IB Physics HL B 4 3 Hl Internal Energy Change Questions

Solve HL questions on internal-energy change by linking temperature, pressure, volume and heat transfer through the first law for ideal gases and thermodynamic paths.

Exam points

Calculate ΔU for an ideal monatomic gas from 3/2 nRΔT or 3/2 Δ(PV), retaining the sign of the temperature or state change.
Relate internal-energy change to average molecular kinetic energy and determine specific heat capacity or energy per molecule from temperature data.
Use process conditions such as constant volume, zero work or adiabatic change to infer heat transfer, internal-energy change and temperature direction.

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Study B.4.3 (HL)—Internal energy change conceptsOpen the corresponding Concept lesson for explanations, examples and exam connections.

[Maximum number: 2]

The p V diagram shows a heat engine cycle consisting of adiabatic, isothermal and isovolumetric parts. The working substance of the engine is an ideal gas.

Figure for Question IB Physics HL B 4 3 Hl Internal Energy Change Questions question 1 — IB Physics HL

The following data are available:

\begin{aligned} & p_{\mathrm{A}}=5.00 \times 10^{5} \mathrm{~Pa} \\ & V_{\mathrm{A}}=2.00 \times 10^{-3} \mathrm{~m}^{3} \\ & T_{\mathrm{A}}=602 \mathrm{~K} \\ & p_{\mathrm{B}}=3.00 \times 10^{4} \mathrm{~Pa} \\ & p_{\mathrm{C}}=4.60 \times 10^{3} \mathrm{~Pa} \end{aligned}

Calculate the thermal energy (heat) taken out of the gas from B to C.

Alternative 1:
$Q=\Delta U=\frac{3}{2}V_C\Delta P$ .
$Q=\frac{3}{2}\times3.33\times10^{-2}\times(3.00\times10^4-4.60\times10^3)=1268.7\approx1270\ \mathrm{J}$ .

Alternative 2:
$Rn=\frac{5\times10^5\times2\times10^{-3}}{602}=1.66$ OR $T_C=\frac{4.6\times10^3\times3.33\times10^{-2}}{1.66}=92.2\ \mathrm{K}$ .
$\Delta U=\frac{3}{2}\times1.661\times(602-92.21)=1270\ \mathrm{J}$ .

Award [2] for BCA. Accept negative values.
Award MP1 if $T_C=92$ is taken from (e).

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IB Physics HL B 4 3 Hl Internal Energy Change Questions question 1