The Ideal Gas Law Modified to Include Gas Density
A Tutorial

Ideal Gas Law (PV = nRT)

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A popular question, which uses PV = nRT, is to determine the molar mass of a gas. Usually, the density of the gas is given and the problem solution process looks like this:

(a) Assume one liter of gas is present
(b) Use PV=nRT to determine moles of gas
(c) Use the density to determine mass of gas
(d) Divide grams by moles to get molar mass.

Many textbooks will show an equation which incorporates the density into the Ideal Gas Law. It looks like this:

MM = $\frac{\mathrm{dRT}}{P}$

where MM is the molar mass and d is the density.

However, textbooks seldom discuss how that equation is arrived at. Below, I show two approches to arriving at the above equation plus an approach that results in a slight (but still correct) variant.

Approach #1: Given the following two equations, combine them to determine the molar mass (MM):

PV = nRT  and  Density = $\frac{m}{V}$ (where m = mass in grams)

Solution:

1) Multiply both sides by the density:

$\left(\frac{\mathrm{m}}{V}\right)$ PV = $\left(\frac{\mathrm{m}}{V}\right)$ nRT

2) V cancels on the left:

mP = $\left(\frac{\mathrm{m}}{V}\right)$ nRT

3) Divide both sides by mol:

$\left(\frac{\mathrm{m}}{\mathrm{mol}}\right)$ P = $\left(\frac{\mathrm{m}}{V}\right)$ RT

I went ahead and cancelled the mol on the right side.

4) $\frac{m}{\mathrm{mol}}$ is the molar mass (MM) and $\frac{m}{V}$ is the density (d):

(MM) P = (d) (RT)

5) Most common arrangement in textbooks:

MM = $\frac{\mathrm{dRT}}{P}$

Approach #2: Given the following two equations, combine them to determine the molar mass (MM):

PV = nRT  and  MM = $\frac{m}{\mathrm{mol}}$

Solution:

1) Rearrange:

mol = $\frac{m}{\mathrm{MM}}$

2) Substitute:

(P) (V) = $\left(\frac{m}{\mathrm{MM}}\right)$ (RT)

3) Multiply both sides by MM

(P) (V) (MM) = (MM) $\left(\frac{m}{\mathrm{MM}}\right)$ (RT)

4) MM will cancel on the right:

(P) (V) (MM) = (m) (RT)

5) Divide both sides by V:

$\left(\frac{1}{V}\right)$ (P) (V) (MM) = $\left(\frac{1}{V}\right)$ (m) (RT)

6) V will cancel on the left. Note that $\left(\frac{m}{V}\right)$ is density (symbol = d). Therefore:

(P) (MM) = (d) (RT)

7) Most common arrangement in textbooks:

MM = $\frac{\mathrm{dRT}}{P}$

Sometimes, as a check, you see the units put in place of the symbols:

$\frac{g}{\mathrm{mol}}$ = $\frac{\mathrm{g\; (L\; atm)\; K}}{\mathrm{L\; (mol\; K)\; atm}}$

L, atm, and K cancel leaving $\frac{g}{\mathrm{mol}}$, the unit associated with molar mass.

Approach #3:

1) Some sources go a slightly difference direction. Here's a repeat of the result of step 4 in the approach just above:

(P) (V) (MM) = (m) (RT)

2) Divide both sides by PV:

MM = $\frac{\mathrm{mRT}}{\mathrm{PV}}$

And that's the final result.

3) It's one small step to the more common arrangement which uses density. However, do not think the equation is wrong. No, no, no. It's perfectly OK since it has a mass in the numerator and a volume in the denominator. That's density. It's just not explicitly shown.

4) In conclusion, this variant is simply a stylistic variation of the most common arrangement and you may find a teacher or some resources using it. Be prepared!

Ideal Gas Law (PV = nRT)

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