Question

An ideal gas with γ=1.4 occupies 3.0 L at 300 K and 120 kPa pressure and is compressed adiabatically until its volume is 2.0 L. It's then cooled at constant pressure until it reaches 300 K, then allowed to expand isothermally back to state A.

Part A

Find the net work done on the gas.

Express your answer using two significant figures.

W =

Part B

Find the minimum volume reached.

Express your answer using two significant figures.

Vmin =

Answer 1

An ideal gas undergoing an adiabatic, reversible process satifies the condtion:
p·V^γ = constant

From this you can calculate the pressure in after the adiabatic compression step:
p₁·V₁^γ = p₂·V₂^γ
<=>
p₂ = p₁·(V₁/V₂)^γ
= 120 kPa · (3.0L/2.0L)^1.4
= 211.69kPa

From ideal gas law you find T₂
p·V = n·R·T
=>
p·V/T = n·R = constant
=>
p₁·V₁/T₁ = p₂·V₂/T₂
<=>
T₂ = T₁ · (p₂/p₁) · (V₂/V₁)
= 300K · (211.69/120) · (3/2)
= 793.84K

The work done on gas from state 1 to 2 is
W₁₂ = - ∫ V₁→V₂ [ p ] dV
use adiabtic relation above to express p in terms of V
p·V^γ = p₁·V₁^γ => p = p₁·V₁^γ·V^(-γ)
=>
W₁₂ = - p₁·V₁^γ · ∫ V₁→V₂ [ V^(-γ) ] dV
= - p₁·V₁^γ · (1/(1-γ) · (V₂^(1-γ) - V₁^(1-γ) )
= - p₁·V₁ · (1/(1-γ) · ( (V₂/V₁)^(1-γ) - 1 )
= p₁·V₁ · (1/(γ-1) · ( (V₁/V₂)^(γ-1) - 1 )
= 120kPa · 0.003m³ · (1/(1.4-1) · ( (3L/2L)^(1.4-1) - 1 )
= 705.6J

Next step (2→3) is isobaric:
p = constant = p₂ = p₃
Hence the work done in this process is
W₂₃ = - ∫ V₂→V₃ [ p ] dV
= - p₂ · ∫ V₂→V₃ dV = - p₂ · (V₃ - V₂)

Use Ideal gas law to compute V₃
p₂·V₂/T₂ = p₃·V₃/T₃
=>
V₃ = V₂ · (T₃/T₂) = 2L · (300K / 793.84K) = 0.755 L

Hence:
W₂₃ = p₂ · (V₂ - V₃)
= 211.69kPa · (0.002m³ - 0.000755m³) = 263.5J

In the last step you go isothermally back to intiial state.
(p₁=120kPa; T₁=300K, V₁=3L)
For isothermal change of state of an ideal gas:
p·V = n·R·T = constant
=>
p·V = p₃·V₃ = p₁·V₁
<=>
p = p₁·V₁/V (or p = p₃·V₃/V)
For any state on the path from 3 to 1.

So the work done on the gas in this step is
W₃₁ = - ∫ V₃→V₁ [ p ] dV
= - p₁·V₁ · ∫ V₃→V₁ [ 1/V ] dV
= - p₁·V₁ · ln(V₁/V₃)
= - 120kPa · 0.003m³ · ln(2L/0.755L)
= -350.70J

Overall work done on the gas is:
W_tot = W₁₂ + W₂₃ + W₃₁
= 705.6J + 263.5J - 350.70J
= 618.4J

The gas has minimum volume after the cooling step. It is
V₃ = 0.755 L