Thermodynamics • Topic 1 of 3

Thermal Equilibrium & the First Law

Thermodynamics is the branch of physics that deals with heat, work and the conversion of one into the other. Unlike kinetic theory, which looks at individual molecules, thermodynamics describes a system using a few large-scale measurable quantities. A thermodynamic system is the definite quantity of matter (often a gas) chosen for study; everything outside it is the surroundings, and the two are separated by a real or imaginary boundary.

The state of a system is fixed by its state variables — pressure $P$, volume $V$, temperature $T$ and amount of substance $n$. These describe the condition of the gas, not how it got there. For an ideal gas they are linked by the equation of state $PV=nRT$.

  • Thermal equilibrium: two systems in contact are in thermal equilibrium when there is no net flow of heat between them, i.e. they are at the same temperature.
  • Zeroth law of thermodynamics: if system A is in thermal equilibrium with system C, and B is also in equilibrium with C, then A and B are in equilibrium with each other. This law defines temperature as the property that decides the direction of heat flow and lets a thermometer (C) compare two bodies.

Internal energy ($U$) is the total energy stored inside a system — the sum of the kinetic and potential energies of all its molecules. For an ideal gas it depends only on temperature, so $U$ rises when $T$ rises. Internal energy is a state function: its change $\Delta U$ depends only on the initial and final states, not on the path taken.

Heat ($\Delta Q$) is energy transferred because of a temperature difference, and work ($\Delta W$) is energy transferred by a force moving through a distance. Both are path-dependent (not state functions). When a gas expands by a small volume $dV$ against pressure $P$, the work done by the gas is $dW=P\,dV$, so over a finite change:

  • $W=\int_{V_1}^{V_2} P\,dV$ — equal to the area under the $P$–$V$ curve.
  • Work done by the gas (expansion) is positive; work done on the gas (compression) is negative.

The first law of thermodynamics is simply the law of conservation of energy applied to heat: the heat supplied to a system goes partly to raise its internal energy and partly to do external work:

  • $\Delta Q=\Delta U+\Delta W$.
  • Sign convention: $\Delta Q$ is positive when heat is added to the system; $\Delta W$ is positive when work is done by the system; $\Delta U$ is positive when the temperature rises.

The first law tells us energy cannot be created or destroyed, but it does not say in which direction a process will go — that is the job of the second law.

First law: heat splits into internal energy and work done by the gasFirst Law: dQ = dU + dWGas (U)dWdQ inPVW = area under P-V
1
Worked Example
A gas absorbs 200 J of heat and does 80 J of work on its surroundings. Find the change in its internal energy.
Solution
  1. Step 1: Use the first law $\Delta Q=\Delta U+\Delta W$, so $\Delta U=\Delta Q-\Delta W$.
  2. Step 2: Substitute $\Delta Q=+200\ \text{J}$ (heat added) and $\Delta W=+80\ \text{J}$ (work by gas): $\Delta U=200-80$.
  3. Step 3: Compute: $\Delta U=120\ \text{J}$.

Answer: $\Delta U=120\ \text{J}$ (internal energy increases).

2
Worked Example
150 J of work is done on a gas while it releases 100 J of heat. Find the change in internal energy.
Solution
  1. Step 1: Apply $\Delta U=\Delta Q-\Delta W$ with correct signs.
  2. Step 2: Heat is released, so $\Delta Q=-100\ \text{J}$; work is done on the gas, so $\Delta W=-150\ \text{J}$.
  3. Step 3: Compute: $\Delta U=(-100)-(-150)=+50\ \text{J}$.

Answer: $\Delta U=+50\ \text{J}$ (internal energy increases).

3
Worked Example
A gas expands at a constant pressure of $2\times10^{5}\ \text{Pa}$ from a volume of $1\times10^{-3}\ \text{m}^3$ to $3\times10^{-3}\ \text{m}^3$. Find the work done by the gas.
Solution
  1. Step 1: At constant pressure $W=P(V_2-V_1)$.
  2. Step 2: Substitute: $W=2\times10^{5}\times(3\times10^{-3}-1\times10^{-3})=2\times10^{5}\times2\times10^{-3}$.
  3. Step 3: Compute: $W=400\ \text{J}$.

Answer: $W=400\ \text{J}$ (done by the gas).

4
Worked Example
In a cyclic process a gas returns to its initial state. If it absorbs 500 J of heat during the cycle, how much net work does it do?
Solution
  1. Step 1: In a cyclic process the system returns to its starting state, so $\Delta U=0$ (internal energy is a state function).
  2. Step 2: From the first law $\Delta Q=\Delta U+\Delta W=0+\Delta W$.
  3. Step 3: Therefore $\Delta W=\Delta Q=500\ \text{J}$.

Answer: Net work done $=500\ \text{J}$.

5
Worked Example
When 1 g of water at 100 °C turns into steam it absorbs 2256 J of heat. The work done in expansion is 169 J. Find the increase in internal energy.
Solution
  1. Step 1: Use $\Delta U=\Delta Q-\Delta W$.
  2. Step 2: Substitute $\Delta Q=+2256\ \text{J}$ and $\Delta W=+169\ \text{J}$: $\Delta U=2256-169$.
  3. Step 3: Compute: $\Delta U=2087\ \text{J}$.

Answer: $\Delta U=2087\ \text{J}$.

6
Worked Example
A system is taken from state A to state B by two paths. Along path 1, $\Delta Q=80\ \text{J}$ and $\Delta W=30\ \text{J}$. Along path 2 the work done is $10\ \text{J}$. Find the heat absorbed along path 2.
Solution
  1. Step 1: $\Delta U$ depends only on the states A and B, so it is the same for both paths.
  2. Step 2: Path 1: $\Delta U=\Delta Q-\Delta W=80-30=50\ \text{J}$.
  3. Step 3: Path 2: $\Delta Q=\Delta U+\Delta W=50+10=60\ \text{J}$.

Answer: Heat absorbed along path 2 $=60\ \text{J}$.

Key Points

  • A thermodynamic system is described by state variables $P$, $V$, $T$, $n$ linked for an ideal gas by $PV=nRT$.
  • Zeroth law: bodies in thermal equilibrium with a third body are in equilibrium with each other — this defines temperature.
  • Internal energy $U$ is a state function; for an ideal gas it depends only on temperature.
  • Heat $\Delta Q$ and work $\Delta W$ are path-dependent; work done by a gas is $W=\int P\,dV$ = area under the $P$–$V$ curve.
  • First law: $\Delta Q=\Delta U+\Delta W$ — conservation of energy; $\Delta Q$ positive if heat added, $\Delta W$ positive if work done by the gas.
Tap an option to check your answer0 / 4
Q1.The first law of thermodynamics is a statement of the conservation of:
Explanation: $\Delta Q=\Delta U+\Delta W$ is the law of conservation of energy applied to heat and work.
Q2.Which of the following is a state function?
Explanation: Internal energy depends only on the state of the system; heat and work depend on the path.
Q3.The zeroth law of thermodynamics introduces the concept of:
Explanation: The zeroth law establishes thermal equilibrium and thereby defines temperature.
Q4.The work done by a gas during expansion equals:
Explanation: $W=\int P\,dV$, which is geometrically the area under the $P$–$V$ curve.