Back to Topic 2.4 — Thermodynamics (HL)
2.4.1Physics HL12 flashcards

First law of thermodynamics

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Card 1 of 122.4.1
2.4.1
Question

Define internal energy U of a gas.

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All 12 Flashcards — First law of thermodynamics

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Card 1definition

Question

Define internal energy U of a gas.

Answer

The **total energy of all the particles**: their random **kinetic energy** + the **potential energy** of the forces between them.

Card 2concept

Question

What does the internal energy of an **ideal gas** depend on?

Answer

**Temperature only** — an ideal gas has no inter-particle PE, so U is fixed by the random KE of the particles.

Card 3formula

Question

State the first law of thermodynamics.

Answer

$Q = \Delta U + W$ — the heat **added** equals the rise in **internal energy** plus the **work done by** the gas.

Card 4formula

Question

Rearrange the first law for ΔU.

Answer

$\Delta U = Q - W$ (heat in **minus** work done by the gas).

Card 5concept

Question

In Q = ΔU + W, what is the sign of Q when heat is **removed**?

Answer

**Negative** — Q is the heat **added** to the gas, so heat leaving makes Q < 0.

Card 6concept

Question

In Q = ΔU + W, what is the sign of W when the gas is **compressed**?

Answer

**Negative** — W is the work done **by** the gas; on compression the surroundings do work on it, so W < 0.

Card 7formula

Question

Work done by a gas at constant pressure?

Answer

$W = P\,\Delta V$ — pressure times the change in volume.

Card 8definition

Question

Units for W = PΔV?

Answer

P in **pascals (Pa)**, ΔV in **cubic metres (m³)**, giving W in **joules (J)**.

Card 9comparison

Question

Internal energy vs heat — what's the difference?

Answer

**Internal energy** is energy a gas **already has** inside; **heat** is energy **flowing** in or out due to a temperature difference.

Card 10concept

Question

For an ideal gas at **constant temperature**, what is ΔU?

Answer

**ΔU = 0** — U depends on temperature alone, so no temperature change means no change in internal energy.

Card 11example

Question

500 J heat added, gas does 200 J work — find ΔU.

Answer

$\Delta U = Q - W = 500 - 200 = 300$ J (the gas warms).

Card 12process

Question

Quick way to handle the signs in the first law?

Answer

Write each sign in **words** first ('heat removed → Q negative', 'gas compressed → W negative'), then plug into $\Delta U = Q - W$.

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