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Proof: Volume ratios in a Carnot cycle

Proof of the volume ratios in a Carnot cycle. Created by Sal Khan.

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• At time, it was glazed over that P is considered to be constant in the quasi-static process. Why is obvious that P needs to be kept constant? What about the system makes it clear that pressure has be considered constant, except to allow the proof to work out?
I am noting that when adiabatic systems were originally discussed in an earlier video, pressure was shown to vary, thus these approximations have been difficult to conjure for the purposes of making thermodynamic proofs. Much thanks.
• Although in an infinitesimal change P is 'almost constant', when Sal substituted
P as a function of V and T he was not keeping P constant. So, when integrating the equation the functionality of P is brought in through P = nRt/V. P is not constant in the total adiabatic process.
Over an infinitesimal interval deltaP approximately equals (nRT/(V+deltaV)-nRT/V) which is much smaller than deltaV
• How can you assume that P is constant?
• As we are taking the process as quasistatic . So for a a very small change we can assume pressure to be constant .
• At , Sal integrated the entire equation. The right hand side of equation is zero. The integration of zero must be a constant as derivative of a constant is zero. But, instead after integrating the equation was written still equal to zero.
Can anyone explain? It will be very helpful. Thanks.
• As Andrew said, it's a constant, but it's the same constant at the beginning and end of the integral, so it subtracts to zero.
(1 vote)
• What is significance of using only adiabatic processes in the relation ignoring isothermal processes occurring in the the carnot's cycle?
• This proof is for a Carnot cycle, which has two adiabatic segments and two isothermal segments. He used the adiabatic processes to do the math, but since V is a state variable, the V ratio of the "four corners" holds true for the CYCLE regardless of how you derive it. It may be possible to derive it using the isothermal segments, but I have not tried. It would have to yield the same ratio, obviously, since it's still the same overall cycle. My guess is that the math is more complicated with the isothermals, or else Sal would have proved it with the isothermals. Not sure. You could try that proof yourself and see if it is easier.
• At , how is (T2/T1)^3/2 the same as (T1/T2)^3/2
• You should probably watch this again — Sal never says those are equal!

It may be helpful to you to write out the manipulations he is showing on paper and think about what is happening at each step ...
• At 2.35 Sal wrote delta U + P delta V = 0 which is true for adiabatic expansion only, not for adiabatic compression. Using this equation he obtained another expression at which should again be true only for adiabatic expansion. Then at how could it be applied for process D to A which is adiabatic compression ?
• Why is it true for expansion but not compression? The only difference between the two is the sign of delta v.
• at why is it (Tf/Ts)^2/3(Vf/Vs)=1????
why is it 1???
• To start with, ln [(Tf/Ts)^3/2(Vf/Vs)]=0. Now, ln (natural log) is logarithm base e (a constant), so log e [(Tf/Ts)^3/2(Vf/Vs)] = 0. From this, and by the property of logarithm, e to the power of 0 (e^0) = [(Tf/Ts)^3/2(Vf/Vs)]. Property of logarithm says whatever number to the 0th power equals 1. So since (e^0) = [(Tf/Ts)^3/2(Vf/Vs)], this whole [(Tf/Ts)^3/2(Vf/Vs)] expression equals 1.
• what is an initial volume and how do you find it on a graph?
• Depending on what your axis are it is usually when x=(0) assuming y is volume