IB Physics HL B.4.10 Adiabatic ideal gas Question Bank

Practise IB Physics HL B.4.10 by applying adiabatic ideal gas 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.4.10 (HL)—Adiabatic ideal gas 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}

Show that the volume at C is $3.33 \times 10^{-2} \mathrm{~m}^{3}$ .

Alternative 1:
Use the adiabatic formula $p_A V_A^{5/3}=p_C V_C^{5/3}$ , so $V_C=\left(\frac{p_A}{p_C}\right)^{3/5}V_A$ .
$V_C=\left(\frac{5.00\times10^5}{4.60\times10^3}\right)^{3/5}\times2.00\times10^{-3}=3.333\times10^{-2}\ \mathrm{m^3}$ .

Alternative 2:
$V_C=V_B$ and $p_AV_A=p_BV_B$ .
$V_C=\frac{5\times10^5\times2\times10^{-3}}{3\times10^4}$ .

Alternative 3:
$V_C=V_B$ and $n=0.2\ \mathrm{mol}$ .
$V_C=\frac{0.2\times8.31\times602}{3\times10^4}$ .

For MP2, working or answer to at least 4 significant figures must be seen.

All question bank results loaded

B.4.10 (HL)—Adiabatic ideal gas question 1