# Question 4760a

Dec 18, 2014

The video shows an application of the ideal gas law, $P V = n R T$. In order to solve for mass, we first need to determine what the number of moles of gas is; keep in mind that $R$'s units are $\frac{L \cdot a t m}{m o l \cdot K}$. So,

n = (PV)/(RT) = (1.80 atm * 3.00L)/(320K * 0.082 (L * atm)/(mol * K)#

Let's isolate the $R$ constant

$n = \frac{1.80 a t m \cdot 3.00 L}{320 K} \cdot \frac{1}{0.082 \frac{L \cdot a t m}{m o l \cdot K}}$

We know that $\frac{1}{\frac{A}{B}} = 1 \cdot {\left(\frac{A}{B}\right)}^{- 1} = \frac{B}{A}$, and we can apply this to units as well

$\frac{1}{0.082} \cdot \frac{1}{\frac{L \cdot a t m}{m o l \cdot K}} = \frac{1}{0.082} \cdot \frac{m o l \cdot K}{L \cdot a t m}$

This is why (mol * K) moves to the numerator. The equation now becomes

$n = \frac{1.80 a t m \cdot 3.00 L}{320 K} \cdot \frac{1 m o l \cdot K}{0.082 L \cdot a t m}$

atm, L, and K are cancelled out since they're both on the numerator, and on the denominator, and the answer becomes

$n = \frac{1.80 \cdot 3.00 m o l}{320 \cdot 0.082} = 0.206$ moles