# In Isothermal expansion of Carnot Cycle, Energy of the system should be constant, Since Temperature is constant. How does the system absorb energy?

In Carnot cycle's Isothermal phase, Heat is said to be absorbed. Energy of the Carnot Engine in Isothermal phase as per Ideal Gas equation should be: $$E=\frac32nRT$$ Since the amount of Gas($$n$$) and the temperature ($$T$$) does not change, The energy of the system should be constant. Keeping that in mind, How does the Carnot engine absorb energy in the isothermal phase without any change in the gas temperature?

• For an isothermal system, the $\textit{internal}$ energy is meant to be constant. If heat is supplied to a system, it can still do work or work can still be done on the system. Aug 25, 2021 at 4:51

Keeping that in mind, How does the Carnot engine absorb energy in the isothermal phase without any change in the gas temperature?

It does so by performing an amount of expansion work exactly equal to the heat added, so that the internal energy remains constant throughout the process. Per the first law,

$$\Delta U=Q-W$$

Then, since $$Q=W$$, $$\Delta U=0$$.

Since the internal energy of an ideal gas depends only on temperature, the temperature remains constant.

In a reversible isothermal process the temperature difference between the system and surroundings is kept infinitesimal so that in the limit they are considered the same, i.e., the system can be considered to always be in thermal equilibrium with the surroundings while the system does work.

To see how this is possible, imagine a reversible isothermal expansion where the hot reservoir and the system are initially at the same temperature, $$T_H$$. The external pressure is then infinitesimally reduced by an amount $$dP$$ (imaging a bag of sand on top of the piston and we remove one grain). The gas undergoes an infinitesimal expansion of $$dV$$ (thereby doing work $$dW$$) which lowers the temperature of the gas by an infinitesimal amount $$dT$$ below the hot reservoir. The hot reservoir now transfers an infinitesimal amount of heat $$dQ$$ which raises the temperature of the system by an amount $$dT$$ back to that of the hot reservoir. Repeat this process over and over again and the system temperature will remain the same as the surroundings while the gas slowly expands doing work.

Hope this helps.

• thanks for the explanation. It is possible to conceive infinitesimal increase of energy, But at the end of the isothermal cycle there is a significant energy gained. Since the energy is not stored in the ideal gas (internal energy being constant), In what form is the energy stored ? Aug 25, 2021 at 21:16
• @physicsEnquiry All the energy added in the form of heat is used to perform expansion work, so that no energy is stored. That's the whole point of the reversible isothermal process (its not a isothermal cycle). Aug 25, 2021 at 21:28

As was mentioned in a comment, the system absorbs heat and does work on the surroundings such that internal energy does not change: $$dU=dQ-PdV=0$$