Sample Problems from Standard 5.2 (Ideal Gas Law) Detailed Solutions Below

Sample Problems from Standard 5.2 (Ideal Gas Law) Detailed solutions below

1. A gas cylinder contains .722 L of hydrogen gas at a pressure of 10.6 atm. If the gas is used to fill a balloon at a pressure of 0.96 atm, what is the volume of the filled balloon?

2. A bubble of carbon dioxide gas in some unbaked bread dough has a volume of 1.15 mL at a temperature of 22.0oC. What volume will the bubble have when the bread is baked and reaches a temperature of 99.0oC?

3. A tank with a volume of 658 mL contains 1.50 g of neon gas. The maximum safe pressure that the tank can withstand is 450 kPa. At what temperature in degrees Celsius will the tank have that pressure?

4. 2 BaO2(s)  2 BaO(s) + O2(g)

What mass of BaO2 reacted if 265 mL of O2 is collected by water displacement at 0.975 atm and 10.0oC?

1. Since the pressure and volume are changing, we start with the relationship

P1V1 / n1RT1 = P2V2 / n2RT2

Since nothing is mentioned about temperature or the number of moles, we can assume they stay

the same. This will cancel them out of the equation along with R to leave us with P1V1 = P2V2.

(10.6 atm)(.722 L) = (0.96 atm)(V2)

Divide both sides by 0.96 atm to isolate the variable.

V2 = (10.6 atm)(.722 L) / (0.96 atm) = 7.97 L

2. In this example, volume and temperature are changing. Everything else is assumed to remain

constant. This time, we are left with V1 /T1 = V2 /T2. Be sure to convert temperature to Kelvins. (1.15 mL) / (295 K) = V2 / 371 K

Cross multiply and divide to determine the new volume.

V2 = (1.15 mL)(371 K) / 295 K = 1.45 mL

3. None of the variables are changing in this problem, so we will use the ideal gas law in its regular form, PV = nRT. Keep in mind, pressure must be in atmospheres, volume in Liters, and the grams will have to be converted to moles by dividing by the molar mass of neon.

P = 450 kPa / 101.3 = 4.44 atm V = .658 L n = 1.50 g / 20.18 = 0.0743 moles

(4.44 atm)(.658 L) = (0.0743 mol)(.0821)(T)

Divide both sides by (0.0743 mol)(.0821)to isolate the variable.

T = (4.44 atm)(.658 L) / (0.0743 mol)(.0821) = 479 K

Convert from K to oC.

479 K – 273 = 206oC

4. From the volume, pressure, and temperature of the gas, O2, we can calculate the number of moles

of O2. Then we can use stoichiometry to calculate moles of BaO2 which can then be converted to grams.

Be sure to convert the measurements to the necessary units required by the ideal gas law.

(0.975 atm)(0.265 L) = n(0.0821)(283 K)

Divide both sides by (0.0821)(283 K) to isolate the variable.

n = (0.975 atm)(0.265 L) / (0.0821)(283 K) = 0.0111 mol O2

Next, we multiply by B / A from the balanced equation to determine moles of BaO2.

0.0111 mol O2 x (2 / 1) = 0.0222 mol BaO2

Multiply by the molar mass of BaO2 to get to grams.

(0.0222 mol)(137 + 16 + 16) = 3.76 g BaO2