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PREVIOUS HSE QUESTIONS AND ANSWERS OF THE CHAPTER “STATES OF MATTER”, 1. According to Kinetic Molecular Theory of gases, the average kinetic energy of gas molecule is directly proportional to, a) Critical temperature, b) Absolute temperature, c) volume, d) no. of moles (1), , Ans: b) Absolute temperature, 2. a) State Gay-Lussac’s law, , (1), , Ans: The law states that at constant volume, the pressure of a fixed mass of gas is directly proportional to, absolute temperature., b) Give the graphical representation of the above law. (1), , Ans:, , 3. a) What are the causes of deviation of real gases from ideal behaviour? (2), b) Identify the property of liquid behind the capillary rise., (1), [December 2020], , Ans: a) This is due to two wrong assumptions of kinetic molecular theory of gases. They are:, i) The actual volume of the molecules is negligible compared to the volume of the gas., ii) There is no force of attraction between the gas particles., b) Surface Tension, 4. The unit of coefficient of viscosity in c.g.s. system is _____., , (1), , Ans: poise, 5. A gas occupies 400 ml volume at 47 °C and 800 mm of Hg pressure. What will be its pressure at a height where the, temperature is 27 °C and volume of the gas is 450 ml ? (2), , Ans: From combined gas law equation, p1V1 = p2V2, T1, T2, Here p1 = 800 mm of Hg, V1 = 400 mL, T1 = 47 + 273 = 320 K, V2 = 450 mL, T2 = 27 + 273 = 300 K, p2 = ?, (800 x 400)/320 = (p2 x 450)/300, p2 = 800 x 400 x 300 = 666.67 mm of Hg, 450 x 320, 6. (a) Write the name of van der Waals force between:, (i) Non-polar molecules (ii) Molecules having permanent dipoles, (1), (b) State Dalton’s law of partial pressures. (1), (c) At higher altitudes, pressure cooker is used for cooking. Give reason. (1), , [March 2020], , Ans: (a) (i) London force or Dispersion force (ii) Dipole – dipole force, , States of Matter - Prepared by ANIL KUMAR K L, GHSS ASHTAMUDI, KOLLAM, Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , Page 1

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(b) The law states that at constant temperature, the total pressure exerted by a mixture of non-reacting, gases is equal to the sum of the partial pressures of the component gases., (c) At higher altitudes, atmospheric pressure is low. So water boils at lower temperature than at sea level. So, a pressure cooker is used for cooking., 7. Which property of liquids is associated with fire polishing of glass? (1), Ans: Surface Tension, 8. Real gases do not obey ideal gas equation under all conditions. Explain. (2), Ans: This is due to two wrong assumptions of kinetic molecular theory of gases. They are:, iii) The actual volume of the molecules is negligible compared to the volume of the gas., iv) There is no force of attraction between the gas particles., These two assumptions become wrong at high pressure and low temperature., 9. (a) State Dalton's law of partial pressures. (1), (b) Calculate the total pressure exerted by a mixture of 8 g of O2, and 4g of H2 enclosed in a vessel of 1 dm3, at 270C. R = 0.083 bar dm3 K-1 mol-I. (2), [July 2019], Ans: (a) This law states that at constant temperature, the total pressure exerted by a mixture of nonreacting gases is equal to the sum of the partial pressures of the component gases., (b) Mass of O2 (wO2) = 8 g, No. of moles of O2 (nO2) = 8/32 = 0.25 mol, Mass of H2 (wH2) = 4 g, No. of moles of H2 (nH2) = 4/2 = 2 mol, Volume of the gas = 1 dm3, Temperature = 270C = 27 + 273 = 300 K, R = 0.083 bar dm3 K-1 mol-1, PTotal = (nO2+nH2) RT/V = (0.25 + 2) x 0.083 x 300/1 = 56.025 bar, 10. The lowest hypothetical temperature at which gases are supposed to occupy zero volume is called, ……………… (1), Ans: Absolute zero of temperature, 11. Give the relation between molar mass of a gas (m) and its density (d). How are the densities of O 2(g) and, CH4(g) related, if they are kept at the same temperature and pressure?, (2), Ans: Molar mass (M) = dRT/P, OR, molar mass (M) α density (d) at constant temperature and pressure., Molar mass of O2 = 32 and that of CH4 = 16. So O2 is two times denser than CH4., OR, MO2/MCH4 = dO2/dCH4, 32/16 = dO2/dCH4, dO2/dCH4 = 2, dO2 = 2 x dCH4., 12. (a) Why do real gases deviate from ideal behaviour?, (1), (b) Write the conditions under which gases deviate from ideality. (1), (c) Define Boyle Point., (1), [March 2019], Ans: (a) See Answer of the question no. 2, States of Matter - Prepared by ANIL KUMAR K L, GHSS ASHTAMUDI, KOLLAM, Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , Page 2

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(b) At high pressure and low temperature., (c) The temperature at which real gases obey ideal gas equation over an appreciable range of pressure, is called Boyle temperature or Boyle point., 13. The critical temperatures of some gases are given in the following table :, Gas, H2, He, O2, N2, Critical Temperature 33.2 5.3, 154.3, 126, (K), If the samples of the above given gases are cooled from 298 K, which one will liquify first by applying, pressure?, (1), Ans: O2, 14. Derive an equation relating molar mass of an ideal gas with its density. (2), Ans: The ideal gas equation is PV = nRT, But n = w/M, So, PV = wRT/M, Or, P = wRT/MV, Or, P= dRT/M, Or, M = dRT/P, 15. Calculate the total pressure in a mixture of 3.5 g of dinitrogen and 16 g of dioxygen confined in a vessel of 2, dm3 at 270C. (R = 0.083 bar dm3 K-1mol-1). (3), [March 2018], Ans: Mass of N2 (wN2) = 3.5 g, No. of moles of N2 (nN2) = 3.5/28 = 0.125 mol, Mass of O2 (wO2) = 16 g, No. of moles of O2 (nO2) = 16/32 = 0.5 mol, Volume of the gas = 2 dm3, Temperature = 270C = 27 + 273 = 300 K, R = 0.083 bar dm3 K-1 mol-1, PTotal = (nN2+nO2) RT/V = (0.125 + 0.5) x 0.083 x 300/2 = 7.78 bar, 16. Examine the following graph and name the gas law corresponding to it., , (1), Ans: Gay Lussac’s law, 17. a) Define 'normal boiling point' of a liquid., b) Give a reason for the use of pressure cooker to cook food, at high altitudes., , (2), , States of Matter - Prepared by ANIL KUMAR K L, GHSS ASHTAMUDI, KOLLAM, Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , Page 3

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Ans: (a) The boiling point at 1 atm pressure is called normal boiling point., (b) At high altitudes, the atmospheric pressure is low. So liquids boil at lower temperatures than at sea, level and hence a pressure cooker is used for cooking food., 18. Write the postulates of kinetic molecular theory of gases. (3), [August 2018], Ans: The important postulates of this theory are:,  Every gas contains a large number of minute and elastic particles (atoms or molecules). The actual, volume of the particles is negligible compared to the volume of the gas.,  There is no force of attraction between the gas particles.,  The particles of a gas are in constant and random motion in straight line. During this motion, they, collide with each other and also with the walls of the container.,  The pressure of a gas is due to the wall collisions of the particles.,  All collisions are perfectly elastic. i.e. the total energy of particles before and after collisions remains, the same.,  The average kinetic energy of gas molecules is directly proportional to absolute temperature., 19. a), , ii), Identify the law represented by the graph given above., iii), State the law. (2), b) Write down the ideal gas equation and explain the terms. (2), Ans (a) i) Boyle’s Law, ii) It states that at constant temperature, the volume of a fixed mass of gas is inversely proportional, to its pressure., (b) The ideal gas equation is PV = nRT., Where P –Pressure, V – Volume, n – No. of moles, R – Universal gas constant and T – Absolute, temperature., 20. a) Give the reason behind the following:, i), The glass window panels of old buildings are thicker at the bottom than at the top., (1), ii), Sharp glass edges are heated for making them smooth., (1), b) Maxwell and Boltzmann have shown that actual distribution of molecular speeds depends on, temperature and molecular mass., i), What do you mean by most probable velocity? (1), ii), At the same temperature which will move faster, N2 or Cl2? (1), [March 2017], Ans: (a) i) Glass is a very viscous liquid. So it has a tendency to flow very slightly. Hence the window panels of, old buildings are thicker at the bottom than at the top., ii) On heating, the glass melts and the surface of the liquid tends to take the rounded shape at the, edges, due to surface tension., States of Matter - Prepared by ANIL KUMAR K L, GHSS ASHTAMUDI, KOLLAM, Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , Page 4

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(b) i) It is the velocity possessed by maximum number of gas molecules., ii) N2, 21. An ideal gas is one which obeys gas laws., a) Derive an ideal gas equation., (2), b) At 270C a gas was compressed to half of its volume. To what temperature it must be heated, so that it, would occupy double its original volume?, (1), c) Liquid drops assume spherical shape. Why?, (1), [September 2016], Ans: (a) According to Boyle’s law: V α 1/P (At constant T and n), According to Charles’ Law: V ∝ T (At constant p and n), According to Avogadro Law: V ∝ n (At constant p and T), On combining these three laws we get:, V ∝ n x T x 1/P, Or, V = R x n x T x 1/P (where R is a constant called universal gas constant), Or, PV = nRT ………………….. (1), This equation is known as ideal gas equation., (b) Let the original volume of the gas be z., Then T1 = 270C = 300K, V1 = z/2, V2 = 2z, T2 = ?, We know that V1/T1 = V2/T2, So, T2 = V2 x T1/V1 = 2z x 300/(z/2) = 300 x 4 = 1200 K, (c) Every liquid tries to reduce their energy by decreasing the surface area. For a given volume sphere, has the minimum surface area. So liquid drops assume spherical shape., 22. Ideal gas equation is true for ideal gases only. There is a modified form of ideal gas equation applicable to, all gases., a) Give the name of the modified form of ideal gas equation and write down it., (2), b) Name the phenomenon behind cleansing action of soap., (1), c) What do you know about Dalton’s law of partial pressures? (1), [March 2016], 2, 2, Ans: (a) van der Waals equation. The equation is (P + n a/V ) (V- nb) = nRT, (b) Surface tension, (c) The law states that at constant temperature, the total pressure exerted by a mixture of non-reacting, gases is equal to the sum of the partial pressures of the component gases., Its mathematical form is:, PTotal = P1+P2+P3+...... (at constant T, V), where PTotal is the total pressure of the mixture of gases and P1,P2,P3 etc. are partial pressures of gases., 23. The Kinetic molecular theory provides a theoretical basis to experimentally observed facts related to gases., a) Which one of the following statements is CORRECT with regard to the gaseous state?, i) Molecules have fixed positions., ii) Molecules are in constant random motion, iii) All molecules have same speed at a given temperature, iv) The average kinetic energy of the gas molecules is inversely proportional to the absolute, temperature. (1), b) A sample of hydrogen gas occupies a volume of 300 ml at 1.2 bar pressure and 5 0 C. Calculate its volume, at 0.45 bar pressure and 700 C. (3), [October 2015], Ans: (a) Molecules are in constant random motion, (b) From combined gas law equation, P1V1 = P2V2, T1, T2, Here P1 = 1.2 bar, V1 = 300 mL, T1 = 5 + 273 = 298 K, P2 = 0.45 bar, T2 = 70 + 273 = 343 K, V2 = ?, States of Matter - Prepared by ANIL KUMAR K L, GHSS ASHTAMUDI, KOLLAM, Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , Page 5

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(1.2 x 300)/298 = (0.45 x V2)/343, V2 = 1.208 x 343/0.45 = 920.8 mL, 24. The gases which obey Gas Laws at all temperatures and pressure are called ideal gases., a) Give reason for the deviation of real gases from the ideal behaviour., (2), b) Calculate the minimum pressure required to compress 500 mL of air at 1 atm pressure to 300 mL at, the same temperature., (2), [March 2015], Ans: (a) Ref. the answer of the qn. No. 2, (b) From Boyle’s law, P1V1 = P2V2, Here P1 = 1 atm, V1 = 500 mL, V2 = 300 mL, P2 = ?, So 1 x 500 = P2 x 300, P2 = 500/300 = 1.67 atm, 25., , a) Name the gas law shown by the above graph. (1), b) State the gas law. (1), c) At 350C and 700mm of Hg pressure, a gas occupies 500 ml volume. What will be its pressure when the, temperature is 150C and the volume of the gas is 450 mL. (2), [August 2014], Ans: (a) Boyle’s law, (b) It states that at constant temperature, the volume of a fixed mass of gas is inversely proportional to its, pressure., (c) From combined gas law equation, P1V1 = P2V2, T1, T2, Here P1 = 700 mm of Hg, V1 = 500 mL, T1 = 35 + 273 = 308 K, P2 = ?, T2 = 15 + 273 = 288 K, V2 = 450 mL, (700 x 500)/308 = (P2 x 450)/288, P2 = 1136.4 x 288/450 = 727.27 mm of Hg, 26. a) Particles of soil at the bottom of a river remain separated, but they stick together when taken out. Name, the property behind this. (2), b) Critical temperatures of ammonia and CO2 are 405.5 K and 304 K respectively. On cooling these gases, from 500 K, which gas will liquify first? (1), c) van der Waals' forces are attractive inter molecular forces. Write the names of any two types of van, der Waals' forces. (2), [March 2014], Ans: a) Surface Tension, b) NH3 will liquify first because its critical temperature will be reached first., c) London forces (Dispersion forces) and Dipole – Dipole forces., 27. a) What is Boyle point or Boyle temperature? (1), b) At high altitudes, a pressure cooker is used for cooking food. Why? (1), c) A neon-dioxygen mixture contains 70.6g dioxygen and 167.5g neon. If the pressure of the mixture of, States of Matter - Prepared by ANIL KUMAR K L, GHSS ASHTAMUDI, KOLLAM, Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , Page 6

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gases in the cylinder is 25 bar, what are the partial pressures of O2 and Ne in the mixture? (2), [March, 2014], Ans: (a) The temperature at which real gases obey ideal gas equation over an appreciable range of pressure is, called Boyle temperature or Boyle point., b) At high altitudes, the atmospheric pressure is low. So liquids boil at lower temperatures than at sea, level and hence a pressure cooker is used for cooking food., c) Mass of O2 (wO2) = 70.6 g, No. of moles of O2 (nO2) = 70.6/32 = 2.21 mol, Mass of Ne (wNe) = 167.5 g, No. of moles of Ne (nNe) = 167.5/20 = 8.375 mol, Mole fraction of O2 = nO2/(nO2 + nNe) = 2.21/(2.21 + 8.375) = 0.21, Mole fraction of Ne = 1 – Mole fraction of O2 = 1 – 0.21 = 0.79, Partial pressures of O2 = Total pressure x mole fraction of O2, = 25 x 0.79 = 19.75 bar, Partial pressures of Ne = Total pressure - partial pressures of O2 = 25 – 19.75 = 5.25 bar, 28. a) van der Waals equation of state explains the behaviour of real gases. What does the van der Waal, constant ‘a’ indicate? (1), b) What is the critical temperature of a gas?, (1), 0, c) At 25 C and 760 mm of Hg pressure, a gas occupies 600 ml volume. What will be its pressure at a height, where temperature is 100C and volume of the gas is 640 ml., (2), [Oct. 2013], Ans: (a) a is a measure of magnitude of inter molecular attractive forces within the gas., (b) The temperature below which a gas can be liquefied by the application of pressure is called, Critical Temperature., (c) From combined gas law equation, P1V1 = P2V2, T1, T2, Here P1 = 760 mm of Hg, V1 = 600 mL, T1 = 25 + 273 = 298 K, P2 = ?, T2 = 10 + 273 = 283 K, V2 = 640 mL, (760 x 600)/298 = (P2 x 640)/283, P2 = 1530.2 x 283/640 = 673.48 mm of Hg, 29. a) What is the thermodynamic scale of temperature., (1), b) Viscosity of liquids decreases as the temperature increases. Why?, (1), c) Gases deviate from ideal behaviour due to the faulty assumptions of Kinetic theory of gases. State those, faulty assumptions., (2), [Oct. 2013], Ans: (a) Kelvin scale, (b) Because at high temperature, molecules have high kinetic energy and can overcome the, intermolecular forces., (c) Ref. the answer of the qn. No. 2, 30. Real gases behave ideally only at certain conditions., a) What is Boyle point of a gas., (1), b) Write the expression for compressibility factor. What is its value for an ideal gas?, (1), States of Matter - Prepared by ANIL KUMAR K L, GHSS ASHTAMUDI, KOLLAM, Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , Page 7

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c) Density of a gas was found to be 5.5 g/L at 2 bar pressure and at 25 0C. Calculate its molar mass., (R=0.083 L bar/K/mol) (2), [March 2013], Ans: (a) The temperature at which real gases obey ideal gas equation over an appreciable range of, pressure is called Boyle temperature or Boyle point., (b) Compressibility factor, z = PV/nRT. For ideal gases, z = 1, (c) Here d = 5.5 g/L, P = 2 bar, T = 25 + 273 = 298 K and R = 0.083 L bar/K/mol, M = ?, Molar mass (M) = dRT/P = 5.5 x 0.083 x 298/2 = 68.02 g/mol, 31. a) The combonation of Boyle’s law, Charle’s law and Avogadro’s law is known as ideal gas equation. But, real gases deviate from ideal behaviour., i) Write the modified gas law equation. (1), ii) Name the above equation., (1), b) Give reason for the following:, i), At hill station, pressure cooker is used for cooking., (1), ii), Window panes of the old buildings become thicker at the bottom than at the top. (1), [Sept. 2012], Ans: (a) i) (P + n2a/V2) (V- nb) = nRT, ii) Van der Waal’s Equation, (b) i) Refer the ans. of the qn. No. 11 (b), ii) Refer the ans. of the qn. No. 14 (a) i, 32. Gas laws are relationships between the measurable properties of gases., a) Name the gas law which gives the relationship between the pressure and temperature of a fixed, amount of gas at constant volume., (1), b) Draw the graph to illustrate the above gas law. (1), c) A definite quantity of an ideal gas is confined in a container of constant volume. When the container is, immersed in a bath of melting ice, the pressure of the gas is 800 mm of Hg. Find the temperature when, the gas pressure is 400 mm of Hg., (2), [March 2012], Ans: (a) Gay Lussac’s law, (b), , (c) P1 = 800 mm of Hg, T1 = 00C = 273 K, P2 = 400 mm of Hg and T2 = ?, From Gay Lussac’s law, P1/T1 = P2/T2, So, T2 = P2.T1/P1 = 400 x 273/800 = 136.5 K, 33. Consider the following isotherms of a gas:, States of Matter - Prepared by ANIL KUMAR K L, GHSS ASHTAMUDI, KOLLAM, Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , Page 8

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P, , P, , V, , 1/V, V, 1/P, a) Which gas law is illustrated by these diagrams? (1), b) Draw the diagram when PV is plotted against P. (1), c) An air filled balloon has a volume of 125L at 760 mm of mercury and 25 0C. What will be its volume, when the pressure is 670 mm of mercury and temperature is 180C? (2), [Oct. 2011], Ans: (a) Boyle’s law, (b), , (c) From combined gas law equation, P1V1 = P2V2, T1, T2, Here P1 = 760 mm of Hg, V1 = 125 L, T1 = 25 + 273 = 298 K, P2 = 670 mm of Hg, T2 = 18 + 273 = 291 K, V2 = ?, (760 x 125)/298 = (670 x V2 )/291, V2 = 318.79 x 291/670 = 138.46 L, 34. In the celsius scale, melting point of ice is 00C. Another scale of temperature is based on absolute zero., a) Identify the scale. (1), b) What is the volume of a gas at absolute zero of temperature? (1), c) Draw a graph showing the relationship between volume and temperature of an ideal gas at constant, pressure. (1), d) Consider a gas at 00C. At what temperature will the volume be doubled if the pressure is kept constant?, (1) [March 2011], Ans: (a) Kelvin scale, (b) zero, (c), , States of Matter - Prepared by ANIL KUMAR K L, GHSS ASHTAMUDI, KOLLAM, Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , Page 9

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(d) T1 = 0+273 = 273 K, V1 = z, V2 = 2z, T2 = ?, From Charle’s law, V1/T1 = V2/T2, So, T2 = V2 x T1/V1 = 2z x 273/z = 546 K, 35. It is found that real gases do not obey ideal gas equation perfectly under all conditions., a) Write the ideal gas equation and mention the terms? (1), b) Why do real gases deviate from ideal behaviour?, (2), c) What are the conditions under which real gases approach ideal behaviour? (1), [Sept. 2010], Ans: (a)The ideal gas equation is PV = nRT, Where P –pressure, V – volume, n – no. of moles, R – Universal gas constant, T – absolute temperature., (b) Refer the ans. of the qn. No. 2, (c) Real gases obey ideal gas equation at low pressure and high temperature., 36. The theory that attempts to explain the behaviour of gases is known as kinetic molecular theory., a) On the basis of this theory, explain the compressible nature of gases and the temperature dependence, on kinetic energy. (2), b) Liquid drops assume spherical shape. Why?, (1), c) How does temperature influence the viscosity of a liquid?, (1), [March 2010], Ans: (a) According to kinetic molecular theory, gas molecules are point masses and there is a large space, between them. So gases are highly compressible. Also on heating a gas, the speed of the molecules and, hence their K.E increases., (b) Refer the ans. of the qn. No.15 (c), (c) When temperature increases, viscosity decreases., 37. Mercury drops are spherical in shape., a) Which property is responsible for the spherical shape of drops? Explain the property., (2), b) How does the above property depend on temperature?, (2), [March 2009], Ans: (a) Surface tension. It is the force acting per unit length perpendicular to the line drawn on the, surface of liquid., (b) When temperature increases, the force of attraction between the molecules decreases and hence, the surface tension decreases., 38. ‘R’ is the universal gas constant in the ideal gas equation., a) Write down the values of R in two different units., (1), 0, b) Calculate the mass of 500 ml of O2 at 27 c and 740 mm of Hg. (2), [June 2008], Ans: (a), , Unit of R, 1. L atm/K/mol, 2. L bar/K/mol, , Value of R, 0.0821, 0.08314 (8.314x10-2), , (b) From ideal gas equation, PV = nRT,, PV = wRT/M. So mass, w = PVM/(RT), Here V = 500 mL = 0.5 L, T = 27+273 = 300K, P = 740 mm of Hg = 740/760 = 0.974 atm, M = 32g/mol, R =, 0.0821 L atm/K/mol, Therefore, w = 0.974 x 0.5 x 32/(0.0821 x 300) = 0.632 g, States of Matter - Prepared by ANIL KUMAR K L, GHSS ASHTAMUDI, KOLLAM, Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , Page 10