internal energy heat and thermodynamic work

The internal energy $U$ of an ideal gas is the total kinetic energy of its molecules; for a monatomic gas it depends only on temperature: $U=\frac{3}{2}\nu RT$, hence $\Delta U=\frac{3}{2}\nu R\,\Delta T$. Heat $Q$ and work $A$ are not energy stored inside the body but two ways of transferring energy through its boundary, where thermodynamic work generalises the mechanical work of mechanics. We use the sign convention $\Delta U=Q+A_{\text{ext}}=Q-A_{\text{gas}},$ where $Q>0$ when heat flows into the gas and $A_{\text{gas}}>0$ when the gas expands and does work itself. Because $pV=\nu RT$, the work of an isobaric process is $A_{\text{gas}}=p\,\Delta V$ and equals the area under the graph in the $p$–$V$ plane. The isoprocesses give useful special cases: isochoric ($A=0$, $\Delta U=Q$), isothermal ($\Delta U=0$, $Q=A_{\text{gas}}$) and adiabatic ($Q=0$, $\Delta U=-A_{\text{gas}}$).

A monatomic ideal gas contains $\nu = 2$ mol. Its temperature rises by $\Delta T = 100$ K. Find the change in internal energy. ($R = 8.31$ J/(mol·K))

A gas is heated at constant pressure $p = 2\cdot 10^{5}$ Pa; its volume grows from $V_1 = 3\cdot 10^{-3}$ m³ to $V_2 = 8\cdot 10^{-3}$ m³. The heat supplied is $Q = 2.4$ kJ. Find the work done by the gas and the change in its internal energy.

A monatomic ideal gas with $\nu = 0.4$ mol is compressed quickly and adiabatically. The external work done on the gas is $A_{\text{ext}} = 750$ J. Find the change in internal energy and the change in temperature, and explain why $Q=0$.

Self-study with a simulation (safe: no heaters, no sealed vessels under pressure). Open the PhET simulation Gas Properties in ideal-gas mode.

1. Fix the amount of substance $\nu$ and record the initial $p$, $V$, $T$.
2. Isothermal run: keep $T$ constant, change the volume and record the pairs $(p,V)$. Check that $pV\approx\text{const}$ and therefore $\Delta U=0$, that is $Q=A_{\text{gas}}$.
3. Isobaric run: choose a constant-pressure mode, measure $\Delta V$ and compute $A_{\text{gas}}=p\,\Delta V$; compare this work with the heat supplied.
4. Isochoric run: lock the volume, change $T$ and verify that $A_{\text{gas}}=0$ and $Q=\Delta U$.
5. Fill in a table: process, $Q$, $A_{\text{gas}}$, $\Delta U$, and check the equality $\Delta U=Q-A_{\text{gas}}$.

Do not use open flames or sealed containers at home: carry out all observations in the simulation only.

To show that the work of a gas is the area under a graph in $p$–$V$ coordinates, and that $Q$, $A_{\text{gas}}$ and $\Delta U$ are linked by the first-law balance for each process.

$p$–$V$ graph. $p$ on the vertical axis, $V$ on the horizontal axis. Segment A–B: a horizontal line from $(V_1,p_0)$ to $(V_2,p_0)$ — isobaric expansion; the shaded rectangle beneath it has area $p_0\,\Delta V=A_{\text{gas}}$. Segment B–C: a vertical line at $V_2$ from $p_0$ to $p_1$ — isochoric cooling, whose area is zero, so $A_{\text{gas}}=0$.

Point sketch. A $(V_1,p_0)$ → B $(V_2,p_0)$ → C $(V_2,p_1)$; arrows show the direction of the process and the shading marks the work.

Energy bar chart. For one process three bars are drawn: $Q$ (heat supplied), $A_{\text{gas}}$ (work done by the gas) and $\Delta U$, with heights chosen so that $\Delta U=Q-A_{\text{gas}}$, i.e. the $\Delta U$ bar equals the difference of the first two.

You inflate a bicycle tyre with a hand pump. In one stroke you push with a force of about $200\ \text{N}$ and move the piston by $0{,}25\ \text{m}$, doing work $A \approx 50\ \text{J}$. After 30 fast strokes the pump barrel becomes distinctly warm — up to $40\text{–}50\ ^\circ\text{C}$ — although the room air was $22\ ^\circ\text{C}$.

The work done on the gas during fast compression does not have time to leave into the surroundings as heat ($Q \approx 0$), so it all goes into increasing the internal energy: $\Delta U = Q + W \approx W > 0$. For an ideal gas $U = \frac{3}{2}\nu RT$, so the temperature rises together with $U$ — which is exactly what the warm pump shows.

An electric kettle of power $2000\ \text{W}$ heats $1{,}5\ \text{kg}$ of water from $20\ ^\circ\text{C}$ to $100\ ^\circ\text{C}$. The specific heat capacity of water is $c = 4200\ \text{J}/(\text{kg}\cdot\text{K})$, so the water needs $Q = cm\Delta T = 4200 \cdot 1{,}5 \cdot 80 = 5{,}04\cdot 10^5\ \text{J}$. With ideal efficiency this would take $t = Q/P \approx 250\ \text{s}$.

Here the energy arrives as heat, not as work: the volume of the water hardly changes on heating, so $W \approx 0$ and all the heat goes into internal energy, $\Delta U \approx Q = cm\Delta T$. The heater power $P = Q/t$ links the heat to the heating time.

A cyclist together with the bicycle, total mass $80\ \text{kg}$, rides down a hill $100\ \text{m}$ high and brakes to a complete stop. The stored mechanical energy is $mgh = 80 \cdot 9{,}8 \cdot 100 \approx 7{,}8\cdot 10^4\ \text{J}$; this energy goes into the brake blocks and the rim, and after a series of braking events the rim heats up by tens of degrees.

The friction force does negative work on the kinetic energy of the wheel but positive work on the material of the brake blocks, so the mechanical energy does not disappear — it turns into internal energy: $\Delta U = |W_{\text{friction}}| = 7{,}8\cdot 10^4\ \text{J}$. Here the internal energy changes through work alone; $Q$ stays small while the braking is short.