In one of the adiabatic stroke in the Carnot cycle, the temperature of the gas reduces from T1(to what it was initially heated) to T2(temperature of the surrounding).

How exactly does the internal energy flow? Because the process is claimed to be adiabatic, which I think, means that there should be no flow of heat.

Also there is a change in the volume (and pressure of course)

Is the internal energy being used up to push the piston, hence increasing the volume?

What is happening?

• Yes, your interpretation is correct. Nov 26, 2019 at 16:57
• @ChetMiller Your comment showed me I failed to make that statement directly. I updated my answer accordingly. Nov 26, 2019 at 21:10

In one of the adiabatic stroke in the Carnot cycle, the temperature of the gas reduces from T1(to what it was initially heated) to T2(temperature of the surrounding).

To be clear, you are describing the reversible adiabatic (isentropic) expansion process that follows the reversible isothermal (constant temperature) expansion process. For the isothermal expansion heat is transferred to the system at temperature $$T_1$$ to perform work. The adiabatic process that follows continues the expansion without heat transfer lowering the temperature to $$T_2$$ which is the temperature of the low temperature reservoir.

How exactly does the internal energy flow? Because the process is claimed to be adiabatic, which I think, means that there should be no flow of heat.

The internal energy does not "flow". It potentially changes in magnitude. There are two types of energy transfers, or "flows" if you wish to call them that, and they are heat and work. The internal energy of the system potentially changes when energy transfers into and/or out of the system. The change in internal energy, $$\Delta U$$ as a function of heat, $$Q$$, and work, $$W$$ is given by the first law:

$$\Delta U=Q-W$$

Where $$Q$$ is positive if heat transfer is into the system and $$W$$ is positive if the system does work on the surroundings.

Also there is a change in the volume (and pressure of course)

Is the internal energy being used up to push the piston, hence increasing the volume?

What is happening?

What is happening in the adiabatic expansion is the system is doing work to expand against external pressure. Since for an adiabatic process $$Q$$ = 0, it follows from the first law that

$$\Delta U=-W$$

which tells us that the expansion work done by the adiabatic process ($$W$$ is positive) comes at the expense of its internal energy (internal energy decreases). So this means, as @Chet Miller stated, you are correct internal energy of the system is being used to push the piston.

If the system consists of an ideal gas, you can determine the effect of using internal energy on the temperature of the system because the the change in internal energy of an ideal gas depends only on change in temperature. So we can write for an ideal gas

$$\Delta U= -W=C_{v}\Delta T=C_{v}(T_{2}-T_{1})$$

Where $$C_v$$ is the specific heat at constant volume for the gas. (Note, for an ideal gas, this specific heat applies even though it is not a constant volume process. But that's a separate proof).

Hope this helps.

• Again, assuming that the internal energy of all the gas molecules of the system are the same (can we?) Why would the internal energy flow? Since the system pushes the piston only using it's kinetic energy, i.e, pressure being applied. What I mean is, is the internal energy used in pushing the Piston? Dec 15, 2019 at 8:59

Yes you are right, the internal energy of the gas is consumed by work against the piston.

$$Q=\Delta U+W=0$$( first law of thermodynamics )

Therefore the work which is done by the gas is at the cost of its internal energy.

Lets first look at the term adiabatic. * An adiabatic process occurs without transfer of heat or mass of substances between a thermodynamic system and its surroundings. In an adiabatic process, energy is transferred to the surroundings only as work.*(Source-Wikipedia) As the definition claims **there is no exchange of heat between the system amd the surrounding.**This leaves us with the internal energy and the work done by the gas to deal with.

In your case there is a temperature drop.That means the system has lost some of its internal energy.This energy is used to do work by the gas,hence the change in volume.

My guess is that you have misunderstood the definition of an adiabatic stroke.It doesn't mean that there is no energy flow.