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# Chapter 5 States of Matter

This chapter explains about the existence of different states of matter in terms of balance between intermolecular forces and thermal energy of particles .The chapter also explains the laws governing behaviour of ideal gases along with their application in various real life situations. It also explains the behaviour of real gases along with the conditions required for real gases. It also provides the difference between gaseous state and vapours. It also explains properties of liquids in terms of intermolecular attractions.

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### Exercise 1 ( Page No. : 159 )

•  Q1 What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to 200 dm3 at 30°C? Ans: Volume of gas(V1) = 500dm3   (given) Pressure of gas (p1) = 1 bar    (given) Volume of compressed gas (V2) = 200dm3  (given) Now ,let P2 be the pressure required to compress the gas. Therefore at constant temperature P1 V1 = P2 V2 (Boyle’s Law) As a result P2 = P1 V1 / V2 = 1 X 500/200 = 2.5 bar Therefore, the minimum pressure required is 2.5 bar. Q2 A vessel of 120 mL capacity contains a certain amount of gas at 35 °C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35 °C. What would be its pressure? Ans: According to Boyle’s Law P1 V1 = P2 V2 Here the temperature is constant. Therefore 1.2 X 120 = P2 X 180 OR P2 = 1.2 X 120 /180 = 0.8 bar Therefore, the pressure would be 0.8 bar. Q3 Using the equation of state pV = nRT; show that at a given temperature density of a gas is proportional to gas pressurep. Ans: The equation of state is given by, pV = nRT ……….. (i) Where, p → Pressure of gas V → Volume of gas n→ Number of moles of gas R → Gas constant T → Temperature of gas From equation (i) we have, p = n RT/V Where n= Mass of gas(m)/ Molar mass of gas(M) Putting value of n in the equation, we have  p = m RT/ MV ------------(ii) Now density(ρ) = m /V ----------------(iii) Putting (iii) in (ii) we get P = ρ RT / M OR ρ = PM / RT Hence, at a given temperature, the density (ρ) of gas is proportional to its pressure (P) Q4 At 0°C, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide? Ans: Using the relationship of density (ρ) of the substance at temperature (T) We have ρ = Mp/RT or p = ρ RT/ M For the given data if M is the molar mass of the gaseous oxide, we have 2 = ρ RT/ M ……………(1) Also for nitrogen 5 = ρ RT/28 ………………….(2) From (1) & (2), we have 5/2 = M/28 Or M = 5 X 28/ 2 = 70g/mol Hence, the molecular mass of the oxide is 70 g/mol. Q5 Pressure of 1 g of an ideal gas A at 27 °C is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at same temperature the pressure becomes 3 bar. Find a relationship between their molecular masses. Ans: Mass of gas A , WA = 1g Mass of gas B,  WB = 2g Pressure exerted by the gas A = 2 bar Total pressure due to both the gases = 3 bar In this case temperature & volume remain constant Now if MA & MB are molar masses of the gases A & B respectively,therefore pA V= WA RT/MA & Ptotal V = (WA/MA + WB/ MB) RT = 2 X V = 1 X RT/MA & 3 X V = (1/MA + 2/MB)RT From these two equations, we get 3/2 = (1/MA + 2/MB) / (1 /MA) = (MB + 2MA) / MB This result in 2MA/ MB = (3/2) -1 = ½ OR MB = 4MA Thus, a relationship between the molecular masses of A and B is given by 4MA = MB Q6 The drain cleaner, Drainex contains small bits of aluminum which react with caustic soda to produce dihydrogen. What volume of dihydrogen at 20 °C and one bar will be released when 0.15g of aluminum reacts? Ans: Aluminium reacts with caustic soda in accordance with the reaction The reaction of aluminium with caustic soda can be represented as: Therefore volume of hydrogen at STP released when 0.15g of Al reactsv =0.15 x 3 x 22.4 /54 = 187ml Now P1 = 1 bar, P2 = 1 bar T1 = 273 K T2 = 20 + 273 = 293 K V1 = 187 ml V2 = x When pressure is held constant,then V2 = P1 V1 T2 / P2 T1 OR x = 1 X 187 X 293 / 0.987 X 273 = 201 ml Therefore, 201 mL of dihydrogen will be released. Q7 What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a 9 dm3 flask at 27 °C ? Ans: Given, Mass of carbon dioxide = 4.4 g Molar mass of carbon dioxide= 44g/mol Mass of methane  = 3.2g Molar mass of methane = 16g/mol Now amount of methane, nCH4 = 3.2/ 16 = 0.2 mol & amount of CO2  nCO2  = 4.4/ 44 = 0.1 mol Also we know , Pv = (nCH4 + nCO2) RT OR P X 9 = (0.2 +0.1) x 0.0821 x 300 Or  p = 0.3 x 0.0821 x 300 / 9 = 0.821 atm Hence, the total pressure exerted by the mixture is 0.821 atm Q8 What will be the pressure of the gaseous mixture when 0.5 L of H2 at 0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in a 1L vessel at 27°C? Ans: From the equation Pv = n RT for the two gases. We can write 0.8 x 0.5 = nH2 x RT  or nH2  = 0.8 x 0.5 / RT ALSO  0.7 x 2.0  = n02 . RT  or n02  = 0.7 x 2 / RT When introduced in 1 L vessel, then Px1L = (n02 + nH2)  RT Putting the values, we get P = 0.4 + 1.4 = 1.8 bar Hence, the total pressure of the gaseous mixture in the vessel is 1.8 bar Q9 Density of a gas is found to be 5.46 g/dm3 at 27 °C at 2 bar pressure. What will be its density at STP? Ans: For an ideal gas Density ρ = p x M/ RT F rom the given data, 5.46 = 2 x M/ 300 x R    ………(1) ALSO  ρSTP = 1 x M/ 273 x R    ………..(2) FROM (1) & (2) ,WE GET ρSTP =  (1 x M/ 273 x R)   x  (300 x R/ 2 x M) x 5.46 = 300 x 5.46 / 273 X 2 = 3.00 g/dm3 Hence, the density of the gas at STP will be 3 g dm–3. Q10 34.05 mL of phosphorus vapour weighs 0.0625 g at 546 °C and 0.1 bar pressure. What is the molar mass of phosphorus? Ans: Given, p = 0.1 bar V = 34.05 mL = 34.05 × 10–3 L = 34.05 × 10–3 dm3 R = 0.083 bar dm3 K–1 mol–1 T = 546°C = (546 + 273) K = 819 K From the gas equation PV = w. RT / M, we get M = w. RT/ Pv    ……….(1) Substituting the given values in the equation (1), we get M = (0.0625 / 0.1 x 34.04) X 82.1 X 819 = 123.46 g/mol Hence, the molar mass of phosphorus is 123.46 g mol–1. Q11 A student forgot to add the reaction mixture to the round bottomed flask at 27 °C but instead he/she placed the flask on the flame. After a lapse of time, he realized his mistake, and using a pyrometer he found the temperature of the flask was 477 °C. What fraction of air would have been expelled out? Ans: Let the volume of  air in the  flask at 27°C be V1 & that of the same amount of the gas at 477°C be V2. According to charle’s law V2/T2  =  V1/ T1   …………………(1) NOW  volume of gas expelled out = V2 – V1, THEN Fraction of the gas expelled out = (V2 – V1 ) / V2 = 1- (V1/ V2)     …………….(2) Also from equation (1)  V1/ V2 = T1/ T2   …………..(3) Substituting the values of (3) in (2), we get Fraction of the air expelled  = 1- T1/ T2 = (T2 – T1)/ T2  = 750- 300 /750 = 0.6 Hence, fraction of air expelled out is 0.6 or 3/5 th Q12 Calculate the temperature of 4.0 mol of a gas occupying 5 dm3 at 3.32 bar. (R = 0.083 bar dm3 K–1 mol–1). Ans: Given, Amount of the gas, n = 4.0 mol Volume of the gas, V = 5 dm3 Pressure of the gas, p = 3.32 bar R = 0.083 bar dm3 K–1 mol–1 From the ideal gas equation, we get pV = nRT ⇒ T = pV / nR       = (3.32x5) /  (4x0.083)       = 50K Hence, the required temperature is 50 K. Q13 Calculate the total number of electrons present in 1.4 g of dinitrogen gas. Ans: Mass of the nitrogen = 1.4 g Molar mass of nitrogen  = 28 g mol–1 Therefore amount of nitrogen = 1.4 / 28 = 0.05 mol Number of nitrogen molecule in 0.05 mol = 6.023 x 1023  x 0.05 = 3.0115 x 1022 Also number of electrons in one molecule of nitrogen  = 14 Therefore, total number of electrons in 1.4 g of nitrogen  = 3.0115 x 1022 x 14  = 4.22 x 1023 Q14 How much time would it take to distribute one Avogadro number of wheat grains, if 1010 grains are distributed each second? Ans: We know , Avogadro number = 6.023 × 1023 Therefore , time taken for distributing the grains = (6.023 × 1023 x 1) / 1010 = 6.023 x 1013 s By Converting these seconds into years, we get = (6.023 x 1013) / (60 x60x24x365) = 1,90,9800 y Hence, the time taken would be 1.909 x 106 years. Q15 Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm3 at 27°C. R = 0.083 bar dm3 K–1 mol–1. Ans: Given, Mass of oxygen  = 8 g, molar mass of oxygen = 32 g/mol Mass of hydrogen = 4 g, molar mass of hydrogen = 2 g/mol Therefore amount of oxygen = 8/ 32 = 0.25 mol And amount of hydrogen = 4/2 = 2 mol From the gas equation PV = n RT, we get,  P X 1  = (0.25 + 2)  X 0.083 X 300 = 56.02 bar Hence, the total pressure of the mixture is 56.02 bar. Q16 Pay load is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the pay load when a balloon of radius 10 m, mass 100 kg is filled with helium at 1.66 bar at 27°C. (Density of air = 1.2 kg m–3 and R = 0.083 bar dm3 K–1 mol–1). Ans: Payload of the ballon  = mass of the displaced air – mass of the ballon Radius of the ballon, r  = 10 m Mass of the ballon, m = 100kg Therefore volume of the ballon = 4/3πr3 = 4/3 x 22/7 x (10)3 = 4190.5 m3 Now volume of the displaced air = 4190.5 m3 Given, Density of air = 1.2 kg m–3 Therefore, the mass of the displaced air = 4190.5 x 1.2  = 5028.6 kg Let w be the mass of helium gas filled into the ballon,then PV = (w/m)RT OR w = PVM/RT = (1.66 X 4190.5 X 103 X 4) / (0.083 X 300)  = 1117 kg (approx) Total mass of the balloon filled with He  = 1117 + 100 = 1217 kg Therefore payload of the balloon  = 5028.6 – 1217 = 3811.6 kg Hence, the pay load of the balloon is 3811.6 kg. Q17 Calculate the volume occupied by 8.8 g of CO2 at 31.1°C and 1 bar pressure. R = 0.083 bar L K–1 mol–1. Ans: From the gas equation, PV = (w/M) RT Where w is the mass of the gas, M is the molar mass of the gas For CO2 , M = 44 g/mol Substituting the given values ,we get 1 x V  = (8.8 / 44) x 0.083 x 304.1  = 5.05 L Hence, the volume occupied is 5.05 L. Q18 2.9 g of a gas at 95 °C occupied the same volume as 0.184 g of dihydrogen at 17 °C, at the same pressure. What is the molar mass of the gas? Ans: From the gas equation, PV = (w/M) RT Substituting the given data in the gas equation, we get PV = (2.9 / M) x R x 368 & PV = (0.184 / 2) x R x 290 From these two equation, we can write (2.9 / M) x R x 368 = (0.184 / 2) x R x 290 By, striking throug R from both side, we get (2.9 / M) x 368 = (0.184 / 2) x 290 Or (2.9 / M)   =  (0.092 X 290) / 368 Or M  = 2.9 x 368 / 0.092 x 290 = 40 g/mol Hence, the molar mass of the gas is 40 g mol–1. Q19 A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen. Ans: Pressure of the gas mixture = 1 bar Let us consider 100g of the mixture So ,mass of hydrogen in the mixture  = 20 g & mass of oxygen in the mixture = 80 g Using the respective molar masses, we get   nH = 20/ 2  = 10 mol . nO = 80 / 32 = 2.5 mol Then, pH = XH x Ptotal = (nH / nH + nO) x P total   = (10 / 10 + 2.5 ) x  1  = 0.8 bar Hence, the partial pressure of dihydrogen is 0.8 bar. Q20 What would be the SI unit for the quantity pV2T 2/n? Ans: The SI unit of the given quantity is obtained by substituting the SI units of all the quantities in its expression. Also The SI unit for pressure, p is Nm–2. The SI unit for volume, V is m3. The SI unit for temperature, T is K. The SI unit for the number of moles, n is mol. Therefore, the SI unit for quantity pV2T2 / n is given by substituting the above values in the given expression pV2T2 / n  = (Nm-2) x (m3)2 x K2 / mol = Nm4K2 mol-1 Q21 In terms of Charles’ law explain why –273°C is the lowest possible temperature. Ans: Charles observed that the volume of certain amount of a gas changes by  VO / 273.15 for each degree rise or fall in temperature.VO being the volume at 0°C.The volume at any temperature t°C is given by Vt  = VO(1 + t°C/ 273.15) Now ,if the temperature is lowered, it becomes clear that at t = -273.15°C Vt  = VO(1+ (-273.15/273.15)) = VO  = (1-1) = 0 And below t = -273.15°C, Vt  becomes negative, which is not possible Therefore the physically significant lowest temperature is -273.15° C Q22 Critical temperature for carbon dioxide and methane are 31.1 °C and –81.9 °C respectively. Which of these has stronger intermolecular forces and why? Ans: The maximum temperature at which a gas can be converted into a liquid by an increase in pressure is called its critical temperature(Tc). This means that the intermolecular forces of attraction between the molecules of a gas are directly proportional to its critical temperature. Hence, intermolecular forces of attraction are stronger in the case of CO2. Q23 Explain the physical significance of Van der Waals parameters. Ans: The van der Waals equation is an equation of state for a fluid composed of particles that have a non-zero volume and a pairwise attractive inter particle force (such as the van der Waals force). The equation is van der Waals equation Physical significance of ‘a’: ‘a’ is a measure of the magnitude of intermolecular attractive forces between the particles Physical significance of ‘b’: ‘b’ is the volume excluded by a mole of particles   p is the pressure of fluid V  is the total volume of container containing the fluid