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A-Level CAIE Physics 15 2 Equation Of State Question Bank

Practice A-Level CAIE Physics 15 2 Equation Of State questions by syllabus topic with past-paper context, marks, difficulty and question previews on Eduninja.

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Question 2

2

+w.

structured1 marks

Question 2(b)

2(b)

Argon-40 \(\left({ }_{18}^{40} \mathrm{Ar}\right)\) may be assumed to be an ideal gas. A mass of 3.2 g of argon- 40 has a volume of \(210 \mathrm{~cm}^{3}\) at a temperature of \(37^{\circ} \mathrm{C}\). Determine, for this mass of argon-40 gas,

structured2 marks

Question 2(b)(ii)

2(b)(ii)

the pressure, pressure = Pa

Mediumstructured2 marks

Answer

p V=n R T \(p \times 210 \times 10^{-6}=0.080 \times 8.31 \times 310\) \(p=9.8 \times 10^{5} \mathrm{~Pa}\) (do not credit if T in \({ }^{\circ} \mathrm{C}\) not K )

Question 2

2

4 marks

Question 2(b)

2(b)

Two containers A and B are joined by a tube of negligible volume, as illustrated in Fig. 2.1. The containers are filled with an ideal gas at a pressure of \(2.3 \times 10^{5} \mathrm{~Pa}\). The gas in container A has volume \(3.1 \times 10^{3} \mathrm{~cm}^{3}\) and is at a temperature of \(17^{\circ} \mathrm{C}\). The gas in container B has volume \(4.6 \times 10^{3} \mathrm{~cm}^{3}\) and is at a temperature of \(30^{\circ} \mathrm{C}\). Calculate the total amount of gas, in mol, in the containers. amount = mol

Mediumstructured4 marks

Answer

p V=n R T amount \(=\left(2.3 \times 10^{5} \times 3.1 \times 10^{-3}\right) /(8.31 \times 290)\) \(+\left(2.3 \times 10^{5} \times 4.6 \times 10^{-3}\right) /(8.31 \times 303)\) =0.296+0.420 \(=0.716 \mathrm{~mol}\) (give full credit for starting equation p V=N k T and \(N=n N_{\mathrm{A}}\) )

Question 2

2

the rate h of thermal energy gained by the ice from the surroundings. W

structured2 marks

Question 2(b)

2(b)

The variation with volume V of the pressure p of an ideal gas as it undergoes a cycle ABCA of changes is shown in Fig. 2.1. The temperature of the gas at A is 290 K . The temperature at B is 870 K . Determine

structured0 marks

Question 2(b)(ii)

2(b)(ii)

the temperature of the gas at C . temperature = K

Mediumstructured0 marks

Answer

\(1.2 \times 10^{5} \times 7.75 \times 10^{-3}=0.20 \times 8.31 \times T\) or \(T=(7.75 / 4.0) \times 290\) C1 \(T=560 \mathrm{~K}\) A1 (Allow tolerance from graph: \(7.7-7.8 \times 10^{-3} \mathrm{~m}^{3}\) )

Question 2

2

the rise in temperature of the gas. temperature rise = K

structured2 marks

Question 2(a)

2(a)

The volume of an ideal gas in a cylinder is \(1.80 \times 10^{-3} \mathrm{~m}^{3}\) at a pressure of \(2.60 \times 10^{5} \mathrm{~Pa}\) and a temperature of 297 K , as illustrated in Fig. 2.1. The thermal energy required to raise the temperature by 1.00 K of 1.00 mol of the gas at constant volume is 12.5 J . The gas is heated at constant volume such that the internal energy of the gas increases by 95.0 J .

structured1 marks

Question 2(a)(i)

2(a)(i)

Calculate

Mediumstructured0 marks

Answer

1. p V=n R T \(1.80 \times 10^{-3} \times 2.60 \times 105=n \times 8.31 \times 297 \quad\) C1 \(n=0.19 \mathrm{~mol}\) A1 2. \(\Delta q=m c \Delta T\) \(95.0=0.190 \times 12.5 \times \Delta T\) B1 \(\Delta T=40 \mathrm{~K}\) A1 (allow 2 marks for correct answer with clear logic shown)

Question 2(a)(ii)

2(a)(ii)

Use your answer in (i) part 2 to show that the final pressure of the gas in the cylinder is \(2.95 \times 10^{5} \mathrm{~Pa}\).

Mediumstructured1 marks

Answer

p / T= constant \(\left(2.6 \times 10^{5}\right) / 297=p /(297+40)\) \(p=2.95 \times 10^{5} \mathrm{~Pa}\)

Question 2

2

A cylinder contains 5.12 mol of an ideal gas at pressure \(5.60 \times 10^{5} \mathrm{~Pa}\) and volume \(3.80 \times 10^{-2} \mathrm{~m}^{3}\).

structured3 marks

Question 2(a)

2(a)

Calculate the thermodynamic temperature of the gas.

Easystructured2 marks

Answer

p V=n R T C1 \[ \begin{aligned} T=\left(5.60 \times 10^{5} \times 3.80 \times 10^{-2}\right) /(5.12 \times 8.31) T=500 \mathrm{~K} \end{aligned} \] A1

Question 2(b)

2(b)

The average kinetic energy \(E_{\mathrm{K}}\) of a molecule of the gas is given by the expression where k is the Boltzmann constant and T is the thermodynamic temperature. The gas is heated at constant pressure so that its temperature rises by 125 K .

structured1 marks

Question 2(b)(i)

2(b)(i)

Show that the new volume of the gas is \(4.75 \times 10^{-2} \mathrm{~m}^{3}\).

Mediumstructured1 marks

Answer

\(V=\left(3.80 \times 10^{-2}\right) \times(500+125) / 500=4.75 \times 10^{-2} \mathrm{~m}^{3}\) A1

Question 2

2

A constant mass of an ideal gas has a volume of \(3.49 \times 10^{3} \mathrm{~cm}^{3}\) at a temperature of \(21.0^{\circ} \mathrm{C}\). When the gas is heated, 565 J of thermal energy causes it to expand to a volume of \(3.87 \times 10^{3} \mathrm{~cm}^{3}\) at \(53.0^{\circ} \mathrm{C}\). This is illustrated in Fig.2.1.

structured2 marks

Question 2(a)

2(a)

Show that the initial and final pressures of the gas are equal.

Mediumstructured2 marks

Answer

use of kelvin temperatures both values of (V / T) correct (11.87), V / T is constant so pressure is constant (allow use of n=1. Do not allow other values of n.)

Question 2

2

2 marks

Question 2(b)

2(b)

A fixed mass of an ideal gas at a temperature of \(20^{\circ} \mathrm{C}\) is sealed in a cylinder by a piston, as shown in Fig. 2.1. The initial volume of the gas is \(1.24 \times 10^{-4} \mathrm{~m}^{3}\). Thermal energy is supplied to the gas and its volume increases by \(5.20 \times 10^{-5} \mathrm{~m}^{3}\).

structured2 marks

Question 2(b)(ii)

2(b)(ii)

Calculate the final thermodynamic temperature T of the gas.

Easystructured2 marks

Answer

\(V \propto T\) or V / T = constant C1 \[ \begin{aligned} 1.24 /(273+20)=(1.24+0.520) / T T=416 \mathrm{~K} \end{aligned} \] A1

Question 2

2

Fig. 2.1 shows a laboratory thermometer that is calibrated to measure temperature in degrees Celsius. The thermometer makes use of the fact that the density of mercury varies with temperature.

structured1 marks

Question 2(c)

2(c)

1 marks

Question 2(c)(ii)

2(c)(ii)

Thermodynamic temperature T may be determined by the behaviour of a type of substance for which T is proportional to the product of pressure and volume. State the name of this type of substance.

Mediumstructured1 marks

Answer

ideal gas B1

Question 2

2

The product of the pressure p and the volume V of an ideal gas is given by the expression where m is the mass of one molecule of the gas.

structured1 marks

Question 2(a)

2(a)

State the meaning of the symbol

structured1 marks

Question 2(a)(i)

2(a)(i)

N,

Easystructured1 marks

Answer

N : (total) number of molecules

Question 2

2

4 marks

Question 2(b)

2(b)

A balloon is filled with helium gas at a pressure of \(1.1 \times 10^{5} \mathrm{~Pa}\) and a temperature of \(25^{\circ} \mathrm{C}\). The balloon has a volume of \(6.5 \times 10^{4} \mathrm{~cm}^{3}\). Helium may be assumed to be an ideal gas. Determine the number of gas atoms in the balloon. number =

Mediumstructured4 marks

Answer

p V=N k T or p V=n R T substitutes temperature as 298 K C1 either \(1.1 \times 10^{5} \times 6.5 \times 10^{-2}=N \times 1.38 \times 10^{-23} \times 298\) or \(\quad 1.1 \times 10^{5} \times 6.5 \times 10^{-2}=n \times 8.31 \times 298\) and \(n=N / 6.02 \times 10^{23} \quad\) C1 \(N=1.7 \times 10^{24}\) A1