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14.3.5 Volume and pressure at constant temperature

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.

  1. Consider a fixed mass of gas held at constant temperature, so the number of particles and their average speed both stay the same.
  2. Reducing the volume brings the walls closer together, so each particle travels a shorter distance between one wall collision and the next.
  3. The particles therefore reach the walls more often, so the number of collisions each second on every square metre of wall increases.
  4. The force of each individual collision is unchanged, because the temperature and therefore the average speed of the particles has not changed.
  5. More collisions of the same strength on each square metre give a greater total force on each square metre, so the pressure increases.
  6. Increasing the volume reverses every step, so the particles hit the walls less often and the pressure decreases.
  7. Squeezing a sealed balloon becomes harder as you go, because the pressure inside climbs as the volume of the trapped gas falls.
  8. 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

  1. For a fixed mass of gas at constant temperature the relationship is p1V1=p2V2p_{1}V_{1}=p_{2}V_{2}p1​V1​=p2​V2​.
  2. 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.
  3. 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.
  4. The equation says that the product of pressure and volume stays constant, so pressure and volume are inversely proportional.
  5. Halving the volume therefore doubles the pressure, and doubling the volume halves it.
  6. Rearranging for the new pressure gives p2=p1V1V2p_{2}=\dfrac{p_{1}V_{1}}{V_{2}}p2​=V2​p1​V1​​.
  7. Rearranging for the new volume gives V2=p1V1p2V_{2}=\dfrac{p_{1}V_{1}}{p_{2}}V2​=p2​p1​V1​​.
  8. 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

  1. A graph of pressure against volume is a curve that falls steeply at small volumes and flattens at large ones, never touching either axis.
  2. That curved shape is the signature of an inverse relationship rather than a proportional one.
  3. 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.
  4. The gradient of that straight line equals the constant product pVpVpV for the sample of gas being tested.
  5. 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}p1​V1​=p2​V2​ 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​=V2​p1​V1​​=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​=p2​p1​V1​​=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}p1​V1​=p2​V2​ 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.
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14.3.5 Volume and pressure at constant temperature Revision Guide

  1. GCSE
  2. /Physics
  3. /14.3.5 Volume and pressure at constant temperature

Revision notes for Edexcel GCSE Physics 14.3.5 Volume and pressure at constant temperature: explanations and worked examples.

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