Oxford Revise AQA A Level Physics | Chapter P26 answers
P26: Thermodynamics and engines
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Question |
Answers |
Extra information |
Mark |
AO |
Spec reference |
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01.1 |
\( \begin{array}{lll} \textit{pV} & = & \textit{nRT}\\ \textit{n} & = & \frac{{pV}}{{RT}}\\ & = & \frac{5.0 \times 10^5\ \rm{Pa} \times 2 \times 10^{-4}\ \rm{m}^3} {8.31\ \rm{J}\ \rm{mol}^{-1}\ \rm{K}^{-1} \times \left( 100 + 273 \right)}\\ & = & 0.032\textrm{ mol}\\ \end{array} \) |
Substitution Answer |
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3.11.2.2 |
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01.2 |
\( \begin{array}{lll} \textit{pV} & = & \textit{nRT}\\ \textit{V} & = & \frac{{nRT}}{p}\\ & = & \frac{0.032 \times 8.31 \times 273}{1.0 \times 10^5\ \rm{Pa}}\\ & = & 7.3 \times 10^{-4}\textrm{ m}^{3}\\ \end{array} \) |
Substitution Answer |
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3.11.2.2 |
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01.3 |
A–B: The gas temperature and pressure decrease as the volume stays the same C–D: The gas temperature and pressure increase as the volume stays the same |
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3.11.2.2 |
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01.4 |
B–C: Work done by the gas = \(p\Delta V\) = 1.0 × 105 Pa × (7.3 – 2) × 10−4 m3 = +53 J D–A: Work done on the gas = \(p\Delta V\) = 5.0 × 105 Pa × (2 – 7.3) × 10−4 m3 = –265 J Magnitude of work done D–A is 5 times work done B–C, and have opposite signs |
1 mark answer, 1 mark sign 1 mark answer, 1 mark sign Comment |
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3.11.2.3 |
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02.1 |
If T is constant, \({p_{\rm{A}}}{V_{\rm{A}}} = {p_{\rm{C}}}{V_{\rm{C}}}\) \(\begin{array}{lll} p_\rm{C} & = & \frac{p_\rm{A} V_\rm{A}}{V_\rm{C}}\\ & = & \frac{3 \times 10^6\ \rm{Pa} \times 2 \times 10^{-3}\ \rm{m}^3}{4 \times 10^{-3}\ \rm{m}^3}\\ & = & 1.5 \times 10^{6}\textrm{ Pa}\\ \end{array} \) |
Explicit statement that T is constant Answer |
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3.11.2.2 |
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02.2 |
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Point A correct and labelled at (2, 3) Point C correct and labelled at (4, 1.5) Curved line from A to C Arrows on lines |
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3.11.2.3 |
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02.3 |
A heat pump Correct reason – one from: Work done by the system is positive/done at a higher temperature Energy is absorbed at a higher temperature and exhausted at a lower temperature |
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3.11.2.2 |
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02.4 |
Isothermal changes can involve the transfer of heat into a system to compensate for the work done by a gas, or the transfer out of a gas to compensate for the work done on it |
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3.11.2.2 |
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02.5 |
\( \begin{array}{l} \textrm{State A: Temperature} = T_\rm{A} = \frac{p_\rm{A}\ V_\rm{A}}{nR} = \frac{3 \times 10^6\ \rm{Pa} \times 2 \times 10^{-3}\ \rm{m}^{3}}{3 \times 8.31} = 241\textrm{K }(240.67)\\ \textrm{State B: Temperature} = T_\rm{B} = \frac{p_\rm{A}V_\rm{B}}{nR} = \frac{3 \times 10^6\ \rm{Pa} \times 4 \times 10^{-3}\ \rm{m}^{3}}{3 \times 8.31} = 481\textrm{K }(481.34)\\ \textrm{Theoretical efficiency} = \frac{T_\rm{H} – T_\rm{C}}{T_\rm{H}} = \frac{481 – 241}{481} = 50\%\\ \end{array} \) |
Calculations (both) Answer |
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3.11.2.5 |
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03.1 |
In an adiabatic change, no heat passes in or out of the gas Q = \(\Delta U\) + W (first law), and here Q = 0 so \(\Delta U\) = −W The work done on the gas increases its temperature |
stated |
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3.11.2.1 |
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03.2 |
\( \begin{array}{l} {p_1} \times V_1^\gamma = {p_2} \times V_2^\gamma\\ {p_2} = \frac{p_1 \times V_1^\gamma}{V_2^\gamma} = \frac{1.00 \times 10^5 \times \left( 240 \times 10^{-6} \right)^{1.4}}{\left( 50.0 \times 10^{-6} \right)^{1.4}}\\ = 8.99 \times 10^{5}\textrm{ Pa}\\ \end{array} \) |
Substitution Answer |
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3.11.2.2 |
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03.3 |
In a petrol engine, the fuel-air mixture is ignited and the pressure increases at constant volume, shown by the vertical section of the graph where heat is added In a diesel engine, fuel is injected and the volume increases at constant pressure, shown by the horizontal section of the graph |
Correct shape for both graphs Correct labels for both graphs Explanation linked to graph Explanation linked to graph |
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3.11.2.4 |
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03.4 |
The mark scheme gives some guidance as to what statements are expected to be seen in a 1 or 2 mark (L1), 3 or 4 mark (L2), and 5 or 6 mark (L3) answer
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The following statements are likely to be present: Bullet point 1 in question (Comparison of frictional losses) 1. Frictional power = indicated power – brake power Indicated power = work done per cycle × no. of cycles sec–1 × no. of cylinders Indicated power of A: 430 × \(\frac{4250}{60}\) × 4 = 121.8kW Indicated power of B: 307 × \(\frac{5360}{60}\) × \(\;{4}\) = 109.7 kW Frictional power A: 121.8 kW – 35 kW = 86.8kW Frictional power B: 109.7 kW – 27 kW = 82.7 kW Frictional losses for Engine A are greater than those for Engine B The statement is incorrect Bullet point 2 in question (Comparison of overall efficiency) 7. Overall efficiency = \(\frac{{{\rm{brake}}\,{\rm{power}}}}{{{\rm{input}}\,{\rm{power}}}}\) A: Input power = 38 × 106 × 2.8 × 10−3 = 106.4 kW B: Input power = 42 × 106 × 2.6 × 10−3 = 109.2 kW Overall efficiency A: 35 × \(\frac{100}{{{106}{.4}}}\)= 33 (32.9)% Overall efficiency B: 27 × \(\frac{100}{{{109}{.2}}}\)= 25 (24.7)% Ratio of efficiencies: \(\frac{33}{25}\)= 1.32 Statement is incorrect |
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3.11.2.4 |
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04.1 |
\( \begin{array}{lll} \textit{P} & = & T \times \omega\\ \omega & = & \frac{P}{T} = \frac{74 \times 10^6}{6.5 \times 10^6} = 11(.4)\ \rm{rad}\ \rm{s}^{-1}\\ & = & \frac{11(.4)\ \rm{rad}}{2\ \pi} \times 60 = 108.7\ \rm{rpm} = 110\ \rm{rpm}\\ \end{array} \) |
Substitution Answer |
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3.11.2.4 |
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04.2 |
Input power = calorific value of fuel × fuel flow rate Container ship: Input power = 38 × 106 × \(\frac{{3.43}}{3600}\) = 36(.2) × 103 W Diesel car: Input power = 46 × 106 × \(\frac{{0.62}}{3600}\) = 7900(7912) W Ratio: 4.6 (4.57) |
Substitution (both) Answer(both) Comparison |
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3.11.2.4 |
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04.3 |
\(\textrm{Overall efficiency} = \frac {\textrm{brake power}}{\textrm{input power}}\)
\( \begin{array}{lll} \textrm{Brake power} & = & \textrm{overall efficiency} \times \textrm{input power}\\ & = & 0.35 \times 7912 = 2800\ (2769)\textrm{ W}\\ \end{array} \) |
Substitution Answer |
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3.11.2.4 |
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04.4 |
\( \begin{array}{l} \frac{P_{\rm{CS}}} {P_{\rm{D}}} = \frac{T_{\rm{CS}}} {T_{\rm{D}}} \frac{\omega_{\rm{CS}}} {\omega _{\rm{D}}}\\ \frac{T_{\rm{CS}}} {T_{\rm{D}}} = \frac{P_{\rm{CS}}} {P_{\rm{D}}} = \frac{74 \times 10^6}{2769} = 26 \times 10^{3}\\ \end{array} \) The mass of the container ship is much larger than that of the diesel car, so a larger torque/force will be needed to produce an acceleration The rotational speed of the crankshaft in the car is probably larger than that in the container ship, so the ratio is smaller |
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3.11.2.4 |
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05.1 |
T1 = 273 + 24 = 297 K, T2 = 273 – 3 = 270 K \(\eta \ = \ 1\ – \ \frac{{{T_2}}}{{{T_1}}}\ = \ 1\ – \ \frac{270}{297}\)= 0.091 Efficiency = 50% of 0.091 = 0.045 |
Calculation of temperatures Answer |
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3.11.2.6 |
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05.2 |
\( \begin{array}{lll} \textrm{COP}_{\textrm{ref}} = \frac{T_{\rm{C}}} {T_{\rm{H}} – T_{\rm{C}}} = \frac{270} {297- 270} = 10\\ \textrm{COP}_{\textrm{ref}} = \frac{Q_{\rm{H}}} {W} = 10\\ \end{array} \) Q = 10% of work done by motor on refrigerator per second = 10 × 1.2 kW = 12 kJ s–1 |
Calculation Calculation Energy per second |
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3.11.2.6 |
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05.3 |
The air inside the refrigerator is being cooled D–A The motor is doing work on the refrigerant A–B The refrigerant is condensing B–C |
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3.11.2.3 |
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05.4 |
Suitable reason, e.g. Energy is transferred due to the flow of the refrigerant |
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3.11.2.6 |
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06.1 |
\( \begin{array}{lll} \rm{COP}_{\rm{hP}} & = & \frac{T_{\rm{H}}} {T_{\rm{H}} – T_{\rm{C}}}\\ & = & \frac{293}{293 – 283} = 29.3\\ \rm{COP}_{\rm{ref}} & = & \frac{T_{\rm{C}}} {T_{\rm{H}} – T_{\rm{C}}}\\ & = & \frac{297}{304 – 297} = 42.4\\ \end{array} \) |
Substitutions Answers |
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3.11.2.6 |
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06.2 |
The device is working between different temperatures in each mode |
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3.11.2.6 |
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06.3 |
\( \begin{array}{lll} \textrm{energy} & = & mc\Delta \theta\\ \textrm{power} & = & \frac{\textrm{mass}} {t} \times c \times \Delta \theta\\ \frac{\textrm{mass}} {t} & = & \frac{\textrm{power}}{c \times \Delta \theta}\\ & = & \frac{230 \times 10^3} {4200 \times (60-10)}\\ & = & 1.095\ldots\textrm{ kg s}^{-1}\\ \end{array} \) \( \begin{array}{l} \textrm{density} = \frac{\textrm{mass}} {\textrm{volume}}\\ \frac {\textrm{volume}}{t} = \frac{ \frac{\textrm{mass}}{t} {\rm{density}}}\\ \end{array} \) \(= \frac{{{1}{.095}..\;{\rm{kg}}\;{{\rm{s}}^{-1}}}}{1000} = {1}{\rm{.1\ \times\ \;1}}{{0}^{-3}}\;{{\rm{m}}^{3}}{{\rm{s}}^{-1}}\) |
Manipulation of SHC Substitution Answer in kg s−1 Answer in m3 s−1 |
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3.6.2.1 |
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06.4 |
A thermistor is connected in series with a resistor and a power supply with a voltmeter/output voltage from across the resistor or thermistor As the temperature increases, the resistance of the thermistor decreases The output can be used to trigger a circuit that will cut off the heater |
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3.5.1.3 |
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07.1 |
Using the conservation of energy: Either: \(\begin{array}{l} \frac{1}{2}\ k\Delta L^2 = \textit{mgh}\\ \textit{h} = \frac{k\Delta L^2}{2mg}\\ \textrm{Column C heading} = \textrm{compression}^{2}\\ \textit{x}\textrm{-axis: }\Delta {L^2}\ \textit{y}\textrm{-axis: }\textit{h}\\ \textrm{Gradient} = \frac{k}{2mg}; \textit{k} = \textrm{gradient} \times 2\textit{mg}\\ \end{array} \) or \(\begin{array}{l} \frac{1}{2}\ k\Delta L^2 = \textit{mgh}\\ \Delta L = \sqrt{ \frac{2mgh}{k} }\\ \textrm{Column C heading} = \sqrt{h}\\ \textit{x}\textrm{-axis:}\ \sqrt{h}\ \textit{y}\textrm{-axis: }\Delta L\\ \textrm{Gradient} = \sqrt{ \frac{2mg}{k} }\\ \textit{k} = \frac{2mg}{\rm{gradient}^2}\\ \end{array} \) |
Explicit or implied Equation of the form y = mx (explicit or implied) Column heading clearly stated x and y labels (both) How to use gradient |
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3.4.1.8 |
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07.2 |
\( \begin{array}{l} \frac{1}{2}\ k\Delta L^2 = \frac{1}{2}\ mv^2_{\textrm{max}}\\ v_{\textrm{max}} = \Delta L \sqrt{ \frac{k}{m} }\\ \end{array} \) Maximum possible velocity for maximum value of \(\Delta L\) = 2.0 cm = 0.02 m \({v_{\max }} = 0.02 \sqrt{ \frac{80}{0.05} } = 0.8\ \rm{m\ s}^{-1}\)The mass is moving fastest at the point when the spring has reached its original length/diagram showing compressed spring and an uncompressed length |
Rearranging k.e. = e.p.e Use of data from table Answer |
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3.4.1.8 |
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07.3 |
\(k = m{\left( {\frac{{{v_{\max }}}}{{\Delta L}}} \right)^2} = 45 \times {\left( {\frac{{0.8}}{{0.08}}} \right)^2} = 4500\ {\rm{N}}\ {{\rm{m}}^{-1}}\) |
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3.4.1.8 |
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07.4 |
\(Q = \Delta U + W = \Delta U + p\Delta V\) = 0/adiabatic change The gas will heat up when compressed due to work done, which will heat the bike cylinder/air Less energy is available for the air to do work on the piston in expansion, so no the piston would not return to its original length |
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3.11.2.2 |
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08.1 |
\({T_1}V_1^{\gamma – 1} = {T_2}V_2^{\gamma – 1}\)
\({\left( {\frac{{{V_1}}}{{{V_2}}}} \right)^{\gamma – 1}} = \frac{{{T_2}}}{{{T_1}}}\)
\(\left( {\frac{{{V_1}}}{{{V_2}}}} \right) = {\left( {\frac{{{T_2}}}{{{T_1}}}} \right)^{\frac{1}{{\gamma – 1}}}}\)
\( = {\left( {\frac{343}{293}} \right)^{\frac{1}{{1.4 – 1}}}}\)
= 1.48 \( \approx 1.5\) |
Substitution Answer |
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3.11.2.2 |
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08.2 |
Any two from:
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3.11.2.4 |
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08.3 |
E = \(\frac{V}{d}\), V = E × d = 10 × 105 × 0.8 × 10−3 = 800 V This is low compared to the pd that is produced because the gas that is conducting is a fuel-air mixture and not just air/fuel molecules are more massive and harder to accelerate/ionise |
Substitution Answer Comment |
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3.7.3.2 |
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08.4 |
The pd from the battery is dc and produces a constant current through a primary coil in a transformer, which produces a constant magnetic field/flux A high secondary pd in a transformer needs a large rate of change of flux Which can be produced when the current is turned on or off |
Reference to constant current/magnetic flux Rate of change of flux and pd Way to produce it |
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3.7.5.6 |
Skills box answers
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Question |
Answer |
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\( \begin{array}{l} \textit{p}_{2} = \frac{{{p_1}V_1^\gamma }}{{V_2^\gamma }}\\ \textit{p}_{2} = \frac{{150 \times {{300}^{1.4}}}}{{{{600}^{1.4}}}} = 56.8\textrm{ kPa}\\ \textit{T}_{1} = (7 + 273)\textrm{ K} = 280\textrm{ K}\\ {T_2} = \frac{{{p_2}{V_2}{T_1}}}{{{p_1}{V_1}}}\\ {T_2} = \frac{{56.8 \times 600 \times 280}}{{150 \times 300}} = 212\textrm{ K}\\ \end{array} \) |
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\(\textit{p}_{1}\textit{V}_{1} = \textit{p}_{2}\textit{V}_{2}\ \textrm{so }{p_2} = \frac{{150 \times 0.06}}{{0.04}} = 225\ {\rm{kPa}}\) |
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\( \begin{array}{l} \textit{p}_{1}\textit{V}_{1}^{\gamma} = \textit{p}_{2}\textit{V}_{2}^{\gamma}\\ {p_2} = \frac{200 \times 50^{1.67}}{5.0^{1.67}} = 9355\textrm{ kPa}\\ \end{array} \) |