How work is done by rapid expansion or compression at adiabatic process?

Question

In adiabatic processes, when a gas is compressed rapidly, the internal energy of the gas increases and work is done on the gas. Similarly in expansion the internal energy decreases and work is done by the gas. Here, how does change the internal energy of gas (kinetic energy of molecules) and how is work done by gas or work done on gas (gas is not inside any cylinder with piston)?

Answer 1

Regardless of how the adiabatic expansion or compression is carried out (slowly or rapidly) the first law applies. For the version of the first law (used primarily in chemistry) where

$$\Delta U=Q+W$$

$$Q=0$$

$$\Delta U=W$$

$W$ is positive if work is done on the system, i.e., compression work, and internal energy increases. $W$ is negative if work is done by the system, i.e., expansion work, and internal energy decreases.

Hope this helps.

Answer 2

This question can be answered by the first law of thermodynamics. If one has a system, and puts heat and work into it, the change in energy is increased by the heat and work put in, or: $$\Delta U = \Delta W + \Delta Q.$$ Adiabatic processes are known to be isenotropic, meaning no heat is transferred between the gas and its surroundings as it is compressed ($\Delta S = \Delta Q/T$), so the internal energy is directly changed by the work done on the gas $$\Delta U = \Delta W.$$ Whenever positive work is done, the internal energy increases and vice versa.

But what determines if work is positive or negative? When a gas is compressed, the atoms are slowly put into a smaller space. This costs energy which is put directly into the gas atoms, so each atom becomes "hotter" and picks up a larger velocity. Therefore, the internal energy must increase when being compressed. When the gas is expanded, the internal energy will decrease and the molecules will reduce in speed.

Answer 3

An adiabatic process doesn't exchange heat with the external environment, and thus, from the first principle of thermodynamics (that holds for closed systems), the variation of internal energy equals the external work done on the system

$\Delta U = \Delta W^{ext}$.

Now, if the system continuously evolves through states of equilibrium, and the dissipation os negligible, the external work equals the opposite of the product pressure of the system and its volume variation,

$\Delta W^{ext} = - p \Delta V$,

and thus the following relation holds for adiabatic processes of a closed system

$\Delta U = - p \Delta V$.

Now, since the pressure is always positive, we can define that:

• net work is done on the system if $\Delta W^{ext} = - p \Delta V \gt 0$, i.e. the volume of the system decreases $\Delta V \lt 0$
• the system is doing net work if $\Delta W^{ext} = - p \Delta V \lt 0$, i.e. the volume of the system increases $\Delta V \gt 0$.

As an example, in a piston engine, we can consider the system as the gas in the combustion chamber. Work is done on the system dyring the compression stroke when $\Delta W^{ext} = - p \Delta V = - pA dh\gt 0$, $dh\lt 0$, while the system does net work during the expansion stroke when $\Delta W^{ext} = - p \Delta V = - pA dh \lt 0$, $dh\gt 0$.

Answer 4

If you are asking about the microscopic view of a gas with no intermolecular interactions (the simplest scenario), then during expansion, each gas particle collides with the piston, transferring some kinetic energy to push the piston upwards. The gas particle rebounds from the piston slower than it approached. The loss of kinetic energy causes the temperature of the gas to decrease, but it decreases through collisions with the piston, not thermal heat loss to the environment.

During compression, it would be the opposite, and now the piston acts as a tennis racket imparting kinetic energy so that the particle leaves the piston with more kinetic energy than it had before the collision. The newly acquired kinetic energy causes the temperature of the gas to increase.

This is very different from the collisions that occur during equilibrium, when each gas particle collides with the piston and rebounds with essentially the same energy, much like hitting a tennis ball against a wall.