# Ideal gas law for open system

The ideal gas law $pV=m\mathcal RT$ can obviously be used to solve for mass.

Can we substitute the volume $V$ for volumetric flow rate $A\cdot\textrm{velocity}$ ($A$ is cross sectional area) through a control volume and solve the for $\textrm{mass flow rate}$?

$$\textrm{ mass flow rate} = (p\cdot \textrm{area}\cdot\textrm{velocity})/(\mathcal RT)$$

Is this valid or do other effects have to be considered?

• This is the molar flow rate, not the mass flow rate. To get the mas flow rate, you need to multiply by the molecular weight. And this would be the mass flow rate into or out of the control volume at cross sectional area A. At steady state, it would also be the the mass flow rate through the control volume. – Chet Miller Sep 30 '16 at 21:38
• @ChesterMiller what you say depends on the $R$ you've settled on - whether it's the universal gas constant or the specific gas constant. Drew, keep in mind the ideal gas law is a state equation. It doesn't necessarily take into account the dynamics of your particular problem. But for steady state I would say Yes, you can do what you want to do. – docscience Oct 1 '16 at 2:36

First you need the assumption that laws of equilibrium thermodynamics apply to a manifestly non-equilibrium case such as a flow. In Modern Thermodynamics by Kondepudi and Prigogine, the authors assert that molecular dynamics simulations show that a fluid element quickly relaxes to thermodynamic equilibrium (locally), and so for most flows we encounter, application of relations from equilibrium thermodynamics is good enough. Once this assumption is made the procedure to find mass flow rate is straight forward:

$$\dot{m}''_\textrm{flow}=\rho u=\frac{p}{\mathcal RT}u$$ gives flow per unit area.

To get mass flow rate over the entire cross-section you integrate above expression over the cross-section:

$$\dot{m}_\textrm{flow}=\int_A ~\mathrm dA ~\frac{p}{\mathcal RT}u$$

in which $p,T,u,$ will depend on position in general. If $p,T,$ are constant over the cross-section, then since flow rate $Q=\int_A ~\mathrm dA ~u$, you get

$$\dot{m}_\textrm{flow}=\frac{p}{\mathcal RT}Q$$

So indeed $Q$ replaces $V$, but only under assumptions made so far.

though the formula is correct, I'm not comfortable with the interpretation. I would derive the mass flow rate as below.

$$\dot m = \rho Q =\rho A \cdot v$$

The density can be written as, $$\rho = \frac{m}{V}=\frac{P}{RT}$$

Then mass flow rate is, $$\dot m=\frac{P}{RT}A\cdot v$$

Here we need to be clear that pressure, temperature, and velocity should be the values at the location where the cross section area is A. To point this out is important because pressure and temperature of some flows changes dramatically along the path.