The work done by the atmosphere

In most real problems the gas is not in contact with a perfect vacuum, but with the atmosphere. When the gas expands, it pushes the atmosphere away; when it is compressed, it is the atmosphere that pushes it. In both cases a second kind of work appears: the work done by the atmosphere.

Principle — Work done by the atmosphere

$\ev{L_\text{atm} = P_\text{atm}\cdot\Delta V_\text{atm}}$ where $\Delta V_\text{atm}$ is the change in volume of the atmosphere and $P_\text{atm} = 1{,}013\cdot 10^5\;\text{Pa}$ is atmospheric pressure.

The delicate point is the sign of $\Delta V_\text{atm}$. In the typical case, where a piston separates the gas from the atmosphere, the two volumes change in opposite ways: if the gas expands, the atmosphere is compressed by the same amount.

Key formula — Gas expanding

$\Delta V_\text{gas} > 0 \qquad \Delta V_\text{atm} = -\Delta V_\text{gas} < 0$ $L_\text{atm} = P_\text{atm}\,\Delta V_\text{atm} < 0$ During expansion the atmosphere does negative work on the gas (the gas must spend energy to make room for it).

When the gas expands by $\Delta V_\text{gas}$, the atmosphere is compressed by the same amount: $\Delta V_\text{atm} = -\Delta V_\text{gas}$. The work done by the atmosphere is therefore $L_\text{atm} = -P_\text{atm}\,\Delta V_\text{gas}$.

When the atmosphere and the gas do not have opposite $\Delta V$

If between the gas and the atmosphere there is a massive piston that rises by $\Delta h$, both volumes change by the same amount: $\Delta V_\text{gas} = A\,\Delta h$ and $\Delta V_\text{atm} = -A\,\Delta h$. But if the piston bends or if the container changes shape, the two $\Delta V$ can be different. One must always ask: by how much does the gas’s volume change? By how much does the atmosphere’s change?

Collegamenti

Argomenti: Teoria cinetica dei gas · Termodinamica Concetti: Pressione · Lavoro di una forza

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