RI 04 The Gaseous State (Ans to self-check)

Answers to Self-check Questions Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 D B D B C D A B C Explanations to Self-check Questions 1 Concept: Manipulating pV = nRT to determine if the equations available are legitimate. Eqn 1: Manipulate ideal gas equation to contain the density term, 𝜌𝜌 Since 𝑛𝑛 = 𝑚𝑚 𝑀𝑀 pV = nRT pV = � 𝑚𝑚 𝑀𝑀�RT Rearrange to give 𝜌𝜌 wh ich is 𝑚𝑚 𝑉𝑉 p = � 𝑚𝑚 𝑉𝑉 � 𝑅𝑅𝑅𝑅 𝑀𝑀 = 𝜌𝜌𝑅𝑅𝑅𝑅 𝑀𝑀 (option 1 is correct) Eqn 2: This is an incorrect expression of ideal gas equation which is pV = nRT, not MRT as shown in Eqn 2. (Option 2 is incorrect) Eqn 3: Manipulate ideal gas equation to contain the concentration term, c pV = nRT p = � 𝑛𝑛 𝑉𝑉� 𝑅𝑅𝑅𝑅 Since c = 𝑛𝑛 𝑉𝑉 , the equation can be written as p = cRT (option 3 is incorrect). 2 Concept: Using pV = nRT to solve for Mr. Involves units conversion from ºC to K and cm3 to m3. pV = nRT = 𝑚𝑚 𝑀𝑀 𝑅𝑅𝑅𝑅 ∴ M = 𝑚𝑚𝑅𝑅𝑅𝑅 𝑝𝑝 𝑉𝑉 = (0.200)(8.31)(30 + 273) (101 × 103)(96.0 × 10−6) = 52.0 g mol−1 ∴ Mr = 52.0 Ans: B Note: 1 cm 3 = 1 x 10–6 m3

3 Concept: Using pV = nRT to express the pressure exerted. Involves unit s conversion from ºC to K and dm3 to m3. pV = nRT p = 𝑛𝑛𝑅𝑅𝑅𝑅 𝑉𝑉 = 1.6 × 10–3 × 8.31 × (273+273) 3.0 × 10–3 Ans: D Note: 1 dm3 = 1 x 10–3 m3 4 Concept: By conservation of mass p1V1 + p2V2 = pTVT (50)(10) + (100)(30) = pT × 40 pT = 87.5 kPa Or applying Boyle’s Law which states that p is inversely proportional to V for a fixed mass of gas under constant T For smaller spacecraft: p1V1 = p2V2 10 x 50 = p2 x 40 p2 = 12.5 kPa For bigger spacecraft: p3V3 = p4V4 30 x 100 = p4 x 40 p4 = 75 kPa pT = 12.5 + 75 = 87.5 kPa Ans: B 5 Concept: Refer to page 11 of notes At low pressure, the gas particles are very far apart and can be considered to have negligible volume compared to the volume of the container. In addition, the intermolecular attractive forces between the widely spaced gas molecules are negligible. At high temperature, the gas molecules possess sufficiently high kinetic energy to overcome the intermolecular attractive forces. Hence the intermolecular attractive forces can be considered negligible. Ans: C 6 Concept: Since mass and volume are fixed, ( 𝑝𝑝1 𝑅𝑅1 = 𝑝𝑝2 𝑅𝑅2 ) can be used to solve for T. Involves units conversion from ºC to K.

𝑝𝑝1 𝑅𝑅1 = 𝑝𝑝2 𝑅𝑅2 𝑝𝑝 27 + 273 = 2𝑝𝑝 𝑅𝑅2 ∴ T2 = (27 + 273) × 2 = 600 K Ans: D 7 Concept: The stronger the intermolecular attractive forces, the greater the deviation from ideal gas. From the given graph, C deviates from ideal gas behaviour more greatly than B. This implies that C has stronger intermolecular forces of attraction compared to B. B C A CH4 (non-polar) • Id-id interactions NH3 (polar) • Pd-pd and id-id interactions and hydrogen bonding • Greater deviation from ideal gas B HF (polar) • Pd-pd, id-id interactions and hydrogen bonding • Greater deviation from ideal gas HBr (polar) • Pd-pd and id-id interactions C CH3CH2Cl (polar) • Pd-pd and id-id interactions • Greater deviation from ideal gas CH3CH3 (non-polar) • Id-id interactions D O2 (non-polar) • Id-id interactions • Larger electron cloud (stronger id-id interactions) • Greater deviation from ideal gas H2 (non-polar) • Id-id interactions • Smaller electron cloud (weaker id-id interactions) Ans: A

8 Concept: Dalton’s law of partial pressure. Calculating partial pressure of a gas in a mixture. Involves units conversion from atm to Pa. nSO2 = 32.1 32.1+16+16 = 0.50078 mol nNH3 = 25.5 14+3 = 1.50 mol pSO2 = xSO2. ptotal = 0.50078 1.50+ 0.50078 . (0.369 X 101325) = 9358.2 Pa = 9.36 kPa (3 s.f.) Ans: B 9 Concept: Rearrange the ideal gas equation pV=nRT in the form y = mx + c, according to what is given in the x and y axis. A p = nRT 1 V . Hence this is a y=kx graph where k=nRT. At higher temperature T2, the gradient of the line should be steeper. (wrong graph) B pV = nRT. Since pV is always a constant (at constant n and T) and not affected by value of p, this is a y=k graph where k=nRT. At T 2, the line should be at a higher value. (wrong graph) C p = nRT 1 V . Hence this is a y= k x graph. There is an inverse relationship between p and V. For a given volume, the pressure should be higher at T2. Hence the graph should shift upwards and outwards. (correct graph) D V = nRT p . Hence this is a y=kx graph where k= RT p . At T2, the gradient of the line should be steeper. (wrong graph) 10(a) At constant T (i) p vs V (ii) p vs 1 V (iii) pV vs p (iv) pV vs V 0 pV p 0 pV V 0 p V 0 1 V p

(b) At constant p (i) V vs T/oC (ii) density vs T/K (c) At constant n (i) pV vs T/K (ii) pV T vs p V T/oC −273 0 density T/K 0 pV T p 0 pV 0 T/K