Answered

A sample of 20.0 moles of a monatomic ideal gas (γ = 1.67) undergoes an adiabatic process. The initial pressure is 400kPa and the initial temperature is 450 K The final temperature of the gas is 320 K. What is the final volume of the gas? Let the ideal-gas constant R = 8.314 J/(mol • K).


310 L

230 L

190 L

270 L

350 L

Verified Answer :

Sanarah

Pressure and volume of an ideal gas undergoing a revisable adiabatic change of state are related as[1]: 
P∙V^γ = C ( a constant ) 

When you substitute pressure using ideal gas law you can derive a relation for temperature and volume for such process: 
P = n∙R∙T/V 
=> 
(n∙R∙T/V)∙V^γ = C 
<=> 
T∙V^(γ - 1) = C/(n∙R∙) (also a constant) 

So volume and temperature in initial (1) and final (2) state are related as: 
T₁∙V₁^(γ - 1) = T₂∙V₂^(γ - 1) 
<=> 
T₁^(1/(γ - 1))∙V₁ = T₂^(1/(^(γ - 1))∙V₂ 

So final volume is given by: 
V₂ = V₁ ∙ (T₁/T₂∙)^(1/(γ - 1)) 

The temperatures are given. The initial volume can be found from ideal gas law: 
V₁ = n∙R∙T₁∙/P₁ 
= 20mol ∙ 8.314 J∙mol⁻¹∙K⁻¹ ∙ 450K / 400×10³ Pa 
= 0.187 m³ 
= 187 L 

Hence, 
V₂ = 187L ∙ (450K / 320∙)^(1/(1.67 - 1)) = 311 L ≈ 310 L

Hope this helps :)

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