"Adiabatic Compression of an Ideal Monatomic Gas"

Question

A 0.500 mol sample of an ideal monatomic gas is compressed reversibly and adiabatically from an initial volume of 10.0 L to a final volume of 2.50 L. The initial temperature of the gas is 400 K.(a) Calculate the final temperature of the gas.(b) Calculate the work done on the gas during the process.(Assume ideal gas behavior and use

γ=53\gamma = \frac{5}{3}

for a monatomic gas.)

JangoTheOfficial · Answer

solution to adiabatic compression of ideal monatomic gasgiven:n = 0.500 mol (moles of gas)v₁ = 10.0 l (initial volume)v₂ = 2.50 l (final volume)t₁ = 400 k (initial temperature)γ = 5/3 (heat capacity ratio for monatomic gas)r = 8.314 j/(mol·k) (universal gas constant)(a) final temperature calculationfor a reversible adiabatic process, the temperature-volume relationship is:
t₁v₁^(γ-1) = t₂v₂^(γ-1)rearranging to solve for t₂:
t₂ = t₁(v₁/v₂)^(γ-1)substituting values:
γ-1 = 5/3 - 1 = 2/3
t₂ = 400 k × (10.0 l / 2.50 l)^(2/3)
= 400 k × (4)^(2/3)
= 400 k × 2.5198
= 1007.9 kfinal answer for (a): t₂ = 1008 k
(b) work done on the gas calculationfor an adiabatic process, work equals the change in internal energy:
w = δu = nc_vδtfor a monatomic ideal gas:
c_v = (3/2)rthus:
w = n(3/2)r(t₂ - t₁)
= 0.500 mol × (3/2) × 8.314 j/(mol·k) × (1007.9 k - 400 k)
= 0.500 × 1.5 × 8.314 × 607.9
= 3793 jfinal answer for (b): w = 3.79 kj (work done on the gas)key notes:1. all calculations assume ideal gas behavior2. the positive work value indicates work done on the system3. temperature increases during compression as expected for adiabatic process4. final answers rounded to 3 significant figuresverification:
the results are physically reasonable:- adiabatic compression should increase temperature- substantial work input required for 4:1 compression ratio- values consistent with monatomic gas properties

emadtutor · Answer

**(a) final temperature calculation**
for an adiabatic reversible process:
t₂ = t₁ × (v₂ / v₁)^(γ - 1)
where:- t₁ = 400 k (initial temperature)- v₁ = 10.0 l- v₂ = 2.50 l- γ (gamma) = 5/3 = 1.667 (for monatomic gas)
t₂ = 400 × (2.50 / 10.0)^(1.667 - 1)  t₂ = 400 × (0.25)^0.667  t₂ ≈ 400 × 0.396  t₂ ≈ **158.4 k**
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**(b) work done on the gas**
work (w) in adiabatic process:w = (n r (t₂ - t₁)) / (1 - γ)
where:- n = 0.500 mol- r = 8.314 j/mol·k- t₁ = 400 k- t₂ = 158.4 k- γ = 1.667
w = (0.5 × 8.314 × (158.4 - 400)) / (1 - 1.667)  w = (4.157 × -241.6) / -0.667  w = (-1004.1) / -0.667  w ≈ **1505 j**
---
**summary:**- final temperature ≈ 158.4 k  - work done on the gas ≈ 1505 j

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