Question #4760a

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Question #4760a

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Answer 1 · 2014-12-18T08:40:18

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

$n = \frac{P V}{R T} = \frac{1.80 a t m \cdot 3.00 L}{320 K \cdot 0.082 \frac{L \cdot a t m}{m o l \cdot K}}$ $n = (PV)/(RT) = (1.80 atm * 3.00L)/(320K * 0.082 (L * atm)/(mol * K)$

Let's isolate the $R$ $R$ constant

$n = \frac{1.80 a t m \cdot 3.00 L}{320 K} \cdot \frac{1}{0.082 \frac{L \cdot a t m}{m o l \cdot K}}$ $n = (1.80 atm * 3.00L)/(320K) * 1/(0.082 (L * atm)/(mol * K))$

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

$\frac{1}{0.082} \cdot \frac{1}{\frac{L \cdot a t m}{m o l \cdot K}} = \frac{1}{0.082} \cdot \frac{m o l \cdot K}{L \cdot a t m}$ $1/(0.082) * 1/((L * atm)/(mol * K)) = 1/(0.082) * (mol * K)/(L * atm)$

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

$n = \frac{1.80 a t m \cdot 3.00 L}{320 K} \cdot \frac{1 m o l \cdot K}{0.082 L \cdot a t m}$ $n = (1.80 atm * 3.00L)/(320K) * ( 1 mol * K)/(0.082 L * atm)$

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

$n = \frac{1.80 \cdot 3.00 m o l}{320 \cdot 0.082} = 0.206$ $n = ( 1.80 * 3.00 mol)/( 320 * 0.082) = 0.206$ moles

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Question #4760a

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