Endothermic To Exothermic

Classical Mechanics Level 5

Source : Resonance AIOT Source : Resonance AIOT

An ideal diatomic gas undergoes process A B AB for which P V PV indicator diagram is shown. At some time during this process, it changes from being endothermic to being exothermic. At that time, the volume of the gas is 5 λ V 0 72 \dfrac{5\lambda V_{0}}{72} . Find λ \lambda .


The answer is 21.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

Equation of the straight line AB is :

P = 5 P 0 − 2 P 0 V V 0 P=5P_{0} - \dfrac{2P_{0}V}{V_{0}}

Q Q (Heat) as a function of volume is :

Q = 35 P 0 V 2 − 6 P 0 V 2 V 0 − 23 P 0 V 0 2 Q=\dfrac{35P_{0}V}{2}-\dfrac{6P_{0} V^2}{V_{0}}-\dfrac{23P_{0}V_{0}}{2}

Above equation is a downward parabola with apex occuring at V ′ = 35 V 0 24 V'=\dfrac{35V_{0}}{24} . After V ′ V' , Q Q is decreasing (Heat must have started evolving after the apex leading to a fall in curve)

Hence V ′ V' must be the required volume .

V ′ = 35 V 0 24 \boxed{V'=\dfrac{35V_{0}}{24}}

2 Helpful 0 Interesting 1 Brilliant 0 Confused

We look for a point on A B AB that is tangent to the adiabatic curve P V γ = k PV^\gamma = k , or P = k V − γ P = kV^{-\gamma} , with γ = 7 / 5 \gamma = 7/5 . The derivative is d P / d V = − γ k V − γ − 1 = − γ P / V dP/dV = -\gamma kV^{-\gamma-1} = -\gamma P/V . This derivative must match the slope of A B AB , which is − 2 P 0 / V 0 -2P_0/V_0 .

Thus d P d V = − 7 5 P V = − 2 P 0 V 0 , \frac{dP}{dV} = -\frac 7 5 \frac{P}{V} = -2\frac{P_0}{V_0}, so that 7 P P 0 − 10 V V 0 = 0 ; 7\frac{P}{P_0} - 10\frac{V}{V_0} = 0; but we also have for line A B AB P P 0 + 2 V V 0 = 5. \frac{P}{P_0} + 2\frac{V}{V_0} = 5. Multiply the last equation by 7 and subtract the first to find 24 V V 0 = 35 , 24\frac{V}{V_0} = 35, so that V = 35 V 0 24 = 5 ⋅ 7 ⋅ 3 ⋅ V 0 24 ⋅ 3 . V = \frac{35V_0}{24} = \frac{5\cdot \boxed{7\cdot 3}\cdot V_0}{24\cdot 3}. Thus the answer is 21 \boxed{21} .

Arjen Vreugdenhil - 3 years, 3 months ago

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Thanks , I commited a silly mistake while solving :)

A Former Brilliant Member - 3 years, 3 months ago

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