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How are the pressure, volume and temperature of a gas related, and how does the kinetic model explain this?

Gas laws and the kinetic model: the relationships between the pressure, volume and temperature of a fixed mass of gas, the kelvin temperature scale, and the kinetic model explanation of gas pressure.

An SQA National 5 Physics answer on the gas laws and the kinetic model, covering the three relationships linking pressure, volume and temperature of a fixed mass of gas, the kelvin temperature scale and absolute zero, and how the kinetic model of moving particles explains gas pressure and each gas law.

Generated by Claude Opus 4.811 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this key area is asking
  2. The kinetic model and gas pressure
  3. The kelvin temperature scale
  4. The three gas laws
  5. Reading gas-law graphs
  6. Try this

What this key area is asking

The SQA wants you to use the three gas-law relationships for a fixed mass of gas, work in the kelvin temperature scale, and use the kinetic model of moving particles to explain gas pressure and each gas law.

The kinetic model and gas pressure

This single idea explains all three gas laws. Anything that makes the particles hit the walls more often or harder (squeezing them into a smaller space, or heating them so they move faster) raises the pressure.

The kelvin temperature scale

For example, 27C27\,{}^{\circ}\text{C} is 27+273=300 K27 + 273 = 300 \text{ K}. Using degrees Celsius in a temperature gas law gives a wrong answer, so always convert first.

The three gas laws

Each law applies to a fixed mass of gas with one quantity held constant. The kinetic model explains each one:

  • Constant temperature: squeezing the gas into a smaller volume means the same particles hit a smaller area more often, so the pressure rises.
  • Constant volume: heating the gas makes the particles move faster, so they hit the walls harder and more often, raising the pressure.
  • Constant pressure: heating the gas makes the particles move faster, so the gas must expand to keep the collision rate, and the pressure, the same.

Reading gas-law graphs

The SQA may show a graph: pressure against volume curves downward (a hotter run sits further out), while pressure against kelvin temperature at constant volume is a straight line through the origin. The fact that the straight line passes through the origin (zero pressure at 0 K0 \text{ K}) is the experimental evidence for absolute zero.

Try this

Q1. Convert 50C50\,{}^{\circ}\text{C} to kelvin. [1 mark]

  • Cue. 50+273=323 K50 + 273 = 323 \text{ K}.

Q2. A gas at 200 kPa200 \text{ kPa} and 0.30 m30.30 \text{ m}^3 is compressed at constant temperature to 0.10 m30.10 \text{ m}^3. Find the new pressure. [2 marks]

  • Cue. p2=200×0.300.10=600 kPap_2 = \frac{200 \times 0.30}{0.10} = 600 \text{ kPa}.

Q3. Use the kinetic model to explain why heating a gas at constant volume raises its pressure. [2 marks]

  • Cue. The particles move faster, so they hit the walls harder and more often.

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SQA N5 style4 marksA fixed mass of gas has a volume of 0.50 cubic metres at a pressure of 100 kPa. The gas is compressed at constant temperature to a volume of 0.20 cubic metres. Calculate the new pressure.
Show worked answer →

At constant temperature, use the pressure-volume relationship (Boyle's law).

Relationship: p1V1=p2V2p_1 V_1 = p_2 V_2.

Substitution: 100×0.50=p2×0.20100 \times 0.50 = p_2 \times 0.20, so 50=0.20p250 = 0.20\,p_2.

p2=500.20=250 kPap_2 = \dfrac{50}{0.20} = 250 \text{ kPa}.

Markers reward using p1V1=p2V2p_1 V_1 = p_2 V_2, correct substitution, and the new pressure. The pressure rises because the same number of particles hit a smaller area more often.

SQA N5 style4 marksA fixed mass of gas at constant volume has a pressure of 120 kPa at 300 K. Calculate its pressure when the temperature is raised to 400 K.
Show worked answer →

At constant volume, use the pressure-temperature relationship, with temperatures in kelvin.

Relationship: p1T1=p2T2\dfrac{p_1}{T_1} = \dfrac{p_2}{T_2}.

Substitution: 120300=p2400\dfrac{120}{300} = \dfrac{p_2}{400}, so p2=120×400300=160 kPap_2 = \dfrac{120 \times 400}{300} = 160 \text{ kPa}.

Markers reward using the pressure-temperature relationship with kelvin temperatures, correct substitution, and the new pressure. The pressure rises because hotter particles move faster and hit the walls harder and more often.

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