Answer: 42.3L
Step-by-step 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.