1 Heat and Gases Chapter 5 Gases

5 Gases The final pressure inside the can is 134 kPa. 11 (a) The marshmallow becomes larger. Practice 5.1 (p.167) (b) Since the gas pressure decreases and the 1 B temperature remains unchanged, by 2 A Boyle’s law, the volume of the gas 3 C trapped inside the marshmallow By pressure law, increases. This makes the marshmallow p p 1 = 2 larger. T T 1 2 12 (a) From the graph, when the pressure is p1 75 1 p2 =  T2 =  600 = 150 kPa 120 kPa, is 100 m–3. T1 300 V 4 B 1  Volume of the gas = = 0.01 m3 5 B 100 3 6 B (b) When p1 = 120 kPa, V1 = 0.01 m .

By pV = nRT, p1V1 = 120  1000  0.01 1 3 pV 105 103  = 1200 Pa m n = = 0.70 = 62 mol 1 RT –3 8.31 (20  273) When p2 = 150 kPa, = 125 m , V2 p1 p2 7 By pressure law, = 1 3 V2 = = 0.008 m . T1 T2 125 T1 T2 =  p2 p2V2 = 150  1000  0.008 p1 = 1200 Pa m3 20  273 3 = 100 10 = p V 0.9 105 1 1 13 (a) By Boyle’s law, = 325.6 K p V = p V = 52.6 C 1 1 2 2 V1 8 The atmospheric pressure supports the water. p2 = p1 V2 9 (a) Increases (b) By pressure law, (b) The pressure in the cooker will be too p2 p3 high and explosion may occur. = T1 T2 10 By the general gas law, T p = 2 p p1V1 p2V2 3 2 = T1 T T 1 2 V T = 1 2 p p1V1 T2 V T 1 p2 =  2 1 T1 V2 (c) By the general gas law, 3 50 10 V1 200  273  p1V1 p3V2 = = 100  273 1.02V1 T1 T2 = 134 kPa

New Senior Secondary Physics at Work (Second Edition) 1  Oxford University Press 2015 1 Heat and Gases Chapter 5 Gases

V1T2 (b) Consider container X and the large p3 = p1 V2T1 container (formed by X and Y). By pV = nRT, 14 By pV = nRT, p V p V X X = X Y X Y p V pYVY n T n T X X = X X X Y X Y n T X X nY TY Since both containers are at the same Since p is constant, temperature T, 1.2 0.9 (0.625p )V p (2V ) = Y X = X Y X nX (20  273) nY (25  273) 1.5 1.5  2.4 n X = 1.36 pX + Y = 0.8125 pY nY ≈ 0.813 pY

Since N  n, NX : NY = nX : nY = 1.36 : 1 The pressure of the two containers is

15 0.813pY after the tap opens. (c) Percentage change p  p = X Y X  100% pX 0.8125p  0.625p = Y Y  100% 0.625pY For a VT of an ideal gas, the intercept on the = 30% temperature axis is always equal to the absolute zero. Practice 5.2 (p.182) nR By pV = nRT, we have V = T. 1 C p 2 C nR The slope of the V–T graph is . 3 D p 4 B When the amount of gas (n) is halved and the 5 Root-mean-square speed pressure (p) is doubled, the slope becomes one 3RT fourth of the original slope. = mN A 16 (a) By pV = nRT, 3 8.31 (100  273) pXVX pYVY = = 3.34 1027  6.02 1023 nX TX nY TY = 2150 m s–1 Since V and T are the same for both 6 (a) The smoke particle move in random containers, motion. nX  1.5  pX = pY =   pY = 0.625pY nY  2.4  (b) It is bombarded by a large amount of air The pressure in container X before the molecules around it. The bombardments come from all sides but not in equal tap opens is 0.625pY. numbers. This results in a random motion

2 New Senior Secondary Physics at Work (Second Edition)  Oxford University Press 2015 1 Heat and Gases Chapter 5 Gases

(c) It move slower. total kinetic energy of one mole of 1 1 3 7 (a) Increases molecules = Nmc2  = RT 2 n 2 (b) When the temperature increases, the air Revision exercise 5 molecules move faster. As the volume is Concept traps (p.185) fixed, the molecules hit the walls more 1 F often. Also, each collision will give a Internal energy of an ideal gas is equal to its greater change in momentum due to the 3 increased speed. Therefore, the air total molecular KE, which is given by nRT. 2 pressure inside the can increases. 2 F 8 The number of air molecules inside the tyre The assumption includes that the molecules increases, so there are more frequent collide among themselves and with the walls bombardments on the tyre wall. As a result, of the container. All these collisions are the pressure of the tyre increases. perfectly elastic. 9 Root-mean-square speed 3RT = Multiple-choice questions (p.185) mN A 3 A 3  8.31 (25  273) = 4 C 0.0337 5 D 1 = 470 m s 3RT 3 crms =  T 10 Since total KE = nRT, mN 2 A increase in total KE the ratio of crms at 80 C to that at 20 C 3 7.28 1024 273 80 =  8.31 (80  25) = 8290 J = 2 6.02 1023 273 20 11 (a) Triples = 1.10 (b) Increases to 9 times the original value 6 A (c) Remains unchanged 7 B 12 (a) m is the mass of a gas molecule. 8 B Nm is the mass of all gas molecules. Suppose the temperature is constant when the 1 (b) Consider pV = Nmc2 and pV = nRT, ideal gas changes from X to Z. We have the 3 following p–V graph. 1 we have Nmc2 = nRT 3 1 3 Nmc2 = nRT 2 2 Since the total kinetic energy of all gas 1 molecules is Nmc2 , 2

16 (a) The air exerts a downward pressure on the newspaper. 1A

p As the area of the newspaper is large, the Y p force exerted on the newspaper due to 1 T = T air pressure is large. 1A X Z The upward force of hitting the ruler on V 1 the newspaper is smaller than the air

At V1, the pressure is p1 when the temperature pressure exerted on the newspaper. 1A

remains at TX. Therefore the piece of newspaper moves By general gas law, pV  T. only very little.

Since pYV1 > p1V1, (b) Hit the ruler abruptly. 1A

TY > TX = TZ Expel as much air under the newspaper 9 B as possible. 1A 10 A 17 (a) By pV = nRT, 1M 1  p  pV By pV = nRT, V = nR    n =  T  RT 100 103 100 106 The volume of gas is inversely proportional to = p 8.31 (25  273) , which is the slope of the line connecting T = 4.04  10–3 mol 1A the point to the origin as shown below. There are 4.04  10–3 mol of molecules pressure / kPa inside the syringe. Number of molecules inside the syringe Y = nNA 1M X = 4.04  10–3  6.02  1023 = 2.43  1021 1A (b) By pV = nRT, temperature / K V  n  N (constant p and T) 1M V V The slope for stage X is greater than that for  1 = 2 N N pX pY 1 2 stage Y, i.e. > , so V > V . 6 T T Y X 100106 10 X Y = 21 N 11 D 2.4310 2 19 12 (HKDSE Practice Paper 2012 Paper 1A Q5) N2 = 2.43  10 1A 13 (HKALE 2012 Paper 2 Q36) The number of gas molecules ejected per 19 14 (HKDSE 2012 Paper 1A Q3) second is 2.43  10 . 15 (HKDSE 2013 Paper 1A Q4) 18 (a) She is incorrect. 1A Different molecules in the gas move at Conventional questions (p.187) different speeds. 1A

1 (b) She is incorrect. 1A  Nmc2 = pV = nRT The speed of an individual molecule is 3 always changing because of collisions. Root-mean-square speed c2 Its speed may increase or decrease. 1A 3RT 3nRT 3RT 19 (a) Atmospheric pressure (100 kPa) 1A = = N = 1A Nm  m mN A (b) The balloons will be inflated. 1A n The volume of air inside the jar (b) Total kinetic energy of the molecules 1 increases when the diagram is pulled = N  mc2 1M 2 downwards. By Boyle’s law, an increase 1 From (a), nRT = Nmc2 in volume implies a decrease in pressure. 3 1A Rearrange the terms, Therefore, air outside will flow into the 1 3 N  mc2  nRT 1A balloons. 1A 2 2 (c) An increase in temperature implies an (c) Total kinetic energy of the molecules 3 increase in pressure inside the bell jar.1A = nRT 1M 2 Therefore, the size of the balloons will 3 decrease. 1A =  3.5  8.31  (70 + 273) 2 20 (a) Since T and V remain unchanged, = 15 000 J 1A by pV = nRT, 1M (d) A real gas behave like an ideal gas at p  n low pressure 1A Percentage decrease in p and high temperature. 1A 220 100 = 100% 220 22 (a) (i) Heat is transferred from the sun to the water at the surface 1A = 54.5 % 1M by radiation. 1A  Percentage of air escape The warm water remains at the = percentage decrease in n surface and the cool water at lower = percentage decrease in p position. No convection current = 54.5% 1A occurs. 1A (b) (i) The root-mean-square speed Besides, water is a poor conductor remains unchanged 1A of heat. Heat is conducted very because the temperature of the type slowly from the water surface to remains unchanged. 1A lower position. 1A (ii) The internal energy decreases 1A Therefore, the temperature of the because the amount of gas water at the surface is higher than molecules decreases. 1A that 10 m below the surface. 21 (a) By general gas law, pV = nRT. 1M (ii) Apply pV = nRT. 1M

Since n remains constant, 3 8.31 (7  273) = p V p V 6.64 1027  6.02 1023 1 1 = 2 2 T T1 2 = 1320 m s–1 1A (2 p )V p V 2 1 = 2 2 24 (a) (i) Decreases 1A 0.9T T 2 2 (ii) Remains unchanged 1A V2 = 2.22V1 1A (b) As the number of air molecules inside  The volume of the gas bubble the carton decreases, the number of would increase by 2.22 times. bombardments on the walls of the carton (b) As a diver ascends, water pressure decreases. 1A decreases, and the air in the lungs As a result, the pressure inside the carton expands. 1A decreases 1A The lungs may get injured if the and the carton collapses due to the larger expansion is too large. 1A pressure outside. 1A 23 (a) By pV = nRT, 1M (c) Heat the sealed carton. 1A number of moles of gas The average kinetic energy of the air pV = molecules inside the carton increases; RT thus the number of bombardments on the 100103 5 = 8.31 (27  273) wall of the carton increases. 1A As a result, the pressure inside the carton = 200.6 mol increases and the carton expands due to ≈ 201 mol 1A the smaller pressure outside. 1A (b) (i) By general gas law, 25 (a) (i) Atmospheric pressure (100 kPa) p1V1 p2V2 = 1A T1 T2 3 3 (ii) As the gas is heated, the gas 100 10  5 = 80 10 V2 27  273 7  273 molecules move faster. The 3 V2 = 5.83 m 1A frequency of collision with the wall The volume of the gas is 5.83 m3. of syringe increases. 1A (ii) Change in internal energy Since the pressure of the gas 3 remains the same as the = nR∆T 1M 2 atmospheric pressure, its volume 3 =  200.6  8.31  (27 – 7) must increase. 1A 2 Therefore, the piston will move to a = 50 000 J 1A higher position. 1A (iii) R.m.s speed (b) As the volume increases, the gas 3RT = 1M molecules hit the wall of syringe less mN A often. 1A

The pressure of the gas decreases as the p / kPa temperature is kept constant. 1A Besides, the average speed of the gas 120 molecules remains unchanged. 1A

The total molecular kinetic energy and 110 hence the internal energy of the gas remains unchanged. 1A 100 (c) The temperature of the gas decreases after the heat source is removed, so the T / K 290 310 330 350 gas molecules move slower and hit the wall of syringe less often. 1A (Correct labelled axes with units) 1A Since the volume is kept constant, the (Correct data points) 1A pressure decreases. 1A (A correct straight line) 1A 26 (HKALE 2004 Paper 2 Q5) (c) 109 kPa 1A 27 (HKDSE 2014 Paper 1B Q2) 29 (a) The pressure of the gas is constant. 1A (b) Atmospheric pressure (100 kPa) 1A Experiment questions (p.190) (c) Keep the upper end of the capillary tube 28 (a) The flask including the neck is not open. 1A completely immersed in water. 1A (d) Stirring the water helps keep the water The rubber tubing that connects the temperature uniform throughout the Bourdon gauge and the flask is too long. beaker. 1A 1A This ensures the air column and the The flask and the thermometer touch the thermometer are at the same bottom of the beaker. 1A temperature. 1A (b) Convert the temperature in Kelvin: (e) The thermometer could break if it is hit violently. 1A p / kPa 100 105 110 115 120 (f) The student is incorrect 1A T / C 20 35 50 65 80 since there is no external factor that T / K 293 308 323 338 353 needs to be removed by a control set-up. 1A 1A 30 (a) Stir the water gently throughout the experiment. 1A (b) The temperature may change rapidly, making the data difficult to record. 1A The temperature inside the syringe cannot be measured. 1A

(c) (i) Physics in article (p.192) V / cm3 32 (a) When the air was pumped out, the pressure inside the sphere became zero. 50 1A 40 The pressure difference between the

30 inside and outside of the sphere was very large. 1A 20 Therefore, it was difficult to separate the 10 two shells. T / C 0 10 30 50 70 (b) Minimum force F = pA 1M 2 (Correct labelled axes with units)   0.5   = (100  103) π   1A   2   (Correct data points) 1A = 1.96  104 N 1A (A correct straight line) 1A (ii) The absolute zero is the intercept on the temperature axis. y-intercept = 37 44  37 Slope m = 50  0 = 0.14 cm3 C–1 1M By y = mx +c, when y = 0, c 37 x = = = –264 C 1A m 0.14 The absolute zero is –264 C. (iii) By pV = nRT, 1M nR V = T p nR Slope of the graph = p Number of mole n slope p = 1M R 0.14 106 100103 = 8.31 = 1.68  10–3 mol 1A 31 (HKDSE 2013 Paper 1B Q2)

8 New Senior Secondary Physics at Work (Second Edition)  Oxford University Press 2015