Gas Law Problem

Question

Boyles Law: 

1. A cylinder of compressed gas has a volume of 3.85 L and a pressure of 463 kPa. What volume would the gas    occupy if it were allowed to escape into a balloon at a pressure of 110.0 kPa? Assume constant temperature.

 

2. A gas at 100.0 kPa takes up a volume of 59.0 cm³. What will the volume be if the pressure is decreased to      20.0 kPa?

 

3. A gas at 120.0 kPa has a volume of 33.0 mL. What is the volume at standard pressure?

 

Answer 1

Hello, Emily,

1. Since the moles of gas do not change, we can use a simplified expression to determine what happens to the gas when conditions are changed:

P1V1/T1 = P2V2/T2

Rearrange to isolate the unknown, V2 in this case.

V2 = V1(T2/T1)(P1/P2)

Note the manner in which I grouped the pressure and temperature values, as ratios of their initial and final states. This makes it easier to understand what is happening.

There is no mention of temperature, so we can assume that T1 = T2, so that ratio cancels to just 1. We are left only with the pressure change (P1/P2). The units cancel, so we can simply divide 463 by 110, for a value of 4.21. This makes sense - the volume will increase when the pressure is decreased. This says it will increase by a factor of 4.21, so the resulting volume, V2, will be 16.2 L (3 sig figs).

2. This is an identical question, with different values and units. Use the same process as above. We can use the units provided in the answer, since the calculations all involve ratios. (Be careful with temperature, however. Absolute temperatures, Kelvin, must be used to obtain the correct ratio. Imagine the answer if a negative Celcius temperature T1 went positive for T2). Enter the data and solve. I found 295 cm³.

3. Same question, same process. The only question (for me) was what is standard pressure in kPa? 1 atm = 101.3 kPa. Now enter the information, as before. My calculation suggests a final volume of 39.1 ml.

Bob

Answer 2

Acording to Boyles law, at constant temperature,

P1V1 = P2V2

1) P1 = 463 kpa, V1 = 3.85 L

P2 = 110 kpa, V2 = ?

V2 = P1V1/P2 = (463)×(3.85)/(110) = 16.21 L

2) P1 = 100 kpa, V1 = 59 cm^3

P2 = 20.0 kpa, V2 = ?

V2 = P1V1/P2 = (100)×(59)/(20) = 295 cm^3

3) P1 = 120 kpa, V1 = 33 mL,

P2 = 101.3 kpa(standard pressure), V2 = ?

V2 = P1V1/V2 = (120)×(33)/(101.3) = 39.09 mL