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

Apply ΔU = 3/2 NkBΔT = 3/2 nRΔT to relate an ideal monatomic system’s internal-energy change to temperature change.
Calculate internal-energy change, amount, particle number or temperature from complete evidence.

[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