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Ideal Gas Law with Molecules and Unit Conversions (Ex1)
Example Problem: A vessel at 25\(^OC\) has a gas of the following concentration: \(5.0 \times {10^9}\frac{{molecules}}{{{m^3}}}\). What is the pressure in the vessel? Use the ideal gas law.
We’ll show rounded decimals, but you should keep as many decimals in your calculator as possible!
Always use the Kelvin temperature in thermodynamics, so 25\(^OC\) is converted to 298K.
Divide the molecules/\({m^3}\) by Avogadro’s number to get mol/\({m^3}\).
\[\frac{{5.0 \times {{10}^9}\frac{{molecules}}{{{m^3}}}}}{{6.022 \times {{10}^{23}}}} = 8.30 \times {10^{ – 15}}\frac{{mol}}{{{m^3}}}\]
Note that 1\({m^3}\) = 1000 L.
The ideal gas law, PV = nRT uses L units if you use \(R = 0.08206\frac{{Latm}}{{molK}}\).
There are different R constants with different units, so be careful which R value you use!
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).
\[P = \frac{n}{v}RT\]
\[P = \left( {8.30 \times {{10}^{ – 18}}\frac{{mol}}{L}} \right)\left( {0.08206\frac{{Latm}}{{molK}}} \right)\left( {298K} \right)\]
\[P = 2.03 \times {10^{ – 16}}atm\]
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. There are many conversion tables you can use.