Temperature, gases and electric current

6 visual steps · 4 worked examples · 6 exercises with solutions

First, let’s see the idea.

In a rigid, sealed container, does doubling the temperature in degrees Celsius double the pressure?

Move forward one step at a time. Change the values and observe what happens.

Heating a sealed gasV = 10 LAbsolute pressure; ideal gas; illustrative particles and arrows.
Drawing description

Keep the container closed. During each heating process the container neither loses gas nor changes volume. The amount and volume controls choose a different experiment; to compare temperatures, keep those two controls fixed. Observe the closed outline and the volume shown in litres. Gas temperature: 20 °C. Fixed container volume: 10 L. Amount of gas: 0,5 mol.

Step 1 of 6

Keep the container sealed

Look at the closed outline and the volume given in litres.

During each heating process, the container neither loses gas nor changes volume. The amount and volume controls let you choose a different experiment; keep these two controls fixed when comparing temperatures.

Gas temperature
20 °C
Rigid volume
10 L
Amount of gas
0,5 mol

What if you change this?

0 °C400 °C
5 L30 L
0,1 mol1 mol

Before moving on

A gas goes from 300 K to 600 K in a rigid, sealed container. How does its absolute pressure change?

Compare with the explanation

It doubles: p₂/p₁ = T₂/T₁ = 600/300 = 2. This ratio requires absolute temperature and constant amount and volume.

You decide when to mark this lesson as studied.

Heating a sealed gas

Slide the drawing to see the details.

Explore the full explanationThe foundations, principles, worked examples and all the exercises, at your own pace.

What you will understand

  • Relate temperature, thermal equilibrium, internal energy and heat.
  • Apply heating, expansion and the ideal gas law with appropriate units.
  • Distinguish absolute pressure from gauge pressure and use kelvin in ratios.
  • Interpret conventional current, voltage and resistance.
  • Calculate power and energy, identifying the limits of the ohmic model.

Before you start

You do not need to know everything by heart. The explanation revisits the necessary ideas and shows how they connect. If a foundation is still unclear, you can return to these lessons:

Temperature and thermal equilibrium

A body in thermal contact with another can transfer energy until both reach the same temperature. Temperature is a state variable that characterises thermal equilibrium and determines the direction of spontaneous transfer due to a temperature difference. Equal temperature does not mean equal internal energy: different masses and materials can have different internal energies.

The zeroth law of thermodynamics states that if two systems are separately in thermal equilibrium with a third, they are also in thermal equilibrium with each other. This underpins using a thermometer as an intermediary. A reading requires time for the sensor to approach the local temperature; it must not be confused with the instantaneous temperature of any region of the body.

The absolute scale uses kelvin. T(K) = θ(°C) + 273,15. Differences in K and °C have the same numerical value, but temperature ratios require kelvin. Doubling from 20 to 40 °C corresponds to going from 293,15 to 313,15 K, which is very far from doubling T. Write K without a degree symbol.

The microscopic model and energy

Matter consists of particles in motion and interaction. Internal energy U includes microscopic kinetic and potential energies relevant to the state. The organised motion of the body as a whole is described separately by macroscopic kinetic energy. A moving stone does not necessarily have a higher temperature than an identical stationary stone.

In a monatomic ideal gas, absolute temperature is linked to the particles’ average translational kinetic energy. This idea helps explain pressure and temperature, but does not justify saying “temperature is energy”: the units and meanings differ. Solids, liquids and gases with complex molecules also have other modes and interactions.

Heat Q is energy transferred because of a temperature difference. Work is another form of transfer, such as compressing a gas or heating a wire electrically. The same increase in internal energy can be achieved through different processes. Therefore Q is not a property stored in the system, whereas U is a function of its state.

Specific heat capacity, phase change and expansion

If there is no change of state and c is approximately constant, Q = mcΔT. m is measured in kg; c in J/(kg·K); Q in J. For a cooling body, ΔT is negative and the Q received is also negative. In a thermally isolated assembly, the energies released and received sum to zero, including containers that cannot be neglected.

For a change of state at constant pressure, use Q = mL, with L in J/kg. For melting or vaporisation, the system receives energy; for the reverse changes, it releases energy. Stages must be separated when a problem includes heating ice, melting it, and then heating the water. The same expression does not describe every stage of a heating curve.

The linear expansion of many solids can be approximated by ΔL = αL₀ΔT over moderate intervals. α is the coefficient of linear expansion, in K⁻¹. The result is a change in length; the final length is L = L₀ + ΔL. The model does not imply that all substances expand in the same way: water behaves differently over certain intervals.

Pressure, volume and amount of gas

A gas exerts pressure because its particles interact with the walls, transferring momentum. Pressure is normal force per area, in Pa. The pressure used in the equation of state is absolute, measured relative to a vacuum. If a gauge shows the difference from atmospheric pressure, local atmospheric pressure must be added to obtain absolute p.

Volume V is the space occupied by the gas at equilibrium and is measured in m³. The amount of substance n is measured in mol. A mole contains a fixed number of entities; you must specify whether those entities are atoms or molecules. In a closed container without leaks, n stays constant, but that does not require V or p to stay constant.

The ideal gas model assumes particles with negligible volume compared with the container and negligible intermolecular interactions outside collisions. It works best for sufficiently dilute gases far from condensation. It is not a universal description of any gas at any pressure, nor a suitable formula for liquids and solids.

Read the equation before substituting numbers

The ideal gas equation is pV = nRT. Here p is in Pa, V in m³, n in mol, T in K and R ≈ 8,314 J/(mol·K). The product Pa·m³ equals J. If you enter volume in litres or pressure in kPa without conversion, you must adapt R consistently; in this lesson we convert everything to SI.

For the same amount of ideal gas, pV/T is constant between states. At constant volume, p is proportional to T: heating increases pressure. At constant pressure, V is proportional to T. At constant temperature, p is inversely proportional to V: halving V doubles p. Every statement requires you to say what stays constant.

This equation connects equilibrium states and does not by itself determine the heat received or work done during a change. Processes with the same initial and final states can transfer energy in different ways. The first law and p(V) diagrams, developed in the Year 11 gases unit, complete this account.

Current and conservation of charge

Electric charge q is measured in C. Average current I = Δq/Δt measures the charge crossing a section per unit time and has unit A = C/s. A current of 0,20 A corresponds to 0,20 C passing per second on average. It does not directly tell us how many metres per second the carriers move.

The conventional direction of current is the direction in which positive charges move. In a metallic conductor, conduction electrons move opposite to the conventional direction. Propagation of a change in the field through the circuit and the slow drift of electrons are different phenomena. The electrical response does not depend on an electron leaving the battery and travelling along the entire wire to the lamp.

In a steady state without charge accumulation, the current entering a junction equals the sum of the currents leaving it. Along a series path without branches, current is the same before and after a resistor. The component transforms energy; it does not consume the charge passing through it. An open circuit interrupts sustained current in the simple model.

Energy per charge and resistance

Potential difference U indicates energy transferred per charge: U = E/q and 1 V = 1 J/C. The source transforms chemical or other energy to maintain a potential difference. A resistor can transform electrical energy into internal energy. Current measures how much charge passes per second; voltage measures how much energy corresponds to each coulomb.

A component’s resistance in a given state is R = U/I, in Ω. Ohm’s law adds a condition: for certain materials and approximately constant conditions, U is proportional to I and R remains constant. A U(I) graph for these components is a straight line through the origin with slope R.

If temperature changes significantly, a metal’s resistance can change. A filament lamp does not necessarily have the same R when cold and when lit. A straight line should not be extrapolated into regimes where the component heats up or changes how it operates. High voltage or high current are not properties independent of the source–circuit assembly.

Series, parallel and measurement in the model

For resistors in series, the same current crosses each component. The potential differences add: total U = I(R₁ + R₂), so equivalent R = R₁ + R₂. The equivalent resistance is the one that would have the same relationship between voltage and current at the same terminals.

In parallel, the branches connect the same two nodes and have the same voltage. The currents add: total I = U/R₁ + U/R₂. Therefore 1/equivalent R = 1/R₁ + 1/R₂. The equivalent resistance is smaller than the smallest branch resistance because there are more paths for current in the model.

An ideal ammeter has negligible resistance and is represented in series with the branch being measured. An ideal voltmeter has very large resistance and is represented between the two points whose potential difference is to be measured. These are descriptions of teaching diagrams, not instructions for work on domestic installations. Study should use appropriate low-voltage educational materials.

Electrical power connects circuits to thermal physics

If each coulomb transfers U joules and I coulombs pass per second, power is P = UI, in W. For an ohmic resistor we can substitute U = RI and obtain P = RI² or P = U²/R. All three forms describe the same case, but comparisons between resistors depend on what the source keeps constant.

With fixed current, increasing R increases RI²; with fixed voltage, increasing R decreases U²/R. There is no contradiction: these are different experiments. For constant power over Δt, E = PΔt. The kilowatt-hour is a unit of energy: 1 kWh = 1000 W × 3600 s = 3,6 × 10⁶ J.

If a resistor heats water, the electrical energy received can be compared with mcΔT. The difference corresponds to heating other parts and transfers to the environment. An unlimited temperature rise must never be predicted using E = PΔt alone: in real systems, losses grow and properties may change, making the initial model unsuitable.

Worked examples, without skipping steps

Follow the reasoning first. Then cover the solution and try to reconstruct it in your own words.

Example 1

Ideal thermal mixing

Mix 0,20 kg of water at 60 °C and 0,30 kg at 20 °C. Neglect the container and surroundings. Determine the final temperature.
  1. The energy released by the hot water equals that received by the cold water: 0,20c(60 − Tf) = 0,30c(Tf − 20).
  2. The same c cancels. Expand: 12 − 0,20Tf = 0,30Tf − 6.
  3. Thus 18 = 0,50Tf and Tf = 36 °C.
Result
Final temperature 36 °C.
Check the result
The result lies between 20 and 60 °C and closer to 20 °C because the cold mass is larger. It is not the simple average of 40 °C.

Example 2

Gas in a rigid container

A closed ideal gas is at 27 °C and an absolute pressure of 120 kPa. It is heated to 57 °C in a rigid container. Calculate the final pressure.
  1. Convert the temperatures: T₁ = 300,15 K and T₂ = 330,15 K.
  2. With n and V constant, p₂/p₁ = T₂/T₁.
  3. p₂ = 120000 × 330,15/300,15 ≈ 131994 Pa.
Result
p₂ ≈ 132 kPa absolute.
Check the result
The increase in absolute temperature is approximately 10%, as is the pressure increase. Using 57/27 would give an incorrect result.

Example 3

Circuit and heating

An ideal 12 Ω resistor receives a constant 6,0 V for 120 s. All the energy heats 0,10 kg of water with c = 4200 J/(kg·K). Determine the temperature change.
  1. I = U/R = 6/12 = 0,50 A. Power is P = UI = 3,0 W.
  2. The energy received is E = 3 × 120 = 360 J.
  3. Without losses, mcΔT = 360. Thus ΔT = 360/(0,10 × 4200) ≈ 0,857 K.
Result
ΔT ≈ 0,86 K, equivalent to a rise of 0,86 °C.
Check the result
The small power produces a small rise in two minutes. The calculation assumes constant resistance and neglects the container.

Example 4

Two resistances in parallel

Ideal 30 Ω and 60 Ω resistors are in parallel across 6,0 V. Determine the currents and equivalent resistance.
  1. Each branch receives 6,0 V. I₁ = 6/30 = 0,20 A and I₂ = 6/60 = 0,10 A.
  2. The total current is 0,30 A.
  3. Equivalent R = U/total I = 6/0,30 = 20 Ω.
Result
Currents 0,20 A and 0,10 A; total 0,30 A; equivalent 20 Ω.
Check the result
1/20 = 1/30 + 1/60. The equivalent is smaller than 30 Ω, as expected for parallel paths.

Now try it yourself

There is no timer. Write what you know, choose a relationship and justify each step. Hints and solutions are available whenever you need them.

Exercise 1

A 2,0 m bar has α = 12 × 10⁻⁶ K⁻¹ and is heated by 50 K. Calculate the expansion.
A hint
The formula gives the change, not the final length.
See the worked solution
ΔL = αL₀ΔT = 12 × 10⁻⁶ × 2 × 50 = 0,0012 m = 1,2 mm. The final length is 2,0012 m. The small ratio ΔL/L₀ = 0,0006 confirms the small-expansion approximation.

Exercise 2

A gauge reads 40 kPa above an atmospheric pressure of 100 kPa. Which pressure goes into pV = nRT?
A hint
The law uses a vacuum reference.
See the worked solution
Absolute p = gauge p + atmospheric p = 140 kPa = 140000 Pa. The 40 kPa is only the difference from the surroundings; using it would underestimate the pressure of the state.

Exercise 3

A closed gas expands isothermally from 0,010 to 0,025 m³. Its initial absolute pressure is 200 kPa. Calculate the final pressure.
A hint
With T constant, the product pV stays constant.
See the worked solution
p₂ = p₁V₁/V₂ = 200000 × 0,010/0,025 = 80000 Pa = 80 kPa absolute. Volume increases by a factor of 2,5 and pressure falls to 1/2,5 of its initial value.

Exercise 4

15 C passes in 10 s through a component with a voltage of 4 V. Calculate I and the energy transferred.
A hint
Use charge per time and energy per charge separately.
See the worked solution
I = 15/10 = 1,5 A. E = Uq = 4 × 15 = 60 J. Alternatively, P = UI = 6 W and E = PΔt = 60 J. Both routes verify the same account.

Exercise 5

A constant power of 50 W acts for 2 h. Express the energy in J and kWh.
A hint
Power times time, with compatible units.
See the worked solution
In SI, 2 h = 7200 s and E = 50 × 7200 = 360000 J. In kWh, 0,050 kW × 2 h = 0,10 kWh. Multiplying 0,10 by 3,6 × 10⁶ J/kWh confirms 360000 J.

Exercise 6

Two resistors of 4 Ω and 8 Ω are in series across 6 V. Calculate the current and voltages.
A hint
In series the same current flows through every component.
See the worked solution
Equivalent R = 12 Ω, so I = 6/12 = 0,50 A. U₁ = 0,50 × 4 = 2 V and U₂ = 0,50 × 8 = 4 V. The voltages sum to 6 V, confirming the result.

Mistakes worth avoiding

Celsius can be used in gas-law ratios.
Ratios require absolute temperature in kelvin.
pV = nRT determines the heat of any process.
The equation connects states; heat and work require analysis of the process.
Current is smaller after a series resistor.
In a steady state charge is conserved; energy per charge changes.
U = RI with constant R applies to any component.
It is an ohmic model that depends on conditions.
kWh is power.
It is energy, the product of power and time.

The essentials to take with you

  • Thermal equilibrium underpins temperature measurement.
  • Internal energy is a state property; heat and work are transfers.
  • Use kelvin and absolute pressure in the gas equation.
  • Every proportional relationship requires the constant variables to be stated.
  • Conventional current and electron motion have opposite directions in a metal.
  • The relationships P = UI and E = PΔt connect circuits to energy conservation.

If you still cannot explain one of these ideas, return to the corresponding section and work through an example again. Recognising the difficulty is part of studying.

References for further study

Text and examples prepared for this learning path. These sources help check the principles and continue studying:

The organisation by school year is a study proposal. Pedagogical review by teachers remains a stage of the project.