IB Physics HL B.4 Thermodynamics Question Bank

[Maximum number: 8]

This question is in two parts

(a)

A gas undergoes a thermodynamic cycle. The P-V diagram for the cycle is shown below.

In the changes of state B to C and D to A , the gas behaves as an ideal gas and the changes in state are adiabatic.

[ 1 ]

(i)

State what is meant by an adiabatic change of state.

[ 1 ]

(b)

With reference to the first law of thermodynamics, explain for the change of state A to B, why energy is transferred from the surroundings to the gas.

[ 4 ]

(c)

Estimate the total work done in the cycle.

[ 3 ]

[Maximum number: 10]

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.

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}

(a)

Suggest why AC is the adiabatic part of the cycle.

[ 2 ]

(b)

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

[ 2 ]

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.

(c)

Suggest, for the change $A \Rightarrow B$ , whether the entropy of the gas is increasing, decreasing or constant.

[ 2 ]

Increasing because thermal energy/heat is being provided to the gas « and temperature is constant, $\Delta S=\frac{\Delta Q}{T} \gg$

(d)

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

[ 2 ]

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).

(e)

The highest and lowest temperatures of the gas during the cycle are 602 K and 92 K .

The efficiency of this engine is about 0.6 . Outline how these data are consistent with the second law of thermodynamics.

[ 2 ]

$e_{\mathrm{C}}=1-\frac{92}{602}=0.847$

this engine has $e<e_{\mathrm{c}}$ as it should

Marking guidance:

Award [0] if no calculation shown.

[Maximum number: 3]

A student is investigating the emf and internal resistance of a cell, using the circuit shown. The ammeter and the voltmeter are ideal.

The graph shows the variation of the voltmeter reading V with the ammeter reading I.

(a)

The student then replaces the variable resistor with an electric heating element in order to investigate the thermal output of the system. The heating element is used to raise the temperature of the water to the boiling point.

After the boiling point is reached, 12 g of the water are vaporized. The specific latent heat of vaporization of water is $2.3 \times 10^{6} \mathrm{Jkg}^{-1}$ .

[ 3 ]

(i)

Calculate the change in entropy of the vaporized water.

[ 2 ]

Use of $\Delta S=\frac{\Delta Q}{T}$ $=\frac{(0.012 \mathrm{~kg})\left(2.3 \times 10^{6}\right)}{100+273}=74 \mathrm{~J} \mathrm{~K}^{-1}$

(ii)

Explain why this entropy change is positive.

[ 1 ]

greater number of possible arrangements (microstates) of particles in steam
greater disorder in particles in steam
greater number of possible ways to distribute energy in steam
Since $\Delta \mathrm{Q}$ is positive, $\Delta S=\frac{\Delta Q}{T}$ is positive

Figure for Answer to Question (ii) — IB Physics HL

[Maximum number: 4]

This question is in two parts

With reference to the first law of thermodynamics, explain for the change of state A to B, why energy is transferred from the surroundings to the gas.

[Maximum number: 3]

This question is in two parts

Estimate the total work done in the cycle.

[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.

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).

[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.

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}

Suggest, for the change $A \Rightarrow B$ , whether the entropy of the gas is increasing, decreasing or constant.

Increasing because thermal energy/heat is being provided to the gas « and temperature is constant, $\Delta S=\frac{\Delta Q}{T} \gg$

[Maximum number: 2]

A student is investigating the emf and internal resistance of a cell, using the circuit shown. The ammeter and the voltmeter are ideal.

The graph shows the variation of the voltmeter reading V with the ammeter reading I.

The student then replaces the variable resistor with an electric heating element in order to investigate the thermal output of the system. The heating element is used to raise the temperature of the water to the boiling point.

After the boiling point is reached, 12 g of the water are vaporized. The specific latent heat of vaporization of water is $2.3 \times 10^{6} \mathrm{Jkg}^{-1}$ .

Calculate the change in entropy of the vaporized water.

Use of $\Delta S=\frac{\Delta Q}{T}$ $=\frac{(0.012 \mathrm{~kg})\left(2.3 \times 10^{6}\right)}{100+273}=74 \mathrm{~J} \mathrm{~K}^{-1}$

[Maximum number: 2]

This question is in two parts. Part 1 is about simple harmonic motion (SHM). Part 2 is about gas in an engine.

Part 1 Simple harmonic motion (SHM)
An object is placed on a frictionless surface. The object is attached by a spring fixed at one end and oscillates at the end of the spring with simple harmonic motion (SHM).

The tension F in the spring is given by F=k x where x is the extension of the spring and k is a constant.

The shaded area WXYZ is 610 J . The total thermal energy transferred out of the gas in one cycle is 1.3 kJ .

Explain why the total thermal energy transferred out of the gas is degraded.

[Maximum number: 1]

A gas is compressed at constant temperature.

What are the changes in entropy of the gas and its surroundings during the compression?

Entropy of the gas

Entropy of surroundings

Increases

Decreases

Increases

Decreases

[Maximum number: 1]

This question is in two parts

A gas undergoes a thermodynamic cycle. The P-V diagram for the cycle is shown below.

In the changes of state B to C and D to A , the gas behaves as an ideal gas and the changes in state are adiabatic.

State what is meant by an adiabatic change of state.

[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.

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.

[Maximum number: 1]

The pressure-volume diagram shows a 3-stage cyclic process of 0.20 moles of an ideal gas in a closed container. A-B is an isothermal process. C-A is an adiabatic process.

The following data are available:

\begin{aligned} \text { Pressure at } A & =2.0 \times 10^{6} \mathrm{~Pa} \\ \text { Volume at } A & =1.0 \times 10^{-3} \mathrm{~m}^{3} \\ \text { Pressure at } B & =2.0 \times 10^{5} \mathrm{~Pa} \\ \text { Volume at } B & =1.0 \times 10^{-2} \mathrm{~m}^{3} \end{aligned}

Outline why this cycle can be used for a heat engine.

[Maximum number: 2]

This question is in two parts. Part 1 is about simple harmonic motion (SHM). Part 2 is about gas in an engine.

Part 1 Simple harmonic motion (SHM)
An object is placed on a frictionless surface. The object is attached by a spring fixed at one end and oscillates at the end of the spring with simple harmonic motion (SHM).

The tension F in the spring is given by F=k x where x is the extension of the spring and k is a constant.

The shaded area WXYZ is 610 J . The total thermal energy transferred out of the gas in one cycle is 1.3 kJ .

Determine the efficiency of cycle WXYZ.

efficiency $=\frac{610}{1300}$ ;
0.47 or 47 %;

Marking guidance:

Award [2] for a bald correct answer.
Award [1] for $\frac{610}{(610+1300)}=0.32$ .
Award [0] for any fraction that would exceed 1 even if later fudged.
Award [0] for efficiency $=\frac{T_{C}-T_{H}}{T_{H}}=38 \%$ , question does not imply that this is a Carnot cycle.

[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.

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}

The highest and lowest temperatures of the gas during the cycle are 602 K and 92 K .

The efficiency of this engine is about 0.6 . Outline how these data are consistent with the second law of thermodynamics.

$e_{\mathrm{C}}=1-\frac{92}{602}=0.847$

this engine has $e<e_{\mathrm{c}}$ as it should

Marking guidance:

Award [0] if no calculation shown.

IB Physics HL B 4 Thermodynamics Questions

Exam points

Covered syllabus points

Covered syllabus points

Question 1

Question (a)

Question (i)

Question (b)

Question (c)

Question 2

Question (a)

Question (b)

Question (c)

Question (d)

Question (e)

Question 3

Question (a)

Question (i)

Question (ii)

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15