First Law and Thermodynamic Processes

Overview

This page focuses on how gases exchange energy through heating and mechanical work. It develops the first law of thermodynamics and applies it to standard gas processes commonly tested at Singapore H2 Physics level.

You should be able to:

• distinguish heat, work and internal energy
• use correct sign conventions
• solve first-law questions
• analyse isochoric, isobaric, isothermal and adiabatic changes
• connect equations to particle behaviour

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

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:

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

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”.

4. Work Done by Gas

General Formula

Work done on gas:

$$W=-\int PdV$$

Meaning of Sign

Compression

Volume decreases:

$$dV<0$$

So:

$$W>0$$

Surroundings transfer energy mechanically into gas.

Expansion

Volume increases:

$$dV>0$$

So:

$$W<0$$

Gas transfers energy to surroundings.

5. Constant Pressure Work

If pressure is constant:

$$W=-P\Delta V$$

Where:

$$\Delta V=V_f-V_i$$

Cases

Expansion

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

Compression

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

6. 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

Any heat supplied is converted into work done by gas.

Any work done on gas leaves as heat.

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

Process	Constant Quantity	Work Done on Gas	Internal Energy
Isochoric	(V)	(0)	(\Delta U=Q)
Isobaric	(P)	(-P\Delta V)	depends
Isothermal (ideal gas)	(T)	non-zero	(\Delta U=0)
Adiabatic	(Q=0)	non-zero	(\Delta U=W)

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

$$\Delta U=Q+W$$

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.