A sample of gas has a volume of 50.0 L at 135*C. Assuming the pressure

remains constant, what is the volume of the gas if the temperature is

decreased to 72.0*C?

42.3 L

O 84.6L

O 423 L

O 26.7L

O 22.4L

Answer: 42.3L

Explanation:

Given that,

Original volume of gas (V1) = 50.0L

Original temperature of gas (T1) = 135°C

[Convert 135°C to Kelvin by adding 273

135°C + 273 = 408K]

New volume of gas (V2) = ?

New temperature of gas (T2) = 72°C

[Convert 72°C to Kelvin by adding 273

72°C + 273 = 345K]

Since volume and temperature are given while pressure is held constant, apply the formula for Charle's law

V1/T1 = V2/T2

50L/408K = V2/345K

To get the value of V2, cross multiply

50L x 345K = 408K x V2

17250L•K = 408K•V2

Divide both sides by 408K

17250L•K / 408K = 408K•V2/408K

42.28L = V2

Round the volume to the nearest tenth, so V2 becomes 42.3L

Thus, the volume of the gas if the temperature is decreased is 42.3 liters.