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When using the ideal gas law and other equations, are we assuming that the gas has uniform pressure (i.e. the pressure does not vary inside the gas)?

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Yes we are assuming this. The gas laws deal with gases in equilibrium, and with macroscopic samples of gas (say of linear dimensions thousands of times greater than the mean separation of the molecules). On this scale any pressure differences will cause bulk movements in the gas so that the pressures quickly equalise.

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  • $\begingroup$ An up vote foy saying a gas law assumes equilibrium $\endgroup$
    – Bob D
    Aug 3, 2020 at 23:27
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The pressure in the ideal gas law, which is an equation of state, is assumed to be mean pressure of the gas.

In statistical mechanics, there are relations between the partition function (also called the sum over states) and mean values of external parameters. So for the average pressure: $$\overline{p}=\frac{1}{\beta}\frac{\partial{Z}}{\partial{V}}$$ where Z is the partition function, V is the volume and $\beta$ is 1/kT (Boltzmann's constant times temperature). Once the Z is calculated for a gas of non-interacting monatomic point particles, this expression leads directly to the equation of state for an ideal gas.

Statistical mechanics assumes there will be fluctuations around this average and those can be calculated as well.

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You don't have to. In the form of the ideal gas law often introduced in Chemistry,

$$ PV=nRT $$

there is an assumption that pressure is constant over some volume $V$, but the equally valid version

$$ P=\frac{\rho k_BT}{\bar{m}}$$

(where $\rho$ is the mass density and $\bar{m}$ is the average mass of a gas particle) makes no such assumption.

An example of when this applies is an ideal gas in a gravitational field. A pressure gradient arises to counteract the force of gravity and keep everything in equilibrium.

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