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Physics 17 Online
OpenStudy (bhollman44265):

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

OpenStudy (bhollman44265):

@gam3rgirl21

OpenStudy (bhollman44265):

@osprey

OpenStudy (osprey):

I'll have a noss, and get back to you ...

OpenStudy (osprey):

Had a noss and was going to apply pV^gamma=nRT. Can't, though, 'cos not enough info in q as far as I can see. http://perendis.webs.com

OpenStudy (gam3rgirl21):

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

OpenStudy (gam3rgirl21):

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

OpenStudy (gam3rgirl21):

So the answer is 310 L

OpenStudy (gam3rgirl21):

Sorry I wasn't able to help sooner

OpenStudy (osprey):

I must be blind, 'cos I canne see any temperatures ! If you've solved it though, good.

OpenStudy (gam3rgirl21):

Yep, thank you. I was I about to do the same thing :)

OpenStudy (gam3rgirl21):

Hope everyone has a great day, bye :)

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