]

By putting values we get:

W_2-3= 2223.30 J

Constant volume process(3-4):

Work done will be zero

Constant temperature process(4-1):

Work done can be given as:

W_4-1 = -P₁V₁ln(V₁/V₄)

= 120kpa×0.005×ln(5L/1.25L)

= -831.78 J

Total work done on gas will be:

W = 2223.30-831.78 =1391.52 J