Thermo Inferno Challenge

A gas possessing isentropic expansion factor γ \gamma is made to undergo a thermodynamic cycle consisting of 2 2 isobaric and 2 2 isothermal processes each.If during the cycle,the ratio of maximum to minimum volume V m a x V m i n = α \dfrac{V_{max}}{V_{min}} = \alpha and ratio of maximum to minimum temperature T m a x T m i n = β \dfrac{T_{max}}{T_{min}} = \beta .Then express the efficiency η \eta of the thermodynamic cycle in terms of α \alpha and β \beta .Also assume α > β > 1 \alpha > \beta > 1 to get a positive efficiency.

If η = ( β a ) l n ( α β ) b ( β c ) γ γ 1 + β l n ( α β ) d \eta = \dfrac{(\beta - a) ln (\dfrac{\alpha}{\beta})^b}{\dfrac{(\beta-c)\gamma}{\gamma - 1}+ \beta ln (\dfrac{\alpha}{\beta})^d} find a + b + c + d a + b + c+ d .

Details and Assumptions :

1) The isentropic expansion factor, also known as the adiabatic index of a gas is the ratio of its molar specific heat capacities at constant pressure and constant volume ; γ = C p C v \gamma = \dfrac{C_{p}}{C_{v}} ; where C p C v = R = C_{p} - C_{v} = R = a constant.

2) The gas is assumed to be ideal.The ideal gas law P V = μ R T PV =\mu RT is always applicable.

3) l n ( x ) ln (x) refers to the natural log of x x ; l o g e ( x ) log_{e} (x) .

4) a , b , c , d a,b,c,d are positive integers in their simplest form.They may or may not be distinct.

5) Just as a sidenote,you may want to investigate the cases when α \alpha and β \beta approach 0 0 or \infty .


This problem is Original and made by me.Hope you liked it ! :D


The answer is 4.

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2 solutions

Ayon Ghosh
Jan 9, 2018

Actually one of my friends on AoPS had given this problem to me possessing the special case α = 4 ; β = 2 ; γ = 5 / 3 \alpha = 4 ; \beta =2 ; \gamma = 5/3 .Monoatomic gas.I solved it and decided it to generalise for α , β , γ \alpha,\beta,\gamma .These are my results.

As of now I dont have time to write out the entire solution ( Cuz of exams and INJSO ) ; So I will just give the expression for the efficiency.Once INJSO ends I will Latex whole solution.

η = ( β 1 ) l n ( α β ) ( β 1 ) γ γ 1 + β l n ( α β ) \large{\boxed{\eta = \dfrac{(\beta - 1) ln (\dfrac{\alpha}{\beta})}{\dfrac{(\beta-1)\gamma}{\gamma - 1}+ \beta ln (\dfrac{\alpha}{\beta})}}}

Thus a = b = c = d = 1 \large{\boxed{a = b = c = d = 1}}

Hint : Use a bit of deduction ; Where will the system possess max volume.Where will it possess min volume ? Same for temperature.

I would strongly suggest drawing a PV plot and make correct deductions.

@Ramon Vicente Marquez beautiful exactly my way !!

Ayon Ghosh - 3 years, 5 months ago

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Deosnþ he seem to have handwriting like A doctor?

Md Zuhair - 3 years, 3 months ago

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