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