Heat, Work and the First Law: Quick Revision

A brief note on Heat, Work and the First Law. Read it before a test. Try each question at the end before you open its answer.

The big ideas

• Thermal equilibrium. Bodies in contact settle at one temperature. Two bodies each in balance with a third are in balance with each other: the zeroth law.
• Internal energy $U$ is the energy of the molecules. For an ideal gas $U = \tfrac{f}{2}nRT$: it depends only on $T$, so it is a state function.
• Heat $Q$ is energy that flows because of a difference of temperature. Work $W$ is energy passed by a force, such as a gas pushing a piston. Both depend on the path.
• The first law: $\Delta U = Q - W$. Heat into the gas and work done by it are positive.
• Work on a graph. $W$ is the area under the path on a pressure-volume graph; around a cycle, the area of the loop.

The facts at a glance

• Steady pressure: $W = p\,\Delta V$
• Steady volume: $W = 0$, $\Delta U = Q$
• Insulated ($Q = 0$): $\Delta U = -W$
• Steady temperature, ideal gas: $\Delta U = 0$, $Q = W = nRT\ln\dfrac{V_2}{V_1}$
• Cycle: $\Delta U = 0$, $Q_\text{net} = W_\text{net}$

A worked example

$0.30\ \text{mol}$ of nitrogen is warmed from $290\ \text{K}$ to $360\ \text{K}$, once under a free piston and once in a rigid tank. Find $W$, $\Delta U$ and $Q$ each time.

1. Internal energy, both ways. $\Delta U = 0.30\ \text{mol} \times 20.8\ \text{J mol}^{-1}\text{K}^{-1} \times 70\ \text{K} = 436.3\ \text{J}$.
2. Piston. $W = nR\,\Delta T = 0.30\ \text{mol} \times 8.31\ \text{J mol}^{-1}\text{K}^{-1} \times 70\ \text{K} = 174.5\ \text{J}$, so $Q = 610.8\ \text{J}$.
3. Tank. $W = 0$, so $Q = \Delta U = 436.3\ \text{J}$.

Practice with new numbers

A program to try

The program works out the example both ways.

Mistakes to avoid

• The sign of $W$. Work done on the gas is negative.
• Heat and temperature. A gas can take in heat with no change of temperature, as in an isothermal expansion.
• Litres. Change to cubic metres, or use kPa times litres for joules.
• Path and state. $\Delta U$ does not depend on the path; $Q$ and $W$ do.
• Mixing up the conventions. Some books write $\Delta U = Q + W$ with $W$ done on the gas. Stick to one.

Check yourself

1. A gas takes in $360\ \text{J}$ and does $145\ \text{J}$ of work. What is $\Delta U$?

2. Why is heat not a state function?

3. A gas at a steady $1.5 \times 10^{5}\ \text{Pa}$ expands from $2.0$ to $2.6$ litres. How much work does it do?

4. What is the internal energy of $0.40\ \text{mol}$ of argon at $350\ \text{K}$?

5. A gas goes round a cycle and does $85\ \text{J}$ of net work. How much net heat does it take in?