Question

Question #4760a

Answer 1

The video shows an application of the ideal gas law, `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`'s units are `(L * atm)/(mol * K)`. So,

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

Let's isolate the `R` constant

`n = (1.80 atm * 3.00L)/(320K) * 1/(0.082 (L * atm)/(mol * K))`

We know that `1/(A/B) = 1 * (A/B)^(-1) = B/A`, and we can apply this to units as well

`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 = (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 = ( 1.80 * 3.00 mol)/( 320 * 0.082) = 0.206` moles