Gas particles exert a pressure by constantly colliding with the walls of the container
Decreasing the volume of a gas causes an increased collision frequency of the gas particles with the container wall
Three graphs that show Boyle’s Law
Increasing the temperature of a gas causes an increased collision frequency of the gas particles with the container wall (a); volume is directly proportional to the temperature in Kelvin (b)
Increasing the temperature of a gas causes an increased collision frequency of the gas particles with the container wall (a); temperature is directly proportional to the pressure (b)
At 25 oC and 100 kPa a gas occupies a volume of 20 dm3. Calculate the new temperature, in oC, of the gas if the volume is decreased to 10 dm3 at constant pressure.
Answer:
Step 1: Rearrange the formula to change the conditions of a fixed amount of gas. Pressure is constant so it is left out of the formula
Step 2: Convert the temperature to Kelvin. There is no need to convert the volume to m3 because the formula is using a ratio of the two volumes
V1 = 20 dm3
V2 = 10 dm3
T1 = 25 + 273 = 298 K
Step 3: Calculate the new temperature
A 2.00 dm3 container of oxygen at a pressure of 80 kPa was heated from 20 oC to 70 oC The volume expanded to 2.25 dm3 . What was the final pressure of the gas?
Answer:
Step 1: Rearrange the formula to change the conditions of a fixed amount of gas
Step 2: Substitute in the values and calculate the final pressure
P1 = 80 kPa
V1 = 2.00 dm3
V2 = 2.25 dm3
T1 = 20 + 273 = 293 K
T2 = 70 + 273 = 343 K
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