In this video blog post, we'll do several gas law practice problems. The first two examples only contain "1 situation."
These are Ideal Gas Law conditions -- only 1 situation.
Ideal Gas Law Examples
ex: A sample of H2 gas occupies a volume of 8.56 L at a temperature of 0°C and a pressure of 1.5 atm. How many moles of hydrogen are present?
➞ PV = nRT ➞ n = PV / RT
➞ n = (1.5 atm) (8.56 L) / (0.0821 L.atm/K.mol) (273 K)
➞ n = 0.57 mol H2
Here's the same solution, but with a natural flow...
ex: Calculate the volume of 0.845 mol of nitrogen gas at a pressure of 699.2 torr and a temperature of 315 K.
➞ we have to convert pressure from torr to atm
n = 0.845 mol N2
T = 315 K
V = ?
P = 699.2 torr
(699.2 torr / 1) . (1 atm / 760 torr) = 0.9200 atm
➞ PV = nRT ➞ V = nRT / P
➞ V = nRT / P = (0.845)(0.0821)(315) / 0.9200 = 23.8 L N2
And, here's an easier-to-follow view of the above process:
PV = nRT problems don't typically have a "situation 1" and "situation 2" occurring. There's usually just one situation.
"Before / After" type problems usually require Boyle's Law, Charles' Law, Avogadro's Law, Gay-Lussac's Law, or the Combined Gas Law.
Let's try some!...
Charles' Law Example
ex: A sample of methane gas with a volume of 38 mL at 5°C is heated to 86°C at constant pressure. Calculate its new volume.
➞ V1 = 38 mL , T1 = 5 + 273 = 278 K
➞ V2 = ?? mL , T2 = 86 + 273 = 359 K
... How come it wasn't necessary to convert V1 and V2 into liters (L) ?
Because "R" (in PV = nRT) is equal to 0.0821 L.atm / K.mol, but "R" is not being used in Charles' Law !! :-)
Combined Gas Law Example
ex: Diborane gas (B2H6) has a pressure of 345 torr at a temperature of -15°C and a volume of 3.48 L. If conditions change such that the temperature is 36°C and the pressure is 468 torr, what will be the volume of the sample?
P1 = 345 torr
T1 = -15 + 273 = 258 K
V1 = 3.48 L
P2 = 468 torr
T2 = 36 + 273 = 309 K
V2 = ?? L
➞ we'll use the Combined Gas Law,
P1V1 / T1 = P2V2 / T2
V2 = 3.07 L B2H6
Ideal Gas Law Practice Problem
ex: A gaseous sample contains 0.35 moles of Argon at a temperature of 13°C and a pressure of 568 torr. If it's heated to 56°C and a pressure of 897 torr, calculate the change in volume that occurs.
➞ this one's tricky because at first we see two pressures...
However, n1 = n2 = remains unchanged and we're asked to find the change in volume (ΔV).
To find ΔV = V2 - V1, we'll need to use PV = nRT twice...
We will need to convert our pressures from torr to atmospheres (atm), and then set-up two separate PV = nRT equations:
➞ P1V1 = nRT1 and P2V2 = nRT2
Here's the work that's involved...
That's it for entry #3 from SECTION 5 - Gases.