11 Temperature and Ideal Gas (H2) 2020 - Tutorial Solution

11-1 TUTORIAL 11: TEMPERATURE & IDEAL GAS SOLUTIONS Level 1 Solutions 1 (a)(i) There is no net transfer of heat between the two bodies. [1] (ii) There is a net transfer of heat from A (the one at the higher temp) to B. [1] (b) The temperatures are quoted in degrees Celsius. In order to compare the temperature of the system, we need to use the abs olute temperature scale, (i.e. Kelvin scale). So in this case, it is not twice since (30 + 273.15) K is not twice that of (15 + 273.15) K. Therefore, one can only say that “it is hotter today than yesterday”. [1] 2 T/ K = /C + 273.15 K = 201.84 + 273.15 K = 474.99 K [1] 3 Absolute zero is the temperature at which all substances have a minimum internal energy (NOT zero energy1). On the thermodynamic scale, it is assigned a value of 0 K. [1] 4 (a)(i) The Avogadro constant, N A, is defined as the number of atoms in 0.012 kg of carbon-12. (NA = 6.02 ×1023 mol-1) [1] (ii) An ideal gas is one that obeys the equation p V = n R T for all values of pressure, volume and temperature, where p: pressure (Pa), V: volume (m 3), n: amount of gas (mol), R: molar gas constant (8.31 J K-1 mol-1), T: thermodynamic temperature (K). [1] [1] (iii) The absolute scale of temperature is a theoretical scale that is independent of the properties of any particular substance. [1] (b) Molecular mass is the mass of one molecule. The molar mass (kg mol -1) is the mass per mole of a substance {NOT: mass of one mole}. The relative atomic mass (no units), Ar, of an atom is defined by the following equation: Ar = atom12carbonaofmassthe12 1 atomanofMass x NB: Students are not required to state the definition of relative atomic mass but are required to know how to deduce the molar mass of a substance given its relative atomic mass. [1] [1] [1] 5 Using V TnpTnpV RpressureInitialR  Final pressure, p’       pV nRT V TRn V Tn 028.1)002.1( )01.1)(02.1( 002.1 01.102.1 ' 'R'   Therefore the percentage increase in p is 2.8 %. [1] [1] 11(a) Mean speed = 1 s m   2605 500300100100300 [1] (b) Mean-square-speed =     2 22222 5 500300100100300   s m 90000 2 [1] 1 Originally, it was thought that all atomic motions would cease at absolute zero. The development of quantum mechanics (see chapter 18) showed that all motion does not cease; the atoms vibrate with the minimum possible motion.

11-2 p /105 Pa A B V / 10-3m3 C Level 2 Solutions 6 From RR n pVTTnpV  Path L → M: When V increases with p & n const, T increases proportionately, since T α V. Hence ans is B. Path M → N: T ∝ pV, i.e. T does not vary linearly with V now since p  constant. Comparing value of pV at M (2.0× 106 × 0.003) with pV at N(0.8× 106 × 0.005) ,  TM > TN {Confirms elimination of (C) & (D)} [1] [1] 7 average kinetic energy of the system  T Over the cycle ABCA, the system returns to the original state, So Tcycle = 0 so net increase in average kinetic energy of the system = 0 Net Increase in U = 0 as it returns to its original state because Tcycle = 0 & U  T. [2] [1] 8 (a)(i) pV = nRT  ∆n = ∆pV RT = (3.23 x 105 - 2.62 x 105)x 0.0120 8.31 x (25 + 273.15) = 0.295 mol [1] [1] [1] (ii) To supply 4 tyres, amount of air to be pumped out from the portable supply, L M N T V 3 1 5 15

11-3 ∆n = 4 x Ans (a)(i) = 1.18 mol. The subsequent decrease in pressure, ∆p, of this supply = ∆nRT V = (1.18)(8.31)(25 + 273.15) 0.0108 = 2.70 x 105 Pa Therefore, the pressure remaining in portable supply = pi - ∆p = 8.72 x 105 – 2.70 x 105 = 6.01 x 105 Pa, without falling below 3.23 x 105 Pa. [1] [1] [1] (b)(i) Average kinetic energy of one molecule of gas = 3 2 kT = 6.17 x 10-21 J. [1] [1] (ii) Average kinetic energy of one mole of gas = NA x Ans (b)(i) = 3710 J [1] (iii) Increase in total kinetic energy = Ans (a)(i) x Ans (b)(ii) = 1100 J [1] [1] 9 (a)(i) From the ideal gas equation, nRTpV  where n = mass molar mass   15.2732731.803.0064.0100.1 5  mass  mass = 0.0770 kg [2] [1] (ii) From pf Vf = nf R Tf (1.0 x 105)(0.064) = nf (8.31)(180 + 273.15) Hence nf = 1.70 mol = (1.70 x 0.030) kg = 0.051 kg = mf m = mi - mf = 0.077 – 0.051 = 0.026 kg [1] [1] 10 PV = nRT n = PV/RT Total amt of gas ntot = n1 + n2 = PV1/RT + PV2/RT = 1.01 x 105x (400 x 10-6 + 200 x 10-6)/(8.31x293.15) = 0.02488 mol ntot = n1 + n2 ntot = Pcom V1/RT1 + Pcom V2/RT2 0.02488 = Pcom x400 x 10-6/(8.31x373.15) + Pcom x200 x 10-6/(8.31x273.15) [1] 0.02488 = 2.171 x 10-7 x Pcom Pcom = 1.15 x 105 Pa [1] [1] 12 As the temperature in the cylinder increases, the average kinetic energy of the gas molecules increases. Thus the root-mean-square speed of the molecules also increases. With greater speed, when a particle collides against the wall, the momentum change per collision increased, exerting a greater force on the wall. As a result, the volume increases causing the collision frequency to decrease and the pressure to stay constant. {From examiner’s report: Any explanation using pV = nRT does not answer the question, since there is no reference to the forces exerted by molecules.} [1] [1] [1]

11-4 13(a) Thermal equilibrium means that X and Y are at the same temperature. Since translational kinetic energy is directly proportional to thermodynamic temperature, X has the same mean translational kinetic energy as Y, 6.0 x 10-21 J [1] (b) 𝑚𝑥<𝑐𝑋 2 > 2 = 𝑚𝑦<𝑐𝑦2> 2 , √<𝑐𝑦2> √<𝑐𝑥2> = √ 𝑚𝑥 𝑚𝑦 = √ 1 2 = 0.707 [1] [1] 14(i) For an ideal gas: 𝑃 = 𝑁𝑚 < 𝑐2 > 3𝑉  𝑃 = 1 3 𝜌 < 𝑐2 > Since 1 2 𝑚 < 𝑐2 > = 3 2 𝑘𝑇 , for the same temperature, < 𝑐2 > is constant.  𝑃 ∝ 𝜌 Since p is proportional to density of gas for a fixed temperature, it behaves like an ideal gas. Alternatively, thinking from pV = nRT pV = 𝑀𝑎𝑠𝑠 𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠RT p = 𝑀𝑎𝑠𝑠 𝑉 1 𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠RT p = 1 𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 R T p ∝  at a fixed temperature [1] (ii) Pick a suitable point from the graph. E.g. ( P = 1.5 x 10 5 and ρ= 1.75 at 300 K) P = 1 3 𝜌 < 𝑐2 > 1.5 x 105 = 1/3 x 1.75 <c2> rms speed = 507 ms-1 [1] [1] [1] (iii) For the same Pressure, the density value at T is at a lower value than at 300K, which means that the mean square speed is higher at T, (P = 1 3 𝜌1 < 𝑐1 2 > = 1 3 𝜌2 < 𝑐2 2 >)) so temperature T is higher. Or for the same density, the Pressure exerted at T is higher which implies that T is higher [1] [1] [1] [1] (iv) At density of 1.0, PT = 1.5 and P300 = 0.85 Since 𝑃𝑇 𝑃300 = 𝑇 300 T = 300 x 1.5/0.85 = 529 K [1] [1] [1] 15.(a) average force = Δp/Δt = 6.0x10-23/1x10-3 = 6 x 10-20N [1] (b) Average during the 1ms contact = 6 x 10-20 N Average force during the remaining 9 ms = 0 Average force during the 10 ms = 6 x10-20/(1+9) = 0.6 x 10-20 N Value is obtained = Area under graph for 1 collision / 10 ms [1] (c) The value is (b) is used. [1]

11-5 16. (a) Since ½ m <c2> α T and they are at the same temperature, ½ 𝑚𝑁 <𝑐𝑁 2> = ½ 𝑚𝑂 <𝑐𝑂 2> Ratio 28 32 =1.07 [1] [1] (b) Since ½ m <c2> α T and they have the same mass,    2 1 : 100 10 2 100 2 10 T T c c  Ratio = 15.373 15.283 = 0.871 [1] [1] 17(a)(i) p = mv – mu = mu – (-mu) = 2mu [1] (