VIDEO
The video shows an application of the ideal gas law , PV = nRT P V = n R T moles of gas is; keep in mind that R R (L * atm)/(mol * K) L ⋅ a t m m o l ⋅ K
n = (PV)/(RT) = (1.80 atm * 3.00L)/(320K * 0.082 (L * atm)/(mol * K) n = P V R T = 1.80 a t m ⋅ 3.00 L 320 K ⋅ 0.082 L ⋅ a t m m o l ⋅ K
Let's isolate the R R
n = (1.80 atm * 3.00L)/(320K) * 1/(0.082 (L * atm)/(mol * K)) n = 1.80 a t m ⋅ 3.00 L 320 K ⋅ 1 0.082 L ⋅ a t m m o l ⋅ K
We know that 1/(A/B) = 1 * (A/B)^(-1) = B/A 1 A B = 1 ⋅ ( A B ) − 1 = B A
1/(0.082) * 1/((L * atm)/(mol * K)) = 1/(0.082) * (mol * K)/(L * atm) 1 0.082 ⋅ 1 L ⋅ a t m m o l ⋅ K = 1 0.082 ⋅ m o l ⋅ K L ⋅ a t m
This is why (mol * K) moves to the numerator. The equation now becomes
n = (1.80 atm * 3.00L)/(320K) * ( 1 mol * K)/(0.082 L * atm) n = 1.80 a t m ⋅ 3.00 L 320 K ⋅ 1 m o l ⋅ K 0.082 L ⋅ 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 = ( 1.80 * 3.00 mol)/( 320 * 0.082) = 0.206 n = 1.80 ⋅ 3.00 m o l 320 ⋅ 0.082 = 0.206
