# A sealed jar has 0.20 moles of gas at a pressure of 300.12 kPa and a temperature of 229 K. What is the volume of the jar?

Jul 19, 2017

$V \cong 1 \cdot L$

#### Explanation:

We assume ideality and solve the Ideal Gas equation for $V$....

$V = \frac{n R T}{P} = \frac{0.20 \cdot m o l \times 8.314 \cdot \frac{L \cdot k P a}{K \cdot m o l} \times 229 \cdot K}{300.12 \cdot k P a}$

=??*L

Note that I had to choose a gas constant $R$ with suitable units. Most of the time these values are quoted on an exam paper so you do not have to memorize them. You still have to be able to use them. If you go thru the calculation, you will find that the various units cancel, and we finally get an answer in $\text{litres}$, as is absolutely required for volume.

i.e.

$V = \frac{n R T}{P} = \frac{0.20 \cdot \cancel{m o l} \times 8.314 \cdot \frac{L \cdot \cancel{k P a}}{\cancel{K} \cdot \cancel{m o l}} \times 229 \cdot \cancel{K}}{300.12 \cdot \cancel{k P a}}$

.........$\text{litres}$ as required..........