Ideal Gases
JUNE 2002
1
(a) (i)
The kinetic theory of gases leads to the equation
m = kT . Explain the significance of the quantity m . ................................................................................................................................... ................................................................................................................................... (ii)
Use the equation to suggest what is meant by the absolute zero of temperature. ................................................................................................................................... ................................................................................................................................... [3]
(b) Two insulated gas cylinders A and B are connected by a tube of negligible volume, as shown sho wn in Fig. 3.1. tap
cylinder A
cylinder B
Fig.. 3.1 Fig
1
Each cylinder has an internal volume of 2.0 × 10–2 m3. Initially, the tap is closed and cylinder A contains contain s 1.2 mol of an ideal gas at a temperature temperature of 37 °C. Cylinder Cylinder B contains the same 5 ideal gas at pressure 1.2 × 10 Pa and temperature 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 fact that the total amount of gas is constant, determine the final pressure of the gas in the cylinders.
pressure = ........................................ Pa [6]
2
Each cylinder has an internal volume of 2.0 × 10–2 m3. Initially, the tap is closed and cylinder A contains contain s 1.2 mol of an ideal gas at a temperature temperature of 37 °C. Cylinder Cylinder B contains the same 5 ideal gas at pressure 1.2 × 10 Pa and temperature 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 fact that the total amount of gas is constant, determine the final pressure of the gas in the cylinders.
pressure = ........................................ Pa [6]
2
NOVEMBER 2003
2
The volume of some air, air, assumed to be an ideal gas, in in the cylinder of a car engine is 3 5 540 540 cm at a pressure of 1.1 × 10 Pa and a temperature of 27 °C. °C. The air is suddenly compressed, so that no thermal energy enters or leaves the gas, to a volume of 30 cm3. The pressure rises to 6.5 × 106 Pa. (a) Determine the temperature of the gas after the compression.
temperature = …………………………… K [3] (b) (i)
State and explain explain the first first law of thermodynamics. ................................................................................................................................... ................................................................................................................................... ................................................................................................................................... .............................................................................................................................. [2]
(ii) Use the law to explain why the temperature of the air changed during the compression.
................................................................................................................................... ................................................................................................................................... ................................................................................................................................... ................................................................................................................................... ...............................................................................................................................[4]
3
JUNE 2004
3
The pressure p of an ideal gas is given by the expression p =
1 Nm 2 . 3 V
(a) Explain the meaning of the symbol .
.......................................................................................................................................... .................................................................................................................................... [2] (b) The ideal gas has a density of 2.4 kg m–3 at a pressure of 2.0 × 105 Pa and a temperature 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 is twice that calculated in (i).
temperature = ......................................... K [3]
4
JUNE 2005
4
(a) State what is meant by an ideal gas.
.......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[2] (b) The product of pressure p and volume V of an ideal gas of density ρ at temperature T is
given by the expressions p = ρ
and
pV = NkT ,
where N is the number of molecules and k is the Boltzmann constant. (i)
State the meaning of the symbol . ...............................................................................................................................[1]
(ii)
Deduce that the mean kinetic energy E K of the molecules of an ideal gas is given by the expression E K = kT .
[2] (c) In order for an atom to escape completely from the Earth’s gravitational field, it must
have 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, assumed to be an ideal gas, could escape from the Earth. The mass of a helium atom is 6.6 × 10 –27 kg.
temperature = ....................................... K [2] (ii)
Suggest why some helium atoms will escape at temperatures below that calculated in (i). ................................................................................................................................... ...............................................................................................................................[1]
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 is 2.9 × 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.
.......................................................................................................................................... .......................................................................................................................................... ..................................................................................................................................... [2] (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 to 3.4 × 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.
.....................................................................................................................................
.......................................................................................................................................... 2.
.....................................................................................................................................
..................................................................................................................................... [2] (b) A gas cylinder contains 4.00 × 104 cm3 of hydrogen at a pressure of 2.50 × 107 Pa and a temperature of 290 K.
The cylinder is to be used to fill balloons. Each balloon, when filled, contains 7.24 × 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.
..........................................................................................................................................
..........................................................................................................................................
..........................................................................................................................................
......................................................................................................................................[3] (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.
..........................................................................................................................................
..........................................................................................................................................
......................................................................................................................................[2] 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. .................................................................................................................................. .................................................................................................................................. ............................................................................................................................ [2]
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 = 13 ρ c 2
where ρ is the density of the gas.
(i)
State the meaning of the symbol c 2. .................................................................................................................................. ..............................................................................................................................[1]
(ii)
Use the expression to show that the mean kinetic energy of the atoms of an ideal gas is given by the expression = 32 kT . Explain any symbols that you use. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. ............................................................................................................................. [4]
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 N A. .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[2]
(b) A balloon is filled with helium gas at a pressure of 1.1 × 105 Pa and a temperature of 25 °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 . .......................................................................................................................................... .......................................................................................................................................... .................................................................................................................................... [2]
(b) Two containers A and B are joined by a tube of negligible volume, as illustrated in Fig. 2.1.
container A
container B
3.1 × 103 cm3
4.6 × 103 cm3
17°C
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. ...................................................................................................................................... .......................................................................................................................................... 2. ...................................................................................................................................... .......................................................................................................................................... [2]
(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 = 13 Nm
where m is the mass of a gas molecule.
(i)
State the meaning of the symbol 1.
N ,
..............................................................................................................................[1] 2.
.
..............................................................................................................................[1]
(ii)
Use the expression to deduce that the mean kinetic energy of a gas molecule at temperature T is given by the equation = 32 kT where k is a constant.
[2]
14
(c) (i)
State what is meant by the internal energy of a substance. .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[2]
(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. .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[2]
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, .................................................................................................................................. ............................................................................................................................. [1]
(ii)
their volume. .................................................................................................................................. .................................................................................................................................. ............................................................................................................................. [2]
16
(b) A cube of volume V contains N molecules of an ideal gas. Each molecule has a component c X of velocity normal to one side S of the cube, as shown in Fig. 2.1. side S
c x
Fig. 2.1 The pressure p of the gas due to the component c X of velocity is given by the expression pV = Nmc X2
where m is the mass of a molecule. Explain how the expression leads to the relation pV = 13 Nm
where 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 = 13 Nm
where m is the mass of a molecule of the gas.
(a) State the meaning of the symbol (i)
N ,
..............................................................................................................................[1]
(ii)
. ..............................................................................................................................[1]
(b) A gas cylinder of volume 2.1 × 10 4 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 . .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[3]
(b) Two cylinders A and B are connected by a tube of negligible volume, as shown in Fig. 2.1.
cylinder A tap T
cylinder B
2.5 × 103 cm3 3.4 × 105 Pa
1.6 × 103 cm3
300K
4.9 × 105 Pa
tube 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.
.......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[3]
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 gas 1.80 × 10–3 m3 2.60 × 105 Pa 297 K
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. .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ..................................................................................................................................... [3]
NOVEMBER 2013
17 The product of the pressure p and the volume V of an ideal gas is given by the expression pV =
1 Nm 3
where m is the mass of one molecule of the gas.
(a) State the meaning of the symbol
(i)
N ,
..............................................................................................................................[1]
(ii)
. ..............................................................................................................................[1]
(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). .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[2]
NOVEMBER 2013
18 (a) (i)
State what is meant by the internal energy of a system. .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[2]
(ii)
Explain why, for an ideal gas, the internal energy is equal to the total kinetic energy of the molecules of the gas. .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[2]
(b) The mean kinetic energy of a molecule of an ideal gas is given by the expression
= 32 kT 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. .................................................................................................................................. ..................................................................................................................................
total energy = ............................................. J [3]
JUNE 2014
19 (a)
Explain what is meant by the Avogadro constant. ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2]
(b) Argon-40 ( 40 18Ar) 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.
3.49 × 103 cm3 21.0°C
3.87×103 cm3 565J
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.
................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [3]
Ideal Gases
1
(a)
(i)
mean kinetic energy ............................................................................................... Ml of the atoms / molecules / particles ...........................................................................A1 at absolute zero, atoms have no kinetic energy ............................................ Bl
[3]
p V = n R T .............................................................................................................. Cl 5 -2 n = (1.2 x 10 x 2.0 x 10 )/(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]
(ii) (b)
2
(i)
(a)
pV/T = constant............................................................................ C1 T = (6.5 x 106 x 30 x 300)/(1.1 x 10 5 x 540)................................. C1 = 985 K .................................................................................... A1 (if uses C, allow 1/3 marks for clear formula)
[3]
°
(b)
(i)
(ii)
3
q + w symbols identified correctly ..........................................................M1 directions correct.......................................................................... A1
(i)
[2]
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’ )
mean (value of the) square of the speeds (velocities) of the atoms/particles/molecules
(a)
(b)
∆U =
p =
1
2
ρ < c
[4]
M1 A1
[2]
C1 >
3 2
(ii)
= 3 x 2 x 10 5/2.4 = 2.5 x 10 5 r.m.s speed = 500 ms -1 new = 1.0 x 106 or < c 2> increases by factor of 4 ∝ T or 3/2 kT = 1/2 m 6 5 T = {(1.0 x 10 ) / (2.5 x 10 )} x 300 = 1200 K
C1 A1 C1 C1 A1 Total
18
[3]
[3] [8]
4 (a)
(b) (i)
obeys the law pV /T = constant or any two named gas laws at all values of p, V and T or two correct assumptions of kinetic theory of ideal gas (B1) third correct assumption (B1)
M1 A1
[2]
mean square speed
B1
[1]
M1 M1 A0
[2]
C1 A1
[2]
B1
[1]
(ii) mean kinetic energy = ½m ρ = Nm/V and algebra leading to [do not allow if takes N = 1] ½m = 3/2 kT (c) (i)
½ × 6.6 × 10-27 × (1.1 ×104)2 = 3/2 × 1.38 × 10-23 ×T 4 T = 1.9 × 10 K
(ii) Not all atoms have same speed/kinetic energy
5
(a)
(b)
obeys the law pV = constant × T ………………………..……………….. at all values of p, V and T ……………………………………………….
M1 A1
[2]
n = (2.9 × 105 × 3.1 × 10–2) / (8.31 × 290) …..………………..………... = 3.73 mol ……………………………………………………………….
C1 A1
[2]
C1 C1 A1
[3]
at new pressure, nn
(c)
= 3.73 ×
3. 4 2.9
290 ×
300
= 4.23 mol ….………………………………………. change = 0.50 mol ……………………………………………………….… number of strokes = 0.50 / 0.012 = 42 (must round up for mark) ……. 6
(a) e.g. fixed mass/ amount of gas ideal gas (any two, 1 each) (b) (i)
n = pV / RT = (2.5 × 107 × 4.00 × 104 x 10-6) / (8.31 × 290) = 415 mol
(ii) volume of gas at 1.85 × 105 Pa = (2.5 × 107 × 4.00 × 104) / (1.85 × 105) = 5.41 × 106 cm3 so, 5.41 × 106 = 4.00 × 104 + 7.24 × 103 N N = 741 (answer 740 or fails to allow for gas in cylinder, max 2/3)
19
B2
[2]
C1 C1 A1
[3]
C1 C1 A1
[3]
7
(a) molecule(s) rebound from wall of vessel / hits walls change in momentum gives rise to impulse / force either (many impulses) averaged to give constant force / pressure the molecules are in random motion or
(b) (i)
p =
� �
2
ρ
1.02 × 105 =
B1
[3]
C1 � �
×
0.900 ×
= 3.4 × 105 c RMS = 580 m s–1 (ii) either ∝ T or = 2 × 3.4 ×105 c RMS = 830 m s–1 (allow 820)
(c) c RMS depends on temperature (alone) so no effect
8
B1 B1
C1 A1
[3]
C1 A1
[2]
B1 B1
[2]
(a) either pV = NkT or pV = nRT and n = N / N .....................................................C1 A 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 (allow 1 mark for calculation of n = 0.467 mol )
(b) (i) volume = (1.2 × 10-10)3 × 2.8 × 1023 or = 4.8 × 10-7 m3
4 3
[2]
r 3 × 2.8 × 1023 ..............................C1
π
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 + 273.2 300.4 K
C1 A1
[2]
A1
[1]
B1
[1]
B1 B1 B1 B1
[4]
C1 A1
[2]
(ii) mean kinetic energy = 3/2 kT = 3/2 × 1.38 × 10–23 × 290 = 6.0 × 10–21 J
C1 A1
[2]
(iii) realisation that total internal energy is the total kinetic energy energy = 6.0 × 10–21 × 68 × 6.02 × 1023 = 2.46 × 105 J
C1 C1 A1
[3]
(ii) 11.6 K (b) (i) ( is the) mean / average square speed (ii)
= Nm/V with N explained so, pV = 1/3 Nm and pV = NkT with k explained so mean kinetic energy / < E K> = ½m = 3/2 kT
ρ
(c) (i) pV = nRT 2.1 × 107 × 7.8 × 10–3 = n × 8.3 × 290 n = 68 mol
10 (a) number of atoms of carbon-12 in 0.012 kg of carbon-12
M1 A1
(b) pV = NkT or pV = nRT substitutes temperature as 298 K either 1.1 × 105 × 6.5 × 10–2 = N × 1.38 × 10–23 × 298 1.1 × 105 × 6.5 × 10–2 = n × 8.31 × 298 and n = N / 6.02 × 1023 or 24 N = 1.7 × 10
C1 C1 C1 A1
11 (a) amount of substance containing same number of particles as in 0.012 kg of carbon-12
M1 A1
(b) pV = nRT amount = (2.3 × 10 5 × 3.1 × 10–3) / (8.31 × 290) + (2.3 × 105 × 4.6 × 10–3) / (8.31 × 303) = 0.296 + 0.420 = 0.716 mol (give full credit for starting equation pV = NkT and N = nN A)
C1
21
[2]
C1 C1 A1
[4]
[2]
[4]
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)
(b) (i) 1. 2.
B2
[2]
number of (gas) molecules
B1
[1]
mean square speed/velocity (of gas molecules)
B1
[1]
(ii) either pV = NkT or pV = nRT and links n and k 2 and = ½m clear algebra leading to =
3 2
M1
kT
(c) (i) sum of potential energy and kinetic energy of molecules/atoms/particles reference to random (distribution) (ii) no intermolecular forces so no potential energy (change in) internal energy is (change in) kinetic energy and this is proportional to (change in ) T
13 (a) (i) either or
random motion constant velocity until hits wall/other molecule
(ii) (total) volume of molecules is negligible compared to volume of containing vessel or radius/diameter of a molecule is negligible compared to the average intermolecular distance
(b) either molecule has component of velocity in three directions or c 2 = c X2 + c Y2 + c Z2 random motion and averaging, so = = = 3 so, pV = ⅓Nm
(c) ∝ T or c rms ∝ � temperatures are 300 K and 373 K c rms = 580ms–1 (Do not allow any marks for use of temperature in units of ºC instead of K)
22
B1
A1
[2]
M1 A1
[2]
B1 B1
[2]
[1]
M1 A1 (M1) (A1)
[2]
M1 M1 A1 A0
[3]
C1 C1 A1
[3]
14 (a) (i)
number of molecules
(ii) mean square speed
(b) (i) 1. pV = nRT n = (6.1 × 10 5 × 2.1 × 104 × 10–6) / (8.31 × 285) n = 5.4 mol 2.
either N = nN A = 5.4 × 6.02 × 1023 = 3.26 × 1024 or pV = NkT N = (6.1 × 105 × 2.1 × 104 × 10–6) / (1.38 × 10 –23 × 285) N = 3.26 × 1024
(ii) either 6.1 × 105 × 2.1 × 10–2 = 1/3 × 3.25 × 1024 × 4 × 1.66 × 10 –27× = 1.78 × 106 c RMS = 1.33 × 103 m s–1 or 1 /2 × 4 × 1.66 × 10 –27 × = 3/2 × 1.38 × 10–23 × 285 = 1.78 × 106 c RMS = 1.33 × 103 m s–1
15 (a) obeys the equation pV = constant × T or pV = nRT p, V and T explained at all values of p, V and T /fixed mass/n is constant
(b) (i) 3.4 × 105 × 2.5 × 103 × 10–6 = n × 8.31 × 300 n = 0.34mol (ii) for total mass/amount of gas 3.9 × 105 × (2.5 + 1.6) × 103 × 10–6 = (0.34 + 0.20) × 8.31 × T T = 360K
(c) when tap opened gas passed (from cylinder B) to cylinder A work done on gas in cylinder A (and no heating) so internal energy and hence temperature increase
B1
[1]
B1
[1]
C1 C1 A1
[3]
C1 A1
(C1) (A1)
[2]
C1 C1 A1 (C1) (C1) (A1)
[3]
M1 A1 A1
[3]
M1 A0
[1]
C1 A1
[2]
B1 M1 A1
[3]
16 (a) (i)
1. pV = nRT 1.80 × 10–3 × 2.60 × 105 = n × 8.31 × 297 n = 0.19 mol
C1 A1
[2]
B1 A1
[2]
M1 A0
[1]
B1 (b) change in internal energy is 120 J / 25 J internal energy decreases / ∆U is negative / kinetic energy of molecules decreases M1 so temperature lower A1
[3]
2.
∆q =
mc ∆T 95.0 = 0.190 × 12.5 × ∆T ∆T = 40 K (allow 2 marks for correct answer with clear logic shown)
(ii) p/T = constant (2.6 × 105) / 297 = p / (297 + 40) p = 2.95 × 105 Pa
N : (total) number of molecules
B1
[1]
(ii) : mean square speed/velocity
B1
[1]
(b) pV = ⅓Nm = NkT (mean) kinetic energy = ½ m algebra clear leading to ½ m = (3/2)kT
C1 A1
[2]
17 (a) (i)
(c) (i) either either or or
energy required = (3/2) × 1.38 × 10–23 × 1.0 × 6.02 × 1023 energy required = 12.5 J (12J if 2 s.f.) energy = (3/2) × 8.31 × 1.0 energy = 12.5 J
(ii) energy is needed to push back atmosphere/do work against atmosphere so total energy required is greater
18 (a) (i)
sum of kinetic and potential energies of the molecules reference to random distribution
(ii) for ideal gas, no intermolecular forces so no potential energy (only kinetic) (b) (i)
either change or
in kinetic energy = 3/2 × 1.38 × 10–23 × 1.0 × 6.02 × 1023 × 180 = 2240J
C1 A1 (C1) (A1)
M1 A1
[2]
[2]
M1 A1
[2]
M1 A1
[2]
C1 A1
[2]
R = kN A
energy = 3/2 × 1.0 × 8.31 = 2240 J
×
180
(ii) increase in internal energy = heat supplied + work done on system 2240 = energy supplied – 1500 energy supplied = 3740J
(C1) (A1) B1 C1 A1
[3]
