Molar Gas Volume (DP IB Chemistry)
Revision Note
Molar Gas Volume
Gases in a container exert a pressure as the gas molecules are constantly colliding with the walls of the container
A particle model of a gas in a container
Gas particles exert a pressure by constantly colliding with the walls of the container
Changing gas volume
Decreasing the volume (at constant temperature) of the container causes the molecules to be squashed together which results in more frequent collisions with the container wall
The pressure of the gas increases
Gas Molecule Collision Frequency with Increasing Pressure Diagram
Decreasing the volume of a gas causes an increased collision frequency of the gas particles with the container wall
The pressure is therefore inversely proportional to the volume (at constant temperature)
This is known as Boyle's Law
Mathematically, we say P ∝ 1 / V or PV = a constant
We can show a graphical representation of Boyle's Law in three different ways:
A graph of pressure of gas plotted against 1 / volume gives a straight line
A graph of pressure against volume gives a curve
A graph of PV versus P gives a straight line
Sketch graphs of Boyles' Law
Three graphs that show Boyle’s Law
Changing gas temperature
When a gas is heated (at constant pressure) the particles gain more kinetic energy and undergo more frequent collisions with the container walls
To keep the pressure constant, the molecules must get further apart and therefore the volume increases
The volume is therefore directly proportional to the temperature in Kelvin (at constant pressure)
This is known as Charles' Law
Mathematically, V ∝ T or V / T = a constant
A graph of volume against temperature in Kelvin gives a straight line
Gas Molecule Collision Frequency with Increasing Temperature Diagram
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)
Changing gas pressure
Increasing the temperature (at constant volume) of the gas causes the molecules to gain more kinetic energy
This means that the particles will move faster and collide with the container walls more frequently
The pressure of the gas increases
The temperature is therefore directly proportional to the pressure (at constant volume)
Mathematically, we say that P ∝ T or P / T = a constant
A graph of temperature in Kelvin of a gas plotted against pressure gives a straight line
Gas Molecule Collision Frequency with Increasing Pressure at Constant Volume Diagram
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)
Pressure, volume and temperature
Combining these three relationships together:
PV = a constant
V / T = a constant
P / T = a constant
We can see how the ideal gas equation is constructed
PV / T = a constant
PV = a constant x T
This constant is made from two components, the number of moles, n, and the gas constant, R, resulting in the overall equation:
PV = nRT
Changing the conditions of a fixed amount of gas
For a fixed amount of gas, n and R will be constant, so if you change the conditions of a gas we can ignore n and R in the ideal gas equation
This leads to a very useful expression for problem solving
Where P1, V1 and T1 are the initial conditions of the gas and P2, V2 and T2 are the final conditions
Worked Example
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
Worked Example
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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