# What is the need of adiabatic compression in Carnot cycle?

Why do we need to compress the gas adiabatically after the isothermal compression of gas? What if we just skip this step? What will be the consequences? By "skipping this step", I mean, after isothermal compression, just make the contact of gas and hot reservoir until both of them get equal temperature (during this we keep the volume of gas constant), and after this we follow the usual steps (i.e., isothermal expansion). Since there is no other heat loss (except the obvious heat damping during isothermal compression), will there be any effect on efficiency? Correct me, if I am wrong.

• Since there is no other heat loss- Work and heat are path functions so the heat does differ in your case. Try plotting your cycle in a graph and calculate it's efficiency manually. Is it more than the carnot cycle? Jun 7, 2021 at 14:03

Why do we need to compress the gas adiabatically after the isothermal compression of gas? What if we just skipped it?

The short answer is it is necessary in order to increase the temperature and internal energy of the system to its original state and complete the cycle. The work done in the compression equals the increase in internal energy (since $$Q=0$$). If we skipped it its internal energy and temperature would be less than at the start of the cycle.

By "skipping this step",I mean, after isothermal compression, just make the contact of gas and hot reservoir until both of them get equal temperature (during this we keep the volume of gas constant),

To do what you say you would have to continue the isothermal compression until you reached the initial volume. That means more heat is discarded leaving less to do work decreasing efficiency. What's more, you will have to add more heat to return to the initial state during the constant volume process, which further reduces the efficiency, since efficiency is the net work done divided by the gross heat added.

Bottom line: Doing what you suggest will result in a cycle efficiency less than the Carnot cycle efficiency.

Hope this helps.

Bob D's answer is good, and I'm no expert (mechanical engineer, not a physicist), but here's how I think about it. Yes, the total energy of the cycle will be greater this way, because the last bit of compression can happen at the low temperature, therefore requiring less work to return to the initial volume while at lower pressure.

The flaw is that when you touch the still cold gas to the hot reservoir to finally return to the initial state, this step now requires heat input, when before it only required mechanical input (compressive force). Furthermore, the input heat required to increase the temperature and pressure of the gas will be more than what would have been required to simply compress it. This is because the heat transfer from the hot reservoir to the cold gas happens across a large temperature difference, which cannot happen statically and will incur loss due to entropy generation.

The reason that this process seemed reasonable at first is because a "high temperature reservoir" makes it sound like the heat that it supplies is free, so getting more net mechanical energy per cycle is appealing, even though in practice that extra input heat has to come from somewhere.