B.4.3 (HL)—Internal energy change

Temperature controls internal energy

For an ideal monatomic gas,

$$\Delta U=\frac32Nk_B\Delta T=\frac32nR\Delta T$$

Only the temperature change and amount of gas are needed.

Use a PV route when useful

With $$PV=nRT$$, a change can sometimes be written as $$\Delta U=\frac32\Delta(PV)$$ for a fixed amount of monatomic ideal gas. Check which variable or path information the question supplies.

Sign

If temperature rises, $$\Delta U>0$$. If temperature falls, $$\Delta U<0$$. Internal energy is a state function, so its change depends only on the initial and final states.

Worked example from the mapped local textbook

An ideal monatomic gas contains $3.4\times10^{24}$ particles and warms from $297\,\mathrm{K}$ to $348\,\mathrm{K}$, so $\Delta T=51\,\mathrm{K}$.

$$\Delta U=\frac32Nk_B\Delta T=\frac32(3.4\times10^{24})(1.38\times10^{-23})(51)=3.6\times10^3\,\mathrm{J}$$

The positive value follows from the temperature rise; the path used to reach the final state does not change this $\Delta U$.

Common trap

Do not add work or thermal transfer into $$\Delta U$$ directly. Use the first law to relate them, but calculate the state-function change from temperature or equivalent state data.