IB Physics HL B.4.3 Internal energy change Question Bank

Practise IB Physics HL B.4.3 by applying internal energy change concepts to exam-style physics questions.

Exam points

identify the relevant physical quantity, model or diagram
apply the correct equation and units to the evidence
justify the result using conservation laws, forces or energy relationships

[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 B.4.3 (HL)—Internal energy change 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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B.4.3 (HL)—Internal energy change question 1