Martha Richmond

2023-03-24

What is the work for the free expansion of a gas into vacuum?

marallocajcyb

Beginner2023-03-25Added 6 answers

A gas does not exert any energy during a free expansion into a vacuum because there is no resistance to the gas' expansion, i.e. $P=0$ and therefore the quantity $-P\Delta V=0$.

To say that $W=-{P}_{\text{atm}}\Delta V$ is to say that the gas (the system) expands against an external pressure (applied by the surroundings), given specifically as atmospheric pressure $P}_{\text{atm}$. This is not an expansion into a vacuum, simply by definition.

If we ignore the part about expanding into a vacuum and just think about the gas expanding, the sign of $W$ should be negative from the perspective of the gas.

For an expansion we know that $\Delta V$ is positive, as the final volume is greater than the initial volume $(\Delta V={V}_{f}-{V}_{i})$ and $P$ is always positive.

The work done by the system should therefore be negative for a regular expansion, as work is done by the system (the gas) onto the surroundings. The opposite is true for a compression ($W>0$ for the gas).

Note that if you define $W=P\Delta V$, then it is from the perspective of the surroundings (around the gas), and not the system (the gas).

To say that $W=-{P}_{\text{atm}}\Delta V$ is to say that the gas (the system) expands against an external pressure (applied by the surroundings), given specifically as atmospheric pressure $P}_{\text{atm}$. This is not an expansion into a vacuum, simply by definition.

If we ignore the part about expanding into a vacuum and just think about the gas expanding, the sign of $W$ should be negative from the perspective of the gas.

For an expansion we know that $\Delta V$ is positive, as the final volume is greater than the initial volume $(\Delta V={V}_{f}-{V}_{i})$ and $P$ is always positive.

The work done by the system should therefore be negative for a regular expansion, as work is done by the system (the gas) onto the surroundings. The opposite is true for a compression ($W>0$ for the gas).

Note that if you define $W=P\Delta V$, then it is from the perspective of the surroundings (around the gas), and not the system (the gas).

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