Volume and the rate of collisions
Definition
Gas pressure
Gas pressure is the force per unit area that gas particles exert on a surface as a result of their collisions with it.
- Consider a fixed mass of gas held at constant temperature, so the number of particles and their average speed both stay the same.
- Reducing the volume brings the walls closer together, so each particle travels a shorter distance between one wall collision and the next.
- The particles therefore reach the walls more often, so the number of collisions each second on every square metre of wall increases.
- The force of each individual collision is unchanged, because the temperature and therefore the average speed of the particles has not changed.
- More collisions of the same strength on each square metre give a greater total force on each square metre, so the pressure increases.
- Increasing the volume reverses every step, so the particles hit the walls less often and the pressure decreases.
- Squeezing a sealed balloon becomes harder as you go, because the pressure inside climbs as the volume of the trapped gas falls.
- An air bubble released at the bottom of a deep tank grows as it rises, because the pressure of the water around it falls and the bubble expands in response.
The pressure and volume equation
- For a fixed mass of gas at constant temperature the relationship is p1V1=p2V2p_{1}V_{1}=p_{2}V_{2}p1V1=p2V2.
- In this equation p1p_{1}p1 and V1V_{1}V1 are the pressure and volume before the change, and p2p_{2}p2 and V2V_{2}V2 are the pressure and volume after it.
- The units only need to be consistent, so both pressures must be in the same unit and both volumes must be in the same unit.
- The equation says that the product of pressure and volume stays constant, so pressure and volume are inversely proportional.
- Halving the volume therefore doubles the pressure, and doubling the volume halves it.
- Rearranging for the new pressure gives p2=p1V1V2p_{2}=\dfrac{p_{1}V_{1}}{V_{2}}p2=V2p1V1.
- Rearranging for the new volume gives V2=p1V1p2V_{2}=\dfrac{p_{1}V_{1}}{p_{2}}V2=p2p1V1.
- The equation fails if the temperature changes or if gas leaks in or out, so check both conditions before using it.
Graphs of pressure and volume
- A graph of pressure against volume is a curve that falls steeply at small volumes and flattens at large ones, never touching either axis.
- That curved shape is the signature of an inverse relationship rather than a proportional one.
- A graph of pressure against 1V\dfrac{1}{V}V1 is a straight line through the origin, which is the standard way of showing that the relationship really is inverse.
- The gradient of that straight line equals the constant product pVpVpV for the sample of gas being tested.
- Multiplying each pressure reading by its volume reading is a quick numerical check, because every pair should give the same product within experimental uncertainty.
Example
Finding the new pressure
- A sealed syringe holds 250 cm3250\ \text{cm}^{3}250 cm3 of gas at a pressure of 1.0×105 Pa1.0\times10^{5}\ \text{Pa}1.0×105 Pa, and the plunger is pushed in until the volume is 100 cm3100\ \text{cm}^{3}100 cm3.
- The temperature is unchanged and no gas escapes, so p1V1=p2V2p_{1}V_{1}=p_{2}V_{2}p1V1=p2V2 applies.
- Rearranging gives p2=p1V1V2=1.0×105×250100p_{2}=\dfrac{p_{1}V_{1}}{V_{2}}=\dfrac{1.0\times10^{5}\times250}{100}p2=V2p1V1=1001.0×105×250.
- The new pressure is p2=2.5×105 Pap_{2}=2.5\times10^{5}\ \text{Pa}p2=2.5×105 Pa.
- The volume was cut to two fifths of its original value, so the pressure rose to two and a half times its original value, as an inverse relationship requires.
Example
Finding the new volume
- A gas occupies 0.50 m30.50\ \text{m}^{3}0.50 m3 at a pressure of 2.0×105 Pa2.0\times10^{5}\ \text{Pa}2.0×105 Pa and is then compressed to a pressure of 5.0×105 Pa5.0\times10^{5}\ \text{Pa}5.0×105 Pa at the same temperature.
- Rearranging gives V2=p1V1p2=2.0×105×0.505.0×105V_{2}=\dfrac{p_{1}V_{1}}{p_{2}}=\dfrac{2.0\times10^{5}\times0.50}{5.0\times10^{5}}V2=p2p1V1=5.0×1052.0×105×0.50.
- The new volume is V2=0.20 m3V_{2}=0.20\ \text{m}^{3}V2=0.20 m3.
- Checking the products gives 2.0×105×0.50=1.0×1052.0\times10^{5}\times0.50=1.0\times10^{5}2.0×105×0.50=1.0×105 and 5.0×105×0.20=1.0×1055.0\times10^{5}\times0.20=1.0\times10^{5}5.0×105×0.20=1.0×105, which confirms the answer.
Example
A bubble rising through water
- A bubble of volume 2.0 cm32.0\ \text{cm}^{3}2.0 cm3 is released where the pressure is 3.0×105 Pa3.0\times10^{5}\ \text{Pa}3.0×105 Pa and rises to where the pressure is 1.0×105 Pa1.0\times10^{5}\ \text{Pa}1.0×105 Pa.
- Assuming the water is at one temperature throughout, p1V1=p2V2p_{1}V_{1}=p_{2}V_{2}p1V1=p2V2 can be used.
- Substituting gives V2=3.0×105×2.01.0×105V_{2}=\dfrac{3.0\times10^{5}\times2.0}{1.0\times10^{5}}V2=1.0×1053.0×105×2.0.
- The bubble reaches the surface with a volume of V2=6.0 cm3V_{2}=6.0\ \text{cm}^{3}V2=6.0 cm3.
- The pressure fell to one third of its starting value, so the volume grew to three times its starting value.
Exam technique
Using the equation confidently
- List the four quantities with their subscripts before substituting, so the before and after values do not get swapped.
- State that the temperature is constant and the mass of gas is fixed, since a question about conditions is looking for exactly that.
- Check the direction of the answer, because a smaller volume must give a larger pressure and a larger volume must give a smaller one.
- Give the reasoning in two steps for an explanation question: the collisions per second change, then the pressure changes.
- Say that the force per collision is unchanged, which is the detail that separates this explanation from the one about heating a gas.
- Use inversely proportional only when the product pVpVpV is constant, and reserve directly proportional for a straight line through the origin.
Common Mistake
- Do not write that the particles move faster when the volume is reduced, because the temperature and the average speed are unchanged.
- Do not use this equation when the temperature also changes, because the relationship only holds at constant temperature.
- Do not mix pressure units within one calculation, because the two pressures must be expressed in the same unit.
- Do not describe pressure and volume as directly proportional, because one rises as the other falls.
- Do not draw a straight line for a pressure against volume graph, because that graph is a curve.
Self review
- Explain why reducing the volume of a fixed mass of gas raises its pressure at constant temperature.
- State what happens to the force of each collision when only the volume changes.
- State the equation linking pressure and volume for a fixed mass of gas at constant temperature.
- Calculate the new pressure when 400 cm3400\ \text{cm}^{3}400 cm3 of gas at 1.0×105 Pa1.0\times10^{5}\ \text{Pa}1.0×105 Pa is compressed to 160 cm3160\ \text{cm}^{3}160 cm3.
- Name the two conditions that must hold before the equation can be used.
- Describe the shape of a graph of pressure against volume and of pressure against one over volume.