### Ideal Gas Law with Molecules and Unit Conversions (Ex1)

A vessel at $${25^o}C$$ has a gas of amount $$5.0 \times {10^9}\frac{{molecules}}{{{m^3}}}$$. What is the pressure in the vessel? Use the ideal gas law.

I'll show rounded decimals, but keep as many decimals in your calculator as possible! Always use the Kelvin temperature in thermodynamics, so $${25^o}C$$ is converted to 298K. Divide the molecules/m3 by Avogadro's number ($${6.022 \times {{10}^{23}}}$$) to get mol/m3.

$\frac{{5.0 \times {{10}^9}\frac{{molecules}}{{{m^3}}}}}{{6.022 \times {{10}^{23}}}} = 8.30 \times {10^{ - 15}}\frac{{mol}}{{{m^3}}}$

Note that 1m3 = 1000L. The ideal gas law, PV=nRT uses L units if you use $$R = 0.08206\frac{{Latm}}{{molK}}$$. Let's convert the units to L first.

$8.30 \times {10^{ - 15}}\frac{{mol}}{{{m^3}}} \times \frac{{1{m^3}}}{{1000L}} = 8.30 \times {10^{ - 18}}\frac{{mol}}{L}$

Now plug into the equation PV=nRT. Note that this quantity of mol/L we found is already n/V (n is moles, V is volume).

$P = \frac{n}{v}RT$

$P = \left( {8.30 \times {{10}^{ - 18}}\frac{{mol}}{L}} \right)\left( {0.08206\frac{{Latm}}{{molK}}} \right)\left( {298K} \right)$

$P = 2.03 \times {10^{ - 16}}atm$

Done. This is not the only way to do it, you can use a different value of R with different units to get P in other units as well.

Please see our other worked-out examples of the ideal gas law below:

Ideal Gas Law Mole Fractions (Ex2)

Ideal Gas Law Two Gases Combined (Ex3)

Ideal Gas Law Partial Pressures and Mole Fractions (Ex4)

Ideal Gas Law Mixing Two Gas Vessels Together (Ex5)