# Internal energy of a reversible adiabatic process

what happens to the heat transfer and the internal energy in a system that undergoes reversible adiabatic process back to its original state. Bearing in mind that heat $$Q$$ was transferred at the initial state.

• the system is undergoing three reversible processes, constant volume, then constant pressure before the final reversible adiabatic process back to its constant volume state where heat Q is initially supplied – Adedayo Adeniji Jun 14 '18 at 8:15
• Let us say we start with T1, P1, V1. Then isochoric heat addition gives us T2, P2, V1 ; T2, P2 > T1, P1 . Then volume, temperature decrease because of heat rejection at constant pressure, giving us T3, P2, V3; such that T1<T3<T2. Now the situation is that pressure P2>P1, volume V3<V1 and temperature T3>T1. Therefore, to complete the cycle, we require adiabatic expansion(work), so that volume increases from V3 to V1 and pressure, temperature drop from P2, T3 to P1, T1 respectively. This is my guess at how the cycle could be possibly executed. – Mohammad Nayef Jul 13 '18 at 18:34
• Regarding heat transfer and internal energy, since the process is adiabatic that means zero heat transfer. Internal energy would decrease in adiabatic expansion. – Mohammad Nayef Jul 13 '18 at 18:37

In an adiabatic process there is no heat transfer between the system undergoing the process and its environment; this is the definition of being adiabatic irrespective of being reversible or irreversible. If the process is adiabatic and reversible then at any stage of the process the system's entropy is constant; if the process is irreversible and adiabatic then during its irreversible stages the entropy of the system increases. In other words, an adiabatic process is such that the interaction between the system and its environment is mechanical, here mechanical is meant to be in its general sense including also electric, magnetic, gravitational, etc., interactions from the other parts of physics.

What happens to the heat transfer and the internal energy in a system that undergoes reversible adiabatic process back to its original state.

Any internal energy increase due to net heat or work into the system prior to the reversible adiabatic process decreases during the adiabatic process due to work out, so that the internal energy of the system returns to its initial state following the cycle.

The following explanation is offered.

The three reversible processes you described in your comments are shown in the Figure below.

Heat is added in the isochoric process (1-2) and no work is done. Internal energy is therefore increased. Heat is rejected in the isobaric process (2-3), which reduces internal energy. However, work is done on the system, which increases internal energy.

The adiabatic process (3-1), which returns the system to its initial state, is an expansion. Since $$Q=0$$, the expansion work done by the system comes at the expense of its internal energy, i.e., a causes a decrease in internal energy.

Since the complete cycle 1-2-3-1 returns the system to its initial state, the total change in internal energy is zero. This tells us that there was an increase in internal energy going from 1-2-3 equal in magnitude to the decrease in internal energy in going from 3-1 as a result of the work done in the adiabatic process.

Hope the helps