AQA A2 Physics P20 Gases Kerboodle Answers

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C20.1 The experimental gas laws AQA A2 Physics P20 Gases Kerboodle Answers : Page No. 319

1 126 kPa 2 79 kPa 3 0.097 m³ 4 2.3 x 10^-5 m³, 1600 kg m^-3

20.2 The ideal gas law AQA A2 Physics P20 Gases Kerboodle Answers  : Page No.  322

1 a the maximum number of moles of gas that can be contained by this cylinder at 50 cc, 1.1 moles b the pressure in the cylinder of this amount of gas at ID oc. 109kPa 2 a the number of moles of gas present, 9.3 x 10^-4 moles b the volume of this gas at OOC and 101 kPa. 2.1 x 10^-5 m³ 3 b The molar mass of the gas in a is 0.03? kg mol-I, Calculate the density of the gas. 1.3 kg m^-3 4 The molar mass of air is 0.029 kg mol-I. a 1.2 kg m^-3 b 2.5 x 10²²

20.3 The kinetic theory of gases AQA A2 Physics P20 Gases Kerboodle Answers : Page No.  326

1 a The pressure law states that for a constant volume of gas in a sealed container the temperature of the gas is directly proportional to its pressure. This can be easily understood by visualising the particles of gas in the container moving with a greater energy when the temperature is increased. This means that they have more collisions with each other and the sides of the container and hence the pressure is increased. If V is a constant, the P/T will be a constant — Pressure Law where V= Volume, P= Pressure and T= Temperature. b i the number of moles of oxygen in the cylinder, 0.97 mol ii the total kinetic energy of all the gas molecules in the container. 4.5 kJ 2 For a hydrogen molecule (molar mass= 0.002 kg mol^-1) at 0⁰C, calculate: a its mean kinetic energy b its root mean square speed. 3 a The temperature would be the same for both types of molecules as they’re in the same sample of air. The mean kinetic energy of any gas molecule = (3/2)kT, where k is a constant and T is the temperature, therefore the mean kinetic energy must be the same for both types of molecule as T is the same b Divide the Crms speed of a nitrogen molecule by that of the oxygen molecule. This will give you a value of 1.069, which will allow to explain that the Crms speed of the nitrogen molecule is therefore around 1.07 times greater than that of the oxygen molecule. 4 a a the number of moles, 1.48mol b the mass of gas present, 4.2 x 10^-2 kg c the root mean square speed of the molecules. 5.2 x 10² ms^-1 Banner 2 Banner 2

Practice questions: Page No.  327-329

1 V_esc = √2gR = (2 × 1.62 ×1740 × 10³)½ = 2400 m s^-1 (2370 m s–1 to 3 sig figs) (b) (i) Mean kinetic energy = 3/2 KT = 1.5 × 1.38 × 10–23 × 400 = 8.3 × 10^-21 J (ii) Mass of an oxygen molecule m= molar mass/NA (iii) Gas molecules have a range of speeds (or kinetic energies).

• The escape speed from the lunar surface is about 4 × the rms speed of the gas molecules.
• Some gas molecules would have sufficient kinetic energy (or speed) to escape from the lunar surface.
• Each day more gas molecules would escape.
• As they would gain kinetic energy from the hot surface.

(c) (i) Why did astronomers conclude that the absorption lines were due to the planet rather than the star?

• Absorption lines are due to the absorption of light of certain wavelengths.
• The lines were observed only when the planet passed across the face of the star.
• If the star had caused the absorption lines, they would have been present all the time (∴ the planet must have absorbed light at these wavelengths).

(ji) Either:

• There would be less absorption by a smaller planet.
• As it would cover a smaller % of the star’s surface when it is in front of the star.
• So the absorption lines would be harder to see.

Or:

• It might not have retained an atmosphere.
• Because the escape velocity of molecules from a smaller planet would be smaller.
• The mean kinetic energy of the molecules (in its atmosphere) would be the same.

2 (a) (i) On the same axes sketch two additional curves A and B, if the following changes are made: (ii) a curve labelled A that is the same shape as but lower than O at all values (iii’ a curve labelled B that is the same shape as but higher than O at all values. (b) (i) the amount of gas in moles, in the cylinder. pV = nRT ∴ 130 × 10³ × 0.20 = n × 8.31 × 290 …from which amount of gas n = 10.8 mol (ii) Average kinetic energy of a molecule (iii) the total kinetic energy 01 the molecules in the cylinder. (5 marks) AQA, 2005 Number of molecules in cylinder = 10.8 × 6.02 × 10²³ = 6.50 × 10²⁴ ∴ Total kinetic energy of the molecules = 6.50 × 10²⁴ × 6.00 × 10²¹ = 3.90 × 10⁴ J 3 (a) The equation of state for an ideal gas is pV = nRT (where p = pressure, V = volume, T = absolute temperature, n = amount of gas in mol, and R = molar gas constant). (b) Sketch a graph 10 show how the pressure, p, of the gas varies with the absolute temperature, T. of the gas. Graph on axes labelled pressure and temperature to show:

• a straight line of positive gradient
• that passes through an origin marked (0, 0).

(c) (b) varies with the absolute temperature. (4 marks)

• When the temperature is increased the average kinetic energy of the molecules (or their speed) increases.
• There are more collisions per second against the walls containing the gas.
• The average change in momentum in each collision is greater…
• leading to a greater rate of change of momentum.
• Force = rate of change of momentum,

and pressure so there is a greater pressure on the walls. (d) Average kinetic energy of a molecule 4 (a) p = pressure and V = volume N = number of molecules m = mass of one molecule (or particle) crms² = mean square speed of molecules (b) One assumption used in the derivation or the equation stated in parl is that molecules are in a state of random motion, (i Molecules have a range of speeds…and no preferred direction of motion. (ii) • the collisions are perfectly elastic

• intermolecular forces are negligible (except during collisions)
• the volume of the molecules is negligible (compared to volume of container)
• the duration of collisions is negligible (compared to the time between collisions)
• the molecules are identical
• a gas consists of a very large number of molecules
• the molecules behave according to the laws of Newtonian mechanics.

(c)

• The gas particles collide with the walls of the container.
• Each collision changes the direction/momentum of the particle.
• The change in momentum causes a force to be exerted on the wall (Newton’s 2nd law)
• Pressure = force/area, so exerting force over an area of the wall results in pressure.
• The smaller the area, the larger the pressure.
• The larger the speed (therefore force) of the particles, the larger the pressure.
• Random motion of particles results in even pressure over the walls of the container.

5 (a) (i) sea level, (3 marks) At sea level: pV = nRT gives 1.0 × 105 × 1.0 = n × 8.31 × 300 from which n = 40.1 mol (ii) 10000m. At 10 000 m: pV = nRT gives 2.2 × 10⁴ × 1.0 = n × 8.31 × 270 from which n = 9.81 mol (b) Total number of gas molecules in 1.0 m³ of air at sea level: = 40.1 × 6.02 × 10²³ = 2.41 × 10²⁵ and at 10 000 m: = 9.81 × 6.02 × 10²³ = 5.91 × 10²⁴ ∴ extra number of molecules at sea level: = (2.41 × 10²⁵) − (5.91 × 10²⁴) = 1.82 × 10²⁵ Number of extra oxygen molecules = 0.23 × 1.82 × 10²⁵ = 4.19 × 10²⁴ 6 (a) (i) Change in momentum of molecule in an elastic collision = mv – (–mv) = 2mv = 2 × 6.8 × 10^-27 × 1300 = 1.77 × 10^-23 N s (or kg m s^-1) (ii) (b) (i) (ii) 7 (a) (b) Taking any pair of values from the line e.g. V = 14.0 cm³ when θ = 100 °C converting to absolute temperature: 100 °C = 100 + 273 = 373 K and substituting values into pV = nRT gives 1.0 × 10⁵ × 14.0 × 10^-6 = n × 8.31 × 373 from which n = 4.52 × 10−4 mol and mass of gas used in experiment = 4.52 × 10^-4 × 0.044 = 1.99 × 10^-5 kg (c) (i) (ii) Pressure is caused by the collisions of the molecules against the container

• pressure is the change in momentum of the molecules per unit area per second.

In (i) at constant temperature:

• when volume increases the molecules have further to travel between collisions
• there are fewer collisions per second against the container walls.

In (ii) at constant volume:

• When the temperature is reduced the molecules have a lower mean speed (or kinetic energy)
• the change in momentum (or impulse) per collision is reduced.
• there are fewer collisions per second with the walls.

8 (a) (i) pV = nRT ∴ 500 × 10³ × V = 15 × 8.31 × 290 gives V = 7.23 × 10^-2 m³ (ii) Average kinetic energy of a gas molecule (b) pV = nRT ∴ 420 × 10³ × 7.23 ×10^-2 = n × 8.31 × 290 gives amount of gas remaining in cylinder n = 12.6 mol (c) Pressure is caused by the collisions of the molecules against the container

• pressure is the change in momentum of the molecules per unit area per second
• when gas is removed there are fewer molecules in the cylinder
• there are fewer collisions per second against the container walls
• the impulse per collision is, on average, the same as it was originally
• since cms² is constant (because the temperature is unchanged).

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