# Relationship between volume and temperature of an ideal gas

I am self-studying thermodynamics, and was reading up Carnot heat engine (Yunus Cengel - thermodynamics book). So, the experiment in the textbook is set up such that, we have an adiabatic piston-cylinder device and the insulation at the cylinder head can be removed to transfer energy from thermal energy reservoir(s) to supply/absorb heat depending upon the where you are in the cycle. It begins with reversible isothermal process by connecting the cylinder to a heat source at temperature T_H. The gas expands, which is also at temperature T_h, and does work on the piston and temperature of the gas drops. This where I have some doubts.

Now, why does the gas expand if both the hot reservoir and system is in thermal equilibrium (same temp. T_h)?. Secondly, from the ideal gas equation Pv = RT, isn't volume and temperature proportional to each other, i.e., as the volume of the gas increases, shouldn't the temperature increase as well?

Thank you for taking the time to read my question.

• you have to reduce the force pushing against the piston that was keeping it at its original volume; i.e. you are decreasing the pressure to allow the volume to expand at constant temperature. Commented Oct 1, 2023 at 14:23

Now, why does the gas expand if both the hot reservoir and system is in thermal equilibrium (same temp. T_h)?

Because during the entire expansion process the external pressure is infinitesimally reduced allowing the gas to expand and do work,

Secondly, from the ideal gas equation Pv = RT, isn't volume and temperature proportional to each other, i.e., as the volume of the gas increases, shouldn't the temperature increase as well?

As the volume increases the pressure decreases so that the temperature remains constant. This is done by slowly reducing the external pressure and keeping it in equilibrium with the gas pressure so that $$PV$$=constant.

The following sequence may be helpful:

1. External pressure is reduced infinitesimally
2. Infinitesimal reduction in external pressure results in gas undergoing infinitesimal expansion (thus doing infinitesimal amount of work)
3. Infinitesimal expansion results in an infinitesimal decrease in gas temperature.
4. Infinitesimal decrease in gas temperature results in infinitesimal temperature difference between gas and surroundings.
5. Infinitesimal difference in temperature results in infinitesimal addition of heat, restoring the temperature of the gas to that of its surroundings.
6. Repeat steps 1-5 until final equilibrium state is reached.

Net result is temperature is constant, total work done by gas equals total heat added to gas and since, for an ideal gas, internal energy depends only on temperature, the change in internal energy is zero per the first law.

See my answer in the following link which includes a diagram with the description of the above steps:Work done by a gas in an isothermal process

Hope this helps.

• @PrajwalKori So just to clarify what? Commented Oct 5, 2023 at 10:30
• just to clarify, the heat transferred from the hot reservoir is used to increase either (or both) the internal energy and do work. So, if we break this process into infinitesimal parts, gas pressure or the force exerted by the gas at the boundary of the system is infinitesimally higher than the external pressure. Is that why the system, in order to maintain equilibrium (and reversibility) reduces the system pressure infinitesimaly while increasing volume, yet keeping the temperature constant with the help of a heat reservoir at the same temperature? Commented Oct 7, 2023 at 19:26
• @PrajwalKori The way I look at it the infinitesimal heat transferred from the hot reservoir restores the infinitesimal decrease in the internal energy that was used in performing the work due to the infinitesimal decrease in external pressure. Ultimately, the energy used to perform the work comes from the energy transferred from the hot reservoir to the system due to an infinitesimal difference in temperature. That mechanism for the transfer of energy from the hot reservoir is what we call "heat". Commented Oct 7, 2023 at 22:09
• Ah, I understand. Thank you for clarifying my doubt! Commented Oct 8, 2023 at 18:46

The process you are looking at is a reversible process so it can in principle go either way. That is, the gas can expand while heat flows into it, or it can contract while heat flows out of it. The direction of the process is fixed in practice by a very small temperature difference. If the gas is slightly colder than the reservoir then heat flows in. If nothing else happens then the gas will warm up until the heat flow stops. But if you allow the gas to expand, by pressing on a piston with a force just a tiny bit smaller than the force exerted by the gas by its pressure, then the piston will move and the gas expands. The "trick" is to do all this while maintaining the applied force just low enough to allow the gas pressure to fall in such a way that the gas temperature stays constant.

This sounds quite difficult to arrange! It is indeed tricky to construct a nearly-ideal Carnot engine, but it can be done. It's not so hard to arrange a mechanical linkage that makes the force applied to the gas relate to its volume in the right way.

In other parts of the cycle, you have two parts with no heat flow and another part which is an isothermal compression. The isothermal compression part is like the above, but now you apply a force just greater than that exerted by the gas, bringing its temperature to just a tiny bit above that of the reservoir, and then heat flows from gas to reservoir.