- Ideal
gas law: PV = nRT = NkT where P = pressure in Pa, V = volume in m3,
n = number of moles of gas, R = 8.31 J K-1 mol-1,
T = temperature in K, k = 1.38 × 10-23 J K-1
- Number
of moles n = mass (g) / molar mass (g mol-1)
- Density
of gas ρ = mass / volume V
- Mass
of gas = N × m where N = number of molecules and m = mass of one
molecule
Using the above equations, you should practice deriving this if it's in the syllabus:
P = 1/3 × ρ<c2>
P = 1/3 × Nm/V × <c2>
PV = 1/3 × Nm × <c2>
NkT = 1/3 × Nm × <c2>
<c2> = 3kT/m
√<c2> = √(3kT/m)
Know
how to work from PV = NkT and P = 1/3 × ρ<c2> to
PV = 1/3 × Nm × <c2> and to √<c2>
= √(3kT/m)...
| (June
2012) (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. [1]
(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.  The pressure p of the gas due to the component cx of velocity is pV = Nmcx2 where m is the mass of a molecule. Explain how the expression leads to the relation pV = 1/3 Nm<c2> where <c2> is the mean square speed of the molecules. [3] (c) The molecules of an ideal gas have a root-mean-square speed of 520 m s-1 at a temperature of 27 ºC. Calculate the rms speed of the molecules at a temperature of 100 ºC. [3] |
(a) (i) The molecules have zero intermolecular forces on one another so do not affect the speeds at which they collide with the walls of the container. [1]
(ii) The molecules have zero volume such that volume of the gas V does not include the volume of the gas. [1]
(b) The mean of the sum of the square of the velocities in the x-direction is
<cx2> = [c1x2 + c2x2 + c3x2 + ... + cNx2]/N [1]
This is one third of the mean of the square of the velocities in directions x-, y- and z-directions. [1]
<cx2> = 1/3 x <c2> [1]
pV = Nm<cx2> = 1/3 × Nm<c2>
(c) pV = 1/3 × Nm<c2>
Also, pV = NkT
NkT = 1/3 × Nm<c2>
<c2> = 3kT/m
√<c2> = √(3kT/m)
At T = 273 + 27 K = 300 K,
√<c2> = 520 m s-1
√(3k x 300/m) = 520
m = 3(1.38 × 10-23) × 300 K / 5202 = 4.993 × 10-26 kg [1]
At T = 273 + 100 = 373 K,
√<c2>
= √(3kT/m)
= √[3(1.38 × 10-23)(373 K)/(4.993 × 10-26)] [1]
= 556 m s-1 [1]
| (Nov
2011) The planet Mars may be considered to be an isolated sphere
of diameter 6.79 × 106 m with its mass of 6.42 x 1023
kg concentrated at its centre. A rock of mass 1.40 kg rests on the
surface of Mars. For this rock, (a) (i) determine its weight [3] (ii) show that its gravitational potential energy is -1.77 × 107 J. [2] (b) Use the information in (a)(ii) to determine the speed at which the rock must leave the surface of Mars so that it will escape the gravitational attraction of the planet. [3] (c) The mean translational kinetic energy of a molecule of an ideal gas is <Ek> = 3/2 × kT where T is the thermodynamic temperature of the gas and k is the Boltzmann constant.
|
(a) (i) F = Gmm/r2 [1]
F = (6.67 × 10-11 N m2 kg-2)(6.42 x 1023 kg)(1.40 kg)/(6.79 × 106 m /2)2 [1]
F = 1.3 N [1]
(ii) GPE = -GMm/r
= -(6.67 × 10-11 N m2 kg-2)(6.42 x 1023 kg)(1.40 kg)/(6.79 × 106 m /2) [1]
= -1.77 × 107 J [1]
(b) 1/2 mv2 = -GPE [1]
v2 = 2 × 1.77 × 107 J / 1.40 kg [1]
v = 5000 m s-1 [1]
(c) (i) <Ek> = 3/2 × kT
1/2 × m<v2> = 3/2 × kT
2u<v2> = 3kT
T = 2u<v2>/3k
T = 2(1.66 × 10-27 kg)(5000 m s-1)2/3(1.38 × 10-23 J K-1) [1]
T = 2000 K [1]
(d) The hydrogen molecules that are near the top of Mar's atmosphere experience low gravitational potential and thus have lower escape speed. [1] Lower temperatures are needed by the molecules to possess lower escape speed. [1]
| (Jun
2008) (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 (rms) speed of neon atoms at (i) 273 K, [3] (ii) 546 K [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 rms speed of changing the density at constant temperature. [2] |
(a) Molecules are in constant motion. [1] They collide with the walls of container. [1] Their change in momentum causes an impact on the walls of container. [1] This gives rise to pressure.
(b) (i) At T = 273 K, P = 1/3 × ρ<c2>
1.02 × 105 Pa = 1/3 × (0.900 kg m-3)<c2> [1]
√<c2> = 583 m s-1 [1]
(ii) PV = 1/3 Nm<c2>
NkT = 1/3 Nm<c2>
<c2> = 3kT/Nm
√<c2> = √(3kT/Nm)
The rms speed is directly proportional to √T.
<c22>/<c12> = T2/T1
<c22>/340000 = 546/273 [1]
√<c22> = 824 m s-1 [1]
(c) P1V1 = P2V2
1/3 × ρ1<c12>V1 = 1/3 × ρ2<c22>V2
ρ2/ρ1 = <c12>V1/<c22>V2
m2/V2 ÷ m1/V1 = <c12>V1/<c22>V2 m2V1/m1V2 = <c12>V1/<c22>V2
m2/m1 = <c12>/<c22>
A change in density due to a change in volume will not have a change in rms speed. [1] A change in density due to an increase in the amount of gas however will increase the rms speed. [1]
Try these questions before you check with the answers below.
- (Nov
2012) An ideal gas has volume V and pressure p.
PV
= 1/3 × Nm<c2>
where
m is the mass of a molecule of the gas.
(a) State the meaning of the symbol (i) N [1] (ii) <c2> [1]
(b) A gas cylinder of volume 2.1 × 104 cm3 contains helium-4 gas at pressure 6.1 × 105 Pa and temperature 12 ºC. Helium-4 may be assumed to be an ideal gas.
- Determine
for the helium gas (1) the amount, in mole, [3] (2) the number of
atoms [2]
- Calculate the root-mean-square speed of the helium atoms. [3]
- (OCR
Jun 2005) The average translational kinetic energy per molecule of
gas 1/2 × m<v2> is related to the absolute
temperature T by 1/2 x m<v2> = 3/2 × kT where k,
the Boltzmann constant, is 1.4 × 10-23 J K-1.
(b) The table shows the root mean square speed of molecules for three gases at 293 K.
| Gas | Mass / 10-27 kg | Rms speed / m s-1 |
| Helium | 6.7 | 1400 |
| Neon | 33 | 600 |
| Carbon dioxide | 47 | 510 |
Propose and carry out an arithmetical test to decide whether the data in the table support the statement that the rms speed is proportional to 1/√m when T is kept constant.
(c) The distribution of molecular speeds in a sample of oxygen gas is as shown.
The area of each bar is proportional to the fraction of molecules having a speed within that range. The chart shows that the molecules have a range of speeds.
Use ideas from kinetic theory to explain why
- gas
molecules have a range of speeds [1]
- gas
molecules are very unlikely to have speeds much greater than the
most common speed. [1]
(i) Use √<c2> = 1.2 × most probable speed to calculate the mean square speed of the molecules in the sample of the gas. [2]
(ii) Use your answer to (d) (i) to estimate the temperature of the gas given the mass of the oxygen molecule is 5.3 × 10-26 kg. [2]
- (OCR Jun 2013) In 2005, the Huygens probe landed on Titan, one of Saturn's Moons. It recorded an atmosphere of nitrogen at a temperature of 94 K. Calculate the root-mean-square speed of a nitrogen molecule at this temperature given that the mass of a nitrogen molecule is 4.7 × 10-26 kg and k = 1.4 × 10-23 J K-1. [3]
- (OCR
Jan 2011) The air in a football has a mass of 1.1 × 10-2
kg.
(b) Calculate the mean square speed <c2> of the particles in the football if the pressure of air in the ball is 1.7 × 105 Pa and volume is 5.4 × 10-3 m3. [2]
(c) The temperature of the air in the football increases during the game. Sketch a graph to show how the mean square speed of air particles in the football varies with absolute temperature T. Assume the air behaves as an ideal gas. [1]
- (Jun
2004) The pressure p of an ideal gas is given by the expression p =
1/3 × Nm/V × <c2>.
(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.
- Determine
the root-mean-square (rms) speed of the gas atoms at 300 K. [3]
- Calculate
the temperature of the gas for the atoms to have an rms speed that
is twice that calculated in (i). [3]
Answers: 1. (a) (i) N is the number of gas molecules. (ii) <c2> is the mean of the square of the speeds of molecules. (b) (i) (1) PV = nRT ==> n = PV/RT = (6.1 × 105 Pa)(2.1 × 104 × 10-6 m3)/(8.31 J K-1 mol-1)(273 + 12 K) = 5.2 moles (2) Number of atoms = number of moles x Avogadro's constant = 5.2 mol × 6.02 × 1023 mol-1 = 3.2 × 1024 (ii) <c2> = 3PV/(Nm) ==> √<c2> = √[3PV/(Nm)] = √[3(6.1 × 105 Pa)(2.1 × 104 × 10-6 m3)/(3.2 × 1024 mol)(4 x 1.66 x 10-27 kg)] = 1300 m s-1
2. (a) 1/2 x m<v2> = 3/2 × kT ==> <v2> = 3kT/m ==> √<v2> = √(3kT/m). Therefore, √<v2> is inversely proportional to √m. (b) √<v2> = √(3kT/m) ==> √<v2> √m = √(3kT). √<v2> √m should be constant if √<v2> is proportional to 1/√m. Helium: √[(6.7 × 10-27 kg)(1400 m s-1)] = 3.1 × 10-12. Neon: √[(633 × 10-27 kg)(600 m s-1)] = 1.9 × 10-11. Carbon dioxide: √[(47 × 10-27 kg)(510 m s-1)] = 4.9 × 10-12. √(<v2> m) are different for each gas. (c) (i) Different molecules possess different kinetic energies. (ii) Gas molecules are not likely to possess kinetic energies much greater than their surrounding molecules. (c) (i) √<c2> = 1.2 × most probable speed = 1.2 × 400 m s-1 = 480 m s-1 ==> <c2> = 230 400 m2 s-2 (ii) 1/2 × m<v2> = 3/2 × kT ==> 1/2 × 5.3 × 10-26 kg × 230 400 m2 s-2 = 3/2 x 1.38 x 10-23 J K-1 x T ==> T = 295 K
3. PV = 1/3 × Nm<c2> ==> NkT = 1/3 x Nm<c2> ==> √<c2> = √(3kT/m) = √[3(1.4 × 10-23 J K-1)(94 K)/(4.7 × 10-26 kg)] = 289 m s-1
4. (a) Number of molecules N = (1.1 × 10-2 kg)(6.02 × 1023 mol-1)/(2.9 × 10-2 kg mol-1) = 2 × 1023 (shown) (b) PV = 1/3 × Nm<c2> ==> <c2> = 3PV/Nm = 3(1.7 × 105 Pa)(5.4 × 10-3 m3)/(1.1 × 10-2 kg) = 250 000 m2 s-2 (c) PV = 1/3 Nm<c2> ==> NkT = 1/3 Nm<c2> ==> <c2> = 3kT/m. Therefore, the graph of <c2> is directly proportional to T.
5. (a) <c2> is the mean of the square of the speeds of particles. <c2> = (c12 + c22 + c32 + ... cN2)/N (b) (i) P = 1/3 × ρ<c2> ==> 1.02 × 105 Pa = 1/3 × (2.4 kg m-3)<c2> ==> √<c2> = 360 m s-1 (ii) PV = 1/3 Nm<c2> ==> NkT = 1/3 Nm<c2> ==> <c2> = 3kT/Nm ==> √<c2> = √(3kT/Nm). The rms speed is directly proportional to √T. √<c22>/√<c12> = √T2/√T1 ==> 2/1 = √T2/√300 ==> √T2 = 2 × √300 ==> T2 = 1200 K
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