# Gas Laws Practice

## Practice

To change a temperature expressed in degrees Celsius to a temperature on the Kelvin scale, what must be done to the Celsius temperature?

Why must we use the Kelvin scale in gas law problems?

so that no temperature has a negative value

To check your solutions to the problems, click the words Hint, Formula and Variables, and Answer.

The volume of a sample of gas is 2.00 L when the temperature is 11.0 °C. While the pressure remains constant, the temperature is changed to a new value, which causes the volume to become 3.00 L. What was the temperature changed to?

Hint

Notice that the temperature given is in degrees Celsius. Kelvin temperatures must be used in all gas laws. This temperature must be converted to Kelvin by adding 273 to it.

Formula and Variables

T1 = 11 °C + 273 = 284 K
V1 = 2.00 L
T2 = ?
V2 = 3.00 L
V1T2 = V2T1
T2 = (V2T1) / V1

T2 = (3.00L x 284 K) / 2.00 L
T2 = 426 K

This is an example of Charles’ law

The volume occupied by a sample of gas is 480 mL when the pressure is 115 kPa. What pressure must be applied to the gas to make its volume become 650 mL?

Formula and Variables

P1 = 115 kPa
V1 = 480 mL
P2 = ?
V2 = 650 mL

P1V1 = P2V2
P2 = (P1V1) / V2

P2 = (P1V1) / V2
P2 = (115 kPa x 480 mL) / 650 mL
P2 = 85 kPa

This is an example of Boyle’s Law.

The volume occupied by a sample of gas is 240.0 mL when the pressure is 1.20 atm. What volume, at constant temperature, will the gas occupy when the pressure is decreased to 0.860 atm?

Formula and Variables

P1 = 1.20 atm
V1 = 240.0 mL
P2 = 0.860 atm
V2 = ?

P1V1 = P2V2
v2 = (P1V1) / P2

V2 = (P1V1) / P2
V2 = (1.20 atm x 240.0 mL) / 0.860 atm
V2 = 335 mL

The volume of a sample of gas is 25.0 mL when the temperature is 270 K. If the temperature is changed to 30.0 °C, what will be the new volume occupied by the gas assuming that the pressure remains constant?

Hint

Notice that the second temperature given is in degrees Celsius. Kelvin temperatures must be used in all gas laws. This temperature must be converted to Kelvin by adding 273 to it.

Formula and Variables

T1 = 270 K
V1 = 25.0 mL
T2 = 30 °C + 273 = 293 K
V2 = ?

V1T2 = V2T1
V2 = (V1T2) / T1

V2 = (V1T2) / T1
V2 = (25.0 mL x 293 K) / 270 K
V2 = 28.1 mL

When the volume of a sample of gas is divided by the temperature of the gas, the result is 1.33 mL / K. The temperature of the gas is changed to a new value, which happens to be 411 K while the pressure is kept constant. What volume does the sample of gas occupy at 411 K?

Note

Don’t confuse lowercase k, which refers to a constant value, with uppercase K, which refers to Kelvin.

Formula and Variables

(V/T) = k
(V/T) = 1.33 mL/K

V = 1.33 mL/K x 411 K
V = 547 mL

When the pressure exerted by a sample of gas is multiplied by the volume occupied by the sample, the result of this multiplication is 1.60 x 105 mm Hg·mL. The pressure exerted by the sample changes to a new value, which happens to be 750 mm Hg. What volume will the sample occupy at this pressure, assuming that temperature remains constant?

Formula and Variables

PV = k