ideal gases
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A-level Physics Problems on Ideal gasesTRANSCRIPT
Ideal Gases
1 (a) (i) The kinetic theory of gases leads to the equation
�� m<c2> = �� kT.
Explain the significance of the quantity �� m<c2>.
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(ii) Use the equation to suggest what is meant by the absolute zero of temperature.
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(b) Two insulated gas cylinders A and B are connected by a tube of negligible volume, asshown in Fig. 3.1.
Fig. 3.1
cylinder A cylinder B
tap
JUNE 2002
1
Each cylinder has an internal volume of 2.0 × 10–2m3. Initially, the tap is closed and cylinder Acontains 1.2 mol of an ideal gas at a temperature of 37 °C. Cylinder B contains the sameideal gas at pressure 1.2 × 105 Pa and temperature 37 °C.
(i) Calculate the amount, in mol, of the gas in cylinder B.
amount = ......................................... mol
(ii) The tap is opened and some gas flows from cylinder A to cylinder B. Using the factthat the total amount of gas is constant, determine the final pressure of the gas inthe cylinders.
pressure = ........................................ Pa[6]
2
NOVEMBER 2003
2 The volume of some air, assumed to be an ideal gas, in the cylinder of a car engine is540 cm3 at a pressure of 1.1 × 105 Pa and a temperature of 27 °C. The air is suddenlycompressed, so that no thermal energy enters or leaves the gas, to a volume of 30 cm3. Thepressure rises to 6.5 × 106 Pa.
(a) Determine the temperature of the gas after the compression.
temperature = …………………………… K [3]
(b) (i) State and explain the first law of thermodynamics.
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(ii) Use the law to explain why the temperature of the air changed during thecompression.
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3
JUNE 2004
3 The pressure p of an ideal gas is given by the expression
p = <c2> .
(a) Explain the meaning of the symbol <c2>.
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(b) The ideal gas has a density of 2.4 kg m–3 at a pressure of 2.0 × 105 Pa and atemperature of 300 K.
(i) Determine the root-mean-square (r.m.s.) speed of the gas atoms at 300 K.
r.m.s. speed = .................................. m s–1 [3]
(ii) Calculate the temperature of the gas for the atoms to have an r.m.s. speed that istwice that calculated in (i).
temperature = ......................................... K [3]
NmV
13
4
JUNE 2005
4 (a) State what is meant by an ideal gas.
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(b) The product of pressure p and volume V of an ideal gas of density ρ at temperature T isgiven by the expressions
and
where N is the number of molecules and k is the Boltzmann constant.
(i) State the meaning of the symbol <c2>.
...............................................................................................................................[1]
(ii) Deduce that the mean kinetic energy EK of the molecules of an ideal gas is givenby the expression
EK = �� kT.
[2]
(c) In order for an atom to escape completely from the Earth’s gravitational field, it musthave a speed of approximately 1.1 × 104 m s–1 at the top of the Earth’s atmosphere.
(i) Estimate the temperature at the top of the atmosphere such that helium, assumedto be an ideal gas, could escape from the Earth. The mass of a helium atom is6.6 × 10–27 kg.
temperature = ....................................... K [2]
(ii) Suggest why some helium atoms will escape at temperatures below that calculatedin (i).
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...............................................................................................................................[1]
p = �� ρ<c2>
pV = NkT,
5
NOVEMBER 2005
5 The air in a car tyre has a constant volume of 3.1 × 10–2 m3. The pressure of this air is2.9 × 105 Pa at a temperature of 17 °C. The air may be considered to be an ideal gas.
(a) State what is meant by an ideal gas.
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(b) Calculate the amount of air, in mol, in the tyre.
amount = ……………………………. mol [2]
(c) The pressure in the tyre is to be increased using a pump. On each stroke of the pump,0.012 mol of air is forced into the tyre.Calculate the number of strokes of the pump required to increase the pressure to3.4 × 105 Pa at a temperature of 27 °C.
number = ……………………………. [3]
6
JUNE 2006
6 (a) The equation
pV = constant × T
relates the pressure p and volume V of a gas to its kelvin (thermodynamic)temperature T.
State two conditions for the equation to be valid.
1. .....................................................................................................................................
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2. .....................................................................................................................................
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(b) A gas cylinder contains 4.00 × 104 cm3 of hydrogen at a pressure of 2.50 × 107 Pa and atemperature of 290 K.
The cylinder is to be used to fill balloons. Each balloon, when filled, contains7.24 × 103 cm3 of hydrogen at a pressure of 1.85 × 105 Pa and a temperature of 290 K.
Calculate, assuming that the hydrogen obeys the equation in (a),
(i) the total amount of hydrogen in the cylinder,
amount = ……………………….. mol [3]
(ii) the number of balloons that can be filled from the cylinder.
number = ……………………….. [3]
7
JUNE 2008
7 (a) Explain qualitatively how molecular movement causes the pressure exerted by a gas.
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(b) The density of neon gas at a temperature of 273 K and a pressure of 1.02 × 105 Pa is 0.900 kg m–3. Neon may be assumed to be an ideal gas.
Calculate the root-mean-square (r.m.s.) speed of neon atoms at
(i) 273 K,
speed = ........................................... m s–1 [3]
(ii) 546 K.
speed = ........................................... m s–1 [2]
(c) The calculations in (b) are based on the density for neon being 0.900 kg m–3. Suggest the effect, if any, on the root-mean-square speed of changing the density at
constant temperature.
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8
NOVEMBER 2009
8 An ideal gas occupies a container of volume 4.5 × 103 cm3 at a pressure of 2.5 × 105 Pa and a temperature of 290 K.
(a) Show that the number of atoms of gas in the container is 2.8 × 1023.
[2]
(b) Atoms of a real gas each have a diameter of 1.2 × 10–10 m.
(i) Estimate the volume occupied by 2.8 × 1023 atoms of this gas.
volume = ......................................... m3 [2]
(ii) By reference to your answer in (i), suggest whether the real gas does approximate to an ideal gas.
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9
JUNE 2010
9 (a) Some gas, initially at a temperature of 27.2 °C, is heated so that its temperature rises to 38.8 °C.Calculate, in kelvin, to an appropriate number of decimal places,
(i) the initial temperature of the gas,
initial temperature = ............................................. K [2]
(ii) the rise in temperature.
rise in temperature = ............................................ K [1]
(b) The pressure p of an ideal gas is given by the expression
p = 13ρ�c 2�
where ρ is the density of the gas.
(i) State the meaning of the symbol �c 2�.
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(ii) Use the expression to show that the mean kinetic energy <EK> of the atoms of an ideal gas is given by the expression
<EK> = 32 kT.
Explain any symbols that you use.
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10
(c) Helium-4 may be assumed to behave as an ideal gas.A cylinder has a constant volume of 7.8 × 103 cm3 and contains helium-4 gas at a pressure of 2.1 × 107 Pa and at a temperature of 290 K.
Calculate, for the helium gas,
(i) the amount of gas,
amount = ......................................... mol [2]
(ii) the mean kinetic energy of the atoms,
mean kinetic energy = .............................................. J [2]
(iii) the total internal energy.
internal energy = .............................................. J [3]
11
JUNE 2011
10 (a) State what is meant by the Avogadro constant NA.
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(b) A balloon is filled with helium gas at a pressure of 1.1 × 105 Pa and a temperature of25 °C.The balloon has a volume of 6.5 × 104 cm3.Helium may be assumed to be an ideal gas.
Determine the number of gas atoms in the balloon.
number = ................................................ [4]
12
JUNE 2011
11 (a) State what is meant by a mole.
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(b) Two containers A and B are joined by a tube of negligible volume, as illustrated in Fig. 2.1.
container A
3.1 × 103 cm3
17 °C
container B
4.6 × 103 cm3
30 °C
Fig. 2.1
The containers are filled with an ideal gas at a pressure of 2.3 × 105 Pa.The gas in container A has volume 3.1 × 103 cm3 and is at a temperature of 17 °C.The gas in container B has volume 4.6 × 103 cm3 and is at a temperature of 30 °C.
Calculate the total amount of gas, in mol, in the containers.
amount = ........................................ mol [4]
13
NOVEMBER 2011
2 (a) One assumption of the kinetic theory of gases is that gas molecules behave as if they are hard, elastic identical spheres.
State two other assumptions of the kinetic theory of gases.
1. ......................................................................................................................................
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2. ......................................................................................................................................
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(b) Using the kinetic theory of gases, it can be shown that the product of the pressure p and the volume V of an ideal gas is given by the expression
pV = 13Nm <c 2>
where m is the mass of a gas molecule.
(i) State the meaning of the symbol
1. N,
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2. <c 2>.
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(ii) Use the expression to deduce that the mean kinetic energy <EK > of a gas molecule at temperature T is given by the equation
<EK> = 32 kT
where k is a constant.
[2]
14
(c) (i) State what is meant by the internal energy of a substance.
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(ii) Use the equation in (b)(ii) to explain that, for an ideal gas, a change in internal energy ΔU is given by
ΔU ∝ ΔT
where ΔT is the change in temperature of the gas.
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15
JUNE 2012
13 (a) The kinetic theory of gases is based on some simplifying assumptions.The molecules of the gas are assumed to behave as hard elastic identical spheres.State the assumption about ideal gas molecules based on
(i) the nature of their movement,
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(ii) their volume.
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16
(b) A cube of volume V contains N molecules of an ideal gas. Each molecule has a component cX of velocity normal to one side S of the cube, as shown in Fig. 2.1.
cx
side S
Fig. 2.1
The pressure p of the gas due to the component cX of velocity is given by the expression
pV = NmcX2
where m is the mass of a molecule.
Explain how the expression leads to the relation
pV = 13Nm<c2>
where <c2> is the mean square speed of the molecules.
[3]
(c) The molecules of an ideal gas have a root-mean-square (r.m.s.) speed of 520 m s–1 at a temperature of 27 °C.
Calculate the r.m.s. speed of the molecules at a temperature of 100 °C.
r.m.s. speed = ....................................... m s–1 [3]
17
NOVEMBER 2012
14 An ideal gas has volume V and pressure p. For this gas, the product pV is given by the expression
pV = 13Nm <c 2>
where m is the mass of a molecule of the gas.
(a) State the meaning of the symbol
(i) N,
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(ii) <c 2>.
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(b) A gas cylinder of volume 2.1 × 104 cm3 contains helium-4 gas at pressure 6.1 × 105 Pa and temperature 12 °C. Helium-4 may be assumed to be an ideal gas.
(i) Determine, for the helium gas,
1. the amount, in mol,
amount = ......................................... mol [3]
2. the number of atoms.
number = .................................................. [2]
(ii) Calculate the root-mean-square (r.m.s.) speed of the helium atoms.
r.m.s. speed = ....................................... m s–1 [3]
JUNE 2013
15 (a) State what is meant by an ideal gas.
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(b) Two cylinders A and B are connected by a tube of negligible volume, as shown in Fig. 2.1.
tap T
tube
cylinder A
cylinder B
2.5 × 103 cm3
1.6 × 103 cm3
4.9 × 105 Pa
3.4 × 105 Pa
300K
Fig. 2.1
Initially, tap T is closed. The cylinders contain an ideal gas at different pressures.
(i) Cylinder A has a constant volume of 2.5 × 103 cm3 and contains gas at pressure 3.4 × 105 Pa and temperature 300 K.
Show that cylinder A contains 0.34 mol of gas.
(ii) Cylinder B has a constant volume of 1.6 × 103 cm3 and contains 0.20 mol of gas. When tap T is opened, the pressure of the gas in both cylinders is 3.9 × 105 Pa. No thermal energy enters or leaves the gas.
Determine the final temperature of the gas.
temperature = .............................................. K [2]
(c) By reference to work done and change in internal energy, suggest why the temperature of the gas in cylinder A has changed.
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JUNE 2013
16 (a) The volume of an ideal gas in a cylinder is 1.80 × 10–3 m3 at a pressure of 2.60 × 105 Pa and a temperature of 297 K, as illustrated in Fig. 2.1.
ideal gas1.80 × 10–3 m3
2.60 × 105 Pa297K
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.
(i) Calculate
1. the amount of gas, in mol, in the cylinder,
amount = ........................................... mol [2]
2. the rise in temperature of the gas.
temperature rise = .............................................. K [2]
(ii) Use your answer in (i) part 2 to show that the final pressure of the gas in the cylinder is 2.95 × 105 Pa.
[1]
(b) The gas is now allowed to expand. No thermal energy enters or leaves the gas.The gas does 120 J of work when expanding against the external pressure.
State and explain whether the final temperature of the gas is above or below 297 K.
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NOVEMBER 2013
17 The product of the pressure p and the volume V of an ideal gas is given by the expression
pV = 13
Nm<c 2>
where m is the mass of one molecule of the gas.
(a) State the meaning of the symbol
(i) N,
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(ii) <c 2>.
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(b) The product pV is also given by the expression
pV = NkT.
Deduce an expression, in terms of the Boltzmann constant k and the thermodynamic temperature T, for the mean kinetic energy of a molecule of the ideal gas.
[2]
(c) A cylinder contains 1.0 mol of an ideal gas.
(i) The volume of the cylinder is constant. Calculate the energy required to raise the temperature of the gas by 1.0 kelvin.
energy = .............................................. J [2]
(ii) The volume of the cylinder is now allowed to increase so that the gas remains at constant pressure when it is heated.
Explain whether the energy required to raise the temperature of the gas by 1.0 kelvin is now different from your answer in (i).
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NOVEMBER 2013
18 (a) (i) State what is meant by the internal energy of a system.
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(ii) Explain why, for an ideal gas, the internal energy is equal to the total kinetic energy of the molecules of the gas.
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(b) The mean kinetic energy <EK> of a molecule of an ideal gas is given by the expression
<EK> = 32kT
where k is the Boltzmann constant and T is the thermodynamic temperature of the gas.
A cylinder contains 1.0 mol of an ideal gas. The gas is heated so that its temperature changes from 280 K to 460 K.
(i) Calculate the change in total kinetic energy of the gas molecules.
change in energy = ............................................. J [2]
(ii) During the heating, the gas expands, doing 1.5 × 103 J of work. State the first law of thermodynamics. Use the law and your answer in (i) to
determine the total energy supplied to the gas.
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total energy = ............................................. J [3]
JUNE 2014
19 (a) Explain what is meant by the Avogadro constant.
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(b) Argon-40 (4018Ar) may be assumed to be an ideal gas.
A mass of 3.2 g of argon-40 has a volume of 210 cm3 at a temperature of 37 °C.
Determine, for this mass of argon-40 gas,
(i) the amount, in mol,
amount = ................................................. mol [1]
(ii) the pressure,
pressure = ................................................... Pa [2]
(iii) the root-mean-square (r.m.s.) speed of an argon atom.
r.m.s. speed = ............................................... m s−1 [3]
JUNE 2014
20 A constant mass of an ideal gas has a volume of 3.49 × 103 cm3 at a temperature of 21.0 °C.When the gas is heated, 565 J of thermal energy causes it to expand to a volume of 3.87 × 103 cm3
at 53.0 °C. This is illustrated in Fig. 2.1.
565 J3.49× 103 cm3
21.0 °C
3.87 ×103 cm3
53.0 °C
Fig. 2.1
(a) Show that the initial and final pressures of the gas are equal.
[2]
(b) The pressure of the gas is 4.20 × 105 Pa.
For this heating of the gas,
(i) calculate the work done by the gas,
work done = ..................................................... J [2]
(ii) use the first law of thermodynamics and your answer in (i) to determine the change in internal energy of the gas.
change in internal energy = ..................................................... J [2]
(c) Explain why the change in kinetic energy of the molecules of this ideal gas is equal to the change in internal energy.
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Ideal Gases
(a) (i) mean kinetic energy ............................................................................................... Ml
of the atoms / molecules / particles ...........................................................................A1 (ii) at absolute zero, atoms have no kinetic energy ............................................ Bl [3]
(b) (i) p V = nRT .............................................................................................................. Cl
n = (1.2 x 105 x 2.0 x 10-2)/(8.31 x 310) ................................................................ Cl
= 0.93 mol ......................................................................................................... A1
(ii) total amount = (1.20 + 0.93) ...................................................................................... Cl
(1.20 + 0.93) = (4.0 x 10-2 x p)/(8.31 x 310) .............................................................. Cl p = 1.37 x 105 Pa .................................................................................................... Al [6]
1
2 (a) pV/T = constant............................................................................ C1T = (6.5 x 106 x 30 x 300)/(1.1 x 105 x 540)................................. C1 = 985 K.................................................................................... A1
(if uses °C, allow 1/3 marks for clear formula)
[3]
(b) (i) ∆U = q + w symbols identified correctly ..........................................................M1directions correct.......................................................................... A1 [2]
(ii) q is zero ....................................................................................... B1
w is positive OR ∆U = w and U increases .................................... B1
∆U is rise in kinetic energy of atoms ............................................M1
and mean kinetic energy ∝ T ....................................................... A1(allow one of the last two marks if states ‘U increases so T rises’)
[4]
3 (a) mean (value of the) square M1 of the speeds (velocities) of the atoms/particles/molecules A1 [2] (b) (i)
p = 3
1
><2
cρ
C1
<c2> = 3 x 2 x 105/2.4 = 2.5 x 105 C1 r.m.s speed = 500 ms-1 A1 [3] (ii) new <c2> = 1.0 x 106 or <c2> increases by factor of 4 C1 <c2> ∝ T or 3/2 kT = 1/2 m<c2> C1
T = {(1.0 x 106) / (2.5 x 105)} x 300 = 1200 K A1 [3] Total [8]
18
4 (a) obeys the law pV/T = constant or any two named gas laws M1at all values of p, V and T A1 [2] or two correct assumptions of kinetic theory of ideal gas (B1)
third correct assumption (B1)
(b) (i) mean square speed B1 [1]
(ii) mean kinetic energy = ½m<c2> M1
ρ = Nm/V and algebra leading to [do not allow if takes N = 1] M1½m<c2> = 3/2 kT A0 [2]
(c) (i) ½ × 6.6 × 10-27× (1.1 ×104)2 = 3/2 × 1.38 × 10-23
×T C1
T = 1.9 × 104 K A1 [2]
(ii) Not all atoms have same speed/kinetic energy B1 [1]
5 (a) obeys the law pV = constant × T ………………………..……………….. M1 at all values of p, V and T ………………………………………………. A1 [2]
(b) n = (2.9 × 105× 3.1 × 10–2) / (8.31 × 290) …..………………..………... C1
= 3.73 mol ………………………………………………………………. A1 [2]
(c) at new pressure, nn = 3.73 ×9.2
4.3×
300
290
= 4.23 mol ….………………………………………. C1 change = 0.50 mol ……………………………………………………….… C1
number of strokes = 0.50 / 0.012 = 42 (must round up for mark) ……. A1 [3]
6 (a) e.g. fixed mass/ amount of gas ideal gas (any two, 1 each) B2 [2]
(b) (i) n = pV / RT C1
= (2.5 × 107 × 4.00 × 104 x 10-6) / (8.31 × 290) C1 = 415 mol A1 [3]
(ii) volume of gas at 1.85 × 105 Pa = (2.5 × 107 × 4.00 × 104) / (1.85 × 105)
= 5.41 × 106 cm3 C1
so, 5.41 × 106 = 4.00 × 104 + 7.24 × 103 N C1 N = 741 A1 [3]
(answer 740 or fails to allow for gas in cylinder, max 2/3)
19
7 (a) molecule(s) rebound from wall of vessel / hits walls B1 change in momentum gives rise to impulse / force B1
either (many impulses) averaged to give constant force / pressure or the molecules are in random motion B1 [3]
(b) (i) p = 3
1 ρ<c2> C1
1.02 × 105 = 3
1 × 0.900 × <c2>
<c2> = 3.4 × 105 C1 cRMS = 580 m s–1 A1 [3]
(ii) either <c2> ∝ T or <c2> = 2 × 3.4 ×105 C1 cRMS = 830 m s–1 (allow 820) A1 [2]
(c) cRMS depends on temperature (alone) B1 so no effect B1 [2]
8 (a) either pV = NkT or pV = nRT and n = N / NA .....................................................C1 clear correct substitution e.g.
2.5 × 105× 4.5 × 103
× 10-6 = N × 1.38 × 10-23× 290 ...............................................M1
N = 2.8 × 1023 .......................................................................................................... A0 [2] (allow 1 mark for calculation of n = 0.467 mol)
(b) (i) volume = (1.2 × 10-10)3 × 2.8 × 1023 or 3
4πr3 × 2.8 × 1023 ..............................C1
= 4.8 × 10-7 m3 2.53 × 10-7 m3 ..................................... A1 [2]
(ii) either 4.5 × 103 cm3 >> 0.48 cm3 or ratio of volumes is about 10-4 ................ B1 justified because volume of molecules is negligible ........................................... B1 [2]
[Total: 6]
20
9 (a) (i) 27.2 + 273.15 or 27.2 + 273.2 C1 300.4 K A1 [2]
(ii) 11.6 K A1 [1]
(b) (i) (<c2> is the) mean / average square speed B1 [1]
(ii) ρ = Nm/V with N explained B1 so, pV = 1/3 Nm<c2> B1 and pV = NkT with k explained B1 so mean kinetic energy / <EK> = ½m<c2> = 3/2 kT B1 [4]
(c) (i) pV = nRT
2.1 × 107 × 7.8 × 10–3 = n × 8.3 × 290 C1 n = 68 mol A1 [2]
(ii) mean kinetic energy = 3/2 kT
= 3/2 × 1.38 × 10–23 × 290 C1
= 6.0 × 10–21 J A1 [2]
(iii) realisation that total internal energy is the total kinetic energy C1
energy = 6.0 × 10–21 × 68 × 6.02 × 1023 C1
= 2.46 × 105 J A1 [3]
10 (a) number of atoms of carbon-12 M1 in 0.012kg of carbon-12 A1 [2]
(b) pV = NkT or pV = nRT C1 substitutes temperature as 298K C1 either 1.1 × 105 × 6.5 × 10–2 = N × 1.38 × 10–23 × 298 or 1.1 × 105 × 6.5 × 10–2 = n × 8.31 × 298 and n = N / 6.02 × 1023 C1 N = 1.7 × 1024 A1 [4]
11 (a) amount of substance M1 containing same number of particles as in 0.012kg of carbon-12 A1 [2]
(b) pV = nRT C1 amount = (2.3 × 105 × 3.1 × 10–3) / (8.31 × 290) + (2.3 × 105 × 4.6 × 10–3) / (8.31 × 303) C1 = 0.296 + 0.420 C1
= 0.716mol A1 [4] (give full credit for starting equation pV = NkT and N = nNA)
21
12 (a) e.g. moving in random (rapid) motion of molecules/atoms/particles no intermolecular forces of attraction/repulsion volume of molecules/atoms/particles negligible compared to volume of
container time of collision negligible to time between collisions (1 each, max 2) B2 [2]
(b) (i) 1. number of (gas) molecules B1 [1]
2. mean square speed/velocity (of gas molecules) B1 [1]
(ii) either pV = NkT or pV = nRT and links n and kand <EK> = ½m<c2> M1
clear algebra leading to <EK> = 2
3kT A1 [2]
(c) (i) sum of potential energy and kinetic energy of molecules/atoms/particles M1 reference to random (distribution) A1 [2]
(ii) no intermolecular forces so no potential energy B1 (change in) internal energy is (change in) kinetic energy and this is proportional to (change in ) T B1 [2]
13 (a) (i) either random motion or constant velocity until hits wall/other molecule B1 [1]
(ii) (total) volume of molecules is negligible M1 compared to volume of containing vessel A1
or radius/diameter of a molecule is negligible (M1) compared to the average intermolecular distance (A1) [2]
(b) either molecule has component of velocity in three directions or c2 = cX
2 + cY2 + cZ
2 M1 random motion and averaging, so <cX
2> = <cY2> = <cZ
2> M1 <c2> = 3<cX
2> A1 so, pV = ⅓Nm<c2> A0 [3]
(c) <c2> ∝ T or crms ∝ T C1 temperatures are 300K and 373K C1 crms = 580 ms–1 A1 [3] (Do not allow any marks for use of temperature in units of ºC instead of K)
22
14 (a) (i) number of molecules B1 [1]
(ii) mean square speed B1 [1]
(b) (i) 1. pV = nRT C1 n = (6.1 × 105 × 2.1 × 104 × 10–6) / (8.31 × 285) C1 n = 5.4mol A1 [3]
2. either N = nNA
= 5.4 × 6.02 × 1023 C1 = 3.26 × 1024 A1
or pV = NkT N = (6.1 × 105 × 2.1 × 104 × 10–6) / (1.38 × 10–23 × 285) (C1) N = 3.26 × 1024 (A1) [2]
(ii) either 6.1 × 105 × 2.1 × 10–2 = 1/3 × 3.25 × 1024 × 4 × 1.66 × 10–27× <c2> C1 <c2> = 1.78 × 106 C1 cRMS = 1.33 × 103 m s–1 A1 or1/2 × 4 × 1.66 × 10–27 × <c2> = 3/2 × 1.38 × 10–23 × 285 (C1) <c2> = 1.78 × 106 (C1) cRMS = 1.33 × 103 m s–1 (A1) [3]
15 (a) obeys the equation pV = constant × T or pV = nRT M1 p, V and T explained A1
at all values of p, V and T/fixed mass/n is constant A1 [3]
(b) (i) 3.4 × 105× 2.5 × 103
× 10–6 = n × 8.31 × 300 M1 n = 0.34mol A0 [1]
(ii) for total mass/amount of gas 3.9 × 105
× (2.5 + 1.6) × 103× 10–6 = (0.34 + 0.20) × 8.31 × T C1
T = 360K A1 [2]
(c) when tap opened gas passed (from cylinder B) to cylinder A B1 work done on gas in cylinder A (and no heating) M1 so internal energy and hence temperature increase A1 [3]
16 (a) (i) 1. pV = nRT
1.80 × 10–3× 2.60 × 105 = n × 8.31 × 297 C1
n = 0.19 mol A1 [2]
2. ∆q = mc∆T
95.0 = 0.190 × 12.5 × ∆T B1 ∆T = 40 K A1 [2] (allow 2 marks for correct answer with clear logic shown)
(ii) p/T = constant (2.6 × 105) / 297 = p / (297 + 40) M1 p = 2.95 × 105 Pa A0 [1]
(b) change in internal energy is 120 J / 25 J B1 internal energy decreases / ∆U is negative / kinetic energy of molecules decreases M1 so temperature lower A1 [3]
17 (a) (i) N: (total) number of molecules B1 [1]
(ii) <c2>: mean square speed/velocity B1 [1]
(b) pV = ⅓Nm<c2> = NkT (mean) kinetic energy = ½ m<c2> C1 algebra clear leading to ½ m<c2> = (3/2)kT A1 [2]
(c) (i) either energy required = (3/2) × 1.38 × 10–23× 1.0 × 6.02 × 1023 C1
either energy required = 12.5 J (12J if 2 s.f.) A1 [2] or energy = (3/2) × 8.31 × 1.0 (C1)
or energy = 12.5 J (A1)
(ii) energy is needed to push back atmosphere/do work against atmosphere M1
so total energy required is greater A1 [2]
18 (a) (i) sum of kinetic and potential energies of the molecules M1 reference to random distribution A1 [2]
(ii) for ideal gas, no intermolecular forces M1 so no potential energy (only kinetic) A1 [2]
(b) (i) either change in kinetic energy = 3/2 × 1.38 × 10–23× 1.0 × 6.02 × 1023
× 180 C1 = 2240J A1 [2] or R = kNA
energy = 3/2 × 1.0 × 8.31 × 180 (C1) = 2240 J (A1)
(ii) increase in internal energy = heat supplied + work done on system B1 2240 = energy supplied – 1500 C1 energy supplied = 3740J A1 [3]
19 (a) the number of atoms M1 in 12 g of carbon-12 A1 [2]
(b) (i) amount = 3.2/40 = 0.080 mol A1 [1]
(ii) pV = nRT
p × 210 × 10–6 = 0.080 × 8.31 × 310 C1
p = 9.8 × 105 Pa A1 [2] (do not credit if T in °C not K)
(iii) either pV = 1/3 × Nm <c2>
N = 0.080 × 6.02 × 1023 (= 4.82 × 1022)
and m = 40 × 1.66 × 10–27 (= 6.64 × 10–26) C1
9.8 × 105× 210 × 10–6 = 1/3 × 4.82 × 1022
× 6.64 × 10–26× <c2> C1
<c2> = 1.93 × 105
cRMS = 440 m s–1 A1 [3]
or Nm = 3.2 × 10–3 (C1)
9.8 × 105× 210 × 10–6 = 1/3 × 3.2 × 10–3
× <c2> (C1)
<c2> = 1.93 × 105
cRMS = 440 m s–1 (A1)
or 1/2 m<c2> = 3/2 kT (C1)
1/2 × 40 × 1.66 × 10–27 <c2> = 3/2 × 1.38 × 10–23× 310 (C1)
<c2> = 1.93 × 105
cRMS = 440 m s–1 (A1)
(if T in °C not K award max 1/3, unless already penalised in (b)(ii))
20 (a) use of kelvin temperatures B1 both values of (V / T) correct (11.87), V / T is constant so pressure is constant M1 [2]
(allow use of n R1. Do not allow other values of n.)
(b) (i) work done R p∆V
R 4.2 × 105× (3.87 – 3.49) × 103
× 10–6 C1 = = = = = R 160 J A1 [2]
(do not allow use of V instead of ∆V)
(ii) increase / change in internal energy R heating of system N work done on system C1 R 565 – 160 = = = = R 405J A1 [2]
(c) internal energy R sum of kinetic energy and potential energy / EK N EP B1 no intermolecular forces M1 no potential energy (so ∆U R ∆EK) A1 [3]