In this section you will gain a deeper understanding of the Ideal Gas Law by performing some ideal gas law calculations.
Remember that the Ideal Gas Law is of the form:
Problem 1: What volume is needed to store 0.050 moles of helium gas at 202.6kPa and 400K?
Solution to Problem 1. We are calculating the volume of an ideal gas V = (nRT) กา P
P = 202.6 kPa
N = 0.050 mol
T = 400K
V = ? L
R = 8.314 J K-1 mol-1
202.6V=0.050x8.314x400
202.6 V = 166.28
V = 166.28 กา 202.6
V = 0.821 L (821mL)
Problem 2: What pressure will be exerted by 20.16g hydrogen gas in a 7.5L cylinder at 20 degree C?
Solution to Problem 2. We are calculating the pressure of an ideal gas: P = (nRT) กา V
+ = ? kPa
V = 7.5L
n = mass กา MM where mass=20.16g, MM(H2)=2x1.008=2.016g/mol
So n = 20.16 กา 2.016=10mol
T = 20o = 20+273=293K
R = 8.314 J K-1 mol-1
Px7.5=10x8.314x293
Px7.5 = 24360.02
P = 24360.02 กา 7.5 = 3248kPa
Problem 3: A 50L cylinder is filled with argon gas to a pressure of 10130.0kPa at 30 degree C. How many moles of argon gas are in the cylinder?
Solution to Problem 3. We are calculating moles of gas n = (PV) กา (RT)
P = 10130.0 kPa
V = 50L
n = ? mol
R = 8.314 J K-1 mol-1
T = 30 degree C = 30+273 = 303K
10130.0x50=nx8.314x303
506500=nx2519.142
n=506500 กา 2519.142=201.1mol
Problem 4. To what temperature does a 250mL cylinder containing 0.40g helium gas need to be cooled in order for the pressure to be 253.25kPa?
Solution to Problem 4: We are calculating gas temperature T = (PV) กา (nR)
P = 253.25kPa
V = 250mL = 250 กา 1000 = 0.250L
n=mass กา MM where mass=0.40g
MM(He)=4.003g/mol
So n = 0.40 กา 4.003 = 0.10mol
R = 8.314 J K mol-1
T = ? K
253.25x0.250=0.10x8.314xT
63.3125 = 0.8314xT
T=63.3125 กา 0.8314=76.15K