First Law and Thermodynamic Processes

Core and enrichment boundary

The 2026 core is internal energy, heat and work as transfer quantities, the work-on sign convention, and $\Delta U=Q+W_{\mathrm{on}}$. Named isochoric, isobaric, isothermal and adiabatic processes and detailed p–V work are enrichment applications.

Overview

This page first develops the 2026 core: how a system’s internal energy changes through heating and mechanical work. It then retains standard gas-process applications from the older anchor as enrichment.

You should be able to:

• distinguish heat, work and internal energy
• use correct sign conventions
• solve first-law questions
• apply the core sign convention consistently
• optionally extend the reasoning to named gas processes

Related hub:

Thermal Physics B

Definition

The first law of thermodynamics is the energy-conservation statement for thermal systems: internal energy changes because of heat transfer and work.

Why It Matters

This topic is where many sign errors happen. Students often know the formula but lose marks by not stating whether work is done on the gas or by the gas, or by confusing isothermal with adiabatic changes.

Key Representations

Core Ideas

• The first law is conservation of energy for a thermal system.
• In this topic, $W$ is usually work done on the gas, so $\Delta U=Q+W$.
• Heat and work are transfer quantities; internal energy is a state property.
• In enrichment applications, named gas processes impose different energy constraints.
• Correct sign convention is as important as correct substitution.

Exam Relevance

For the syllabus core, use this branch for internal-energy definitions, first-law calculations and sign-convention explanations. Treat named-process identification and p–V work as enrichment after the core is secure.

1. Internal Energy Revisited

Internal energy $U$ is the total microscopic energy stored in a system.

It consists of:

• random kinetic energy of particles
• intermolecular potential energy

For an ideal gas:

• intermolecular forces are negligible
• potential energy is approximately zero

Hence:

$$U=\text{total kinetic energy}$$

Enrichment

The following formula applies to a monatomic ideal gas and is beyond the explicit 2026 first-law outcome.

For a monatomic ideal gas:

$$U=\frac{3}{2}nRT$$

So internal energy depends only on temperature.

2. Heat, Work and Internal Energy

Heat $Q$

Heat is energy transferred because of a temperature difference.

• energy into system: $Q>0$
• energy out of system: $Q<0$

Work $W$

In this chapter, $W$ means work done on the gas.

• compression by surroundings: $W>0$
• expansion by gas: $W<0$

Internal Energy Change

$$\Delta U=U_f-U_i$$

• increase in internal energy: $\Delta U>0$
• decrease in internal energy: $\Delta U<0$

3. First Law of Thermodynamics

Statement

The increase in internal energy of a system equals heat supplied to the system plus work done on the system.

$$\Delta U=Q+W$$

This is an application of conservation of energy.

Rearranged Forms

$$Q=\Delta U-W$$ $$W=\Delta U-Q$$

Use whichever form is most convenient.

Figure: The first law links internal energy change to heat transfer and work, using the sign convention $\Delta U=Q+W$ with work done on the gas taken as positive.

Use this sign convention consistently before substituting numbers; most first-law mistakes come from mixing “work done on gas” and “work done by gas”.

Sign-Convention Bridge

Many questions describe work using ordinary language rather than the symbol $W$. Translate the wording before using the formula.

Wording in question	Meaning	Sign of $W$ in $\Delta U=Q+W$
heat supplied to the gas	energy enters by heating	$Q>0$
heat lost by the gas	energy leaves by heating	$Q<0$
work done on the gas	surroundings compress or push the gas	$W>0$
gas does work on surroundings	gas expands and transfers energy out mechanically	$W<0$

If a question gives work done by the gas, convert it using:

$$W_{\text{on}}=-W_{\text{by}}$$

Then apply:

$$\Delta U=Q+W_{\text{on}}$$

For an ideal gas, $\Delta U$ depends only on temperature. If temperature is unchanged, $\Delta U=0$ even when heat and work are both non-zero.

4. Enrichment: pressure–volume boundary work

Scope

The 2026 core requires $\Delta U=Q+W_{\mathrm{on}}$. Integral p–V work and the named process catalogue below are useful extensions rather than explicit outcomes.

General Formula

For a quasistatic process, work done on the gas is

$$W_{\mathrm{on}}=-\int p_{\mathrm{gas}}dV.$$

More generally, boundary work is set by the external pressure. For constant external pressure, $W_{\mathrm{on}}=-p_{\mathrm{ext}}\Delta V$.

Meaning of Sign

Compression

Volume decreases:

$$dV<0$$

So, using $W\equiv W_{\mathrm{on}}$:

$$W_{\mathrm{on}}>0$$

Surroundings transfer energy mechanically into gas.

Expansion

Volume increases:

$$dV>0$$

So:

$$W_{\mathrm{on}}<0$$

Gas transfers energy to surroundings.

5. Constant Pressure Work

If the relevant external pressure is constant:

$$W_{\mathrm{on}}=-p_{\mathrm{ext}}\Delta V$$

Where:

$$\Delta V=V_f-V_i$$

Cases

Expansion

$$\Delta V>0\Rightarrow W<0$$

Compression

$$\Delta V<0\Rightarrow W>0$$

6. Enrichment: standard thermodynamic processes

Figure: The four standard ideal-gas processes compared on a p-V diagram: isochoric, isobaric, isothermal, and adiabatic.

This comparison is most useful as a quick process-identification map: first state what stays constant, then simplify the first law accordingly.

6.1 Isochoric Process (Constant Volume)

Condition

$$V=\text{constant}$$

Therefore:

$$\Delta V=0$$

No boundary movement, so:

$$W=0$$

Hence first law becomes:

$$\Delta U=Q$$

Interpretation

All heat supplied changes internal energy.

For Ideal Gas

• temperature rises if heat is supplied
• pressure rises because molecules move faster

6.2 Isobaric Process (Constant Pressure)

Condition

$$P=\text{constant}$$

Work done:

$$W=-P\Delta V$$

First law:

$$\Delta U=Q-P\Delta V$$

Interpretation

Heat supplied may be used for:

• increasing internal energy
• doing expansion work

6.3 Isothermal Process (Constant Temperature)

Condition

$$T=\text{constant}$$

For ideal gas:

$$U\propto T$$

Hence:

$$\Delta U=0$$

Therefore:

$$Q=-W$$

Interpretation

For an isothermal ideal-gas process, net heat supplied equals work done by the gas in magnitude because $\Delta U=0$. Conversely, net work done on the gas is balanced by energy transferred out by heating.

Shape on p-V Graph

Rectangular hyperbola:

$$PV=\text{constant}$$

6.4 Adiabatic Process

Condition

No heat exchange:

$$Q=0$$

Hence:

$$\Delta U=W$$

Interpretation

Compression

• $W>0$
• internal energy rises
• temperature rises

Expansion

• $W<0$
• internal energy falls
• temperature falls

How Achieved

• thermal insulation
• very rapid process (little time for heat transfer)

7. Process Summary Table

Use this table as an exam shortcut after identifying the process type.

Process	Condition	Work done on gas	First-law form	Exam interpretation
Isochoric	$V=\text{constant}$	$W=0$	$\Delta U=Q$	all heat supplied changes internal energy
Isobaric	$P=\text{constant}$	$W=-P\Delta V$	$\Delta U=Q-P\Delta V$	heat supplied may both raise temperature and do expansion work
Isothermal ideal gas	$T=\text{constant}$	non-zero if volume changes	$\Delta U=0$, so $Q=-W$	heat transfer exactly balances mechanical work
Adiabatic	$Q=0$	non-zero if volume changes	$\Delta U=W$	work changes internal energy and temperature

Direction and Temperature Checks

• Isobaric expansion: $\Delta V>0$, so $W<0$; if heat is supplied, some energy leaves as work done by the gas.
• Isothermal expansion: $\Delta U=0$; heat supplied equals work done by the gas in magnitude.
• Adiabatic expansion: $Q=0$ and $W<0$, so $\Delta U<0$; the gas cools.
• Adiabatic compression: $Q=0$ and $W>0$, so $\Delta U>0$; the gas warms.

These sign checks are often faster than calculation and help catch algebra mistakes.

8. Particle Explanations

Heating at Constant Volume

• molecules move faster
• more energetic wall collisions
• pressure increases

Compression

• walls push molecules inward
• work transferred into gas
• temperature often rises

Expansion

• gas pushes piston outward
• gas loses energy as work
• temperature may fall if no heat enters

9. Worked Examples

Example 1: Heat Supplied + Compression

A gas receives $120\text{J}$ heat.
Work done on gas is $40\text{J}$.

Find $\Delta U$.

$$\Delta U=Q+W$$ $$=120+40=160\text{J}$$

Internal energy increases by $160\text{J}$.

Example 2: Gas Expands Doing Work

A gas absorbs $200\text{J}$ heat and does $70\text{J}$ work.

Since gas does work:

$$W=-70$$

Then:

$$\Delta U=200-70=130\text{J}$$

Example 3: Constant Volume Heating

Gas at constant volume absorbs $90\text{J}$.

Since:

$$W=0$$

Then:

$$\Delta U=90\text{J}$$

Example 4: Isothermal Expansion

Ideal gas expands isothermally and does $50\text{J}$ work.

So:

$$W=-50$$

Since:

$$\Delta U=0$$

Then:

$$0=Q-50$$ $$Q=50\text{J}$$

Heat absorbed is $50\text{J}$.

Example 5: Adiabatic Compression

Gas compressed adiabatically with work done on gas $75\text{J}$.

Since:

$$Q=0$$ $$\Delta U=75\text{J}$$

Temperature increases.

10. Strategy for First-Law Questions

Step 1: Identify Sign Convention

Use:

• heat into gas positive
• work on gas positive

Step 2: Determine Process Type

Look for:

• constant volume
• constant pressure
• constant temperature
• insulated or adiabatic

Step 3: Apply Simplification

Examples:

• constant volume → $W=0$
• isothermal ideal gas → $\Delta U=0$
• adiabatic → $Q=0$

Step 4: Substitute Carefully

$$\Delta U=Q+W$$

Step 5: Check the Physical Meaning

After calculating, ask:

• Did an expansion make $W$ negative under the work-on-gas convention?
• Did a compression make $W$ positive?
• Does an isothermal ideal-gas process have $\Delta U=0$?
• Does an adiabatic process have $Q=0$?
• If the gas returns to its starting state, is the net $\Delta U=0$?

The numerical answer should match the energy story. If the sign does not match the process, the likely error is usually the work convention.

Links

• Thermal Physics B
• Kinetic Theory and Ideal Gases
• p-V Diagrams and Cycles
• Thermal Physics B Common Exam Traps
• Work, Energy and Power

Summary

The first law becomes much easier once the process type is clear. The real skill is identifying what is zero, what changes sign, and which quantity is a state property.