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In Feynman's treatment of the Carnot cycle, he considers a perfect gas in a cylinder+piston in which we are injecting heat $Q_{1}$ with the help of a heat reservoir at constant temperature $T_{1}$. At the same time, we are expanding the piston by ourselves. He said,

Suppose that we have a gas in a cylinder equipped with a frictionless piston. The gas is not necessarily a perfect gas. The fluid does not even have to be a gas, but to be specific let us say we do have a perfect gas. Also, suppose that we have two heat pads, $T_{1}$ and $T_{2}$—great big things that have definite temperatures, $T_{1}$ and $T_{2}$. We will suppose in this case that $T_{1}$ is higher than $T_{2}$.

Let us first heat the gas and at the same time expand it, while it is in contact with the heat pad at $T_{1}$. As we do this, pulling the piston out very slowly as the heat flows into the gas, we will make sure that the temperature of the gas never gets very far from $T_{1}$. If we pull the piston out too fast, the temperature of the gas will fall too much below $T_{1}$ and then the process will not be quite reversible, but if we pull it slowly enough, the temperature of the gas will never depart much from $T_{1}$.

So if the gas is being kept at a constant temperature $T_{1}$, and at the same time we are expanding the piston ourselves and injecting heat into the gas, then the injected heat $Q_{1}$ must be flowing to the outside, isn't it?

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    $\begingroup$ With all due respect to Feynman, the word "pulling" is really problematic in this quote. We don't have to pull on the piston to get it to do work. All we need to do is let it push on our hand as it moves outward as a result of adding heat to the gas in the cylinder. Pulling was a really bad choice of words (in my humble opinion). $\endgroup$ Jan 2, 2020 at 12:23
  • $\begingroup$ @Chet Miller So it's the heat $Q_{1}$ that's being used as work to move the piston, and we're just observers? $\endgroup$
    – Hilbert
    Jan 2, 2020 at 12:28
  • $\begingroup$ I agree with @Bob D's answer. $\endgroup$ Jan 2, 2020 at 15:20

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So if the gas is being kept at a constant temperature 𝑇1, and at the same time we are expanding the piston ourselves and injecting heat into the gas, then the injected heat 𝑄1 must be flowing to the outside, isn't it?

No heat leaves the system in this expansion process. The injected heat equals the expansion work because, for an ideal gas, any change in internal energy is proportional to the change in temperature of the gas. Since the temperature of the gas does not change, the change in internal energy has to be zero. Then, from the first law for a closed system

$$\Delta U=Q-W$$

$$\Delta U=0$$

$$Q=W$$

The process Feynman is describing is a reversible isothermal expansion. To carry it out the external pressure has to be slowly reduced so that the gas pressure is always in equilibrium with the external pressure and the temperature of the gas remains constant. As @Chet Miller pointed out, that doesn't require the piston to be "pulled" as Feynman says, but that the external pressure be allowed to slowly decrease.

Hope this helps.

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