RI 2025 Tut 4 The Gaseous State (Suggested Ans)

3 Raffles Institution Year 5 H2 Chemistry 2025 Tutorial 4 – The Gaseous State (Suggested Answers for Discussion Questions) Discussion Questions 11 Mass of CO2 in 500 cm3 of saturated solution = 0.500  1.50 = 0.750 g Mass of CO2 released = 2.00 – 0.750 = 1.25 g Amount of CO2 released = 1.25/molar mass = 1.25/44.0 = 0.0284 mol Volume of CO2 released = 0.0284  24 = 0.682 dm 3 = 682 cm3 12 (a) For a fixed amount of ideal gas at constant T, pV remains constant as p increases. (b) Using the ideal gas equation, pV = nRT 𝑉ൌ ሺ0.40ሻሺ8.31ሻሺ300ሻ 12.0 ൈ 10ହ = 8.31 ൈ 10-4 m3 = 0.831 dm3 (c) pressure, p / Pa volume, V / dm3 pressure x volume, pV / Pa dm3 5.0 ൈ 105 1.924 9.62 ൈ 105 10.0 ൈ 105 0.926 9.26 ൈ 105 15.0 ൈ 105 0.592 8.88 ൈ 105 When p = 12.0 ൈ 105 Pa, estimated pV = 9.26 ൈ 105 – ଶ ହ (9.26ൈ 105 – 8.88 ൈ 105) = 9.11 ൈ 105 Pa dm3 When estimated 9.11 ൈ 105 Pa dm3, V = 9.11 ൈ 105 Pa dm3 ൊ 12.0 ൈ 105 Pa = 0.759 dm3 (d) Value of V in (c) is smaller than that in (b). This implies that the gas occupies a volume smaller than what it would were it ideal. As the pressure exerted on the system is moderately high, the polar HC l gas molecules come closer together and the intermolecular attractive forces (id-id and pd-pd) between the HCl gas molecules become significant, causing the volume to be smaller than what it would were it ideal.

4 13 (a) 2 main assumptions are:  The gas particles exert negligible in termolecular attractive forces on one another.  The volume of gas particles is negligib le compared to the volume of container. Reject: volume of gas is negligible (missing ‘particles’) Other assumptions:  The gas particles are in constant random motion.  Collision between gas particles are perfectly elastic.  The average kinetic energy of gas particles is proportional to absolute temperature. (b) 8.31pV RnT  J K–1mol–1 (c) Under low pressure and high temperature  At low pressure, the gas particles are very far apart. Hence the volume of the gas particles can be considered negligible compar ed to the volume of the container. In addition, the intermolecular attractive forces between widely spaced gas particles are also negligible.  At high temperature, the gas particles pos sess sufficiently high kinetic energy to overcome the intermolecular attractive forces. Hence, the intermolecular attractive forces can be considered negligible. (d) At moderately high pressure:  O 2 molecules come closer together and the intermolecular attractive forces between O 2 molecules become significant. This causes the gas to occupy a volume smaller than that of an ideal gas. This results in a smaller V and pV/nT < 8.31 (negative deviation). At very high pressure:  The molecular size of O 2 gas particles is significant, resulting in the gas to occupy a larger V than expected and and pV/nT > 8.31 (positive deviation).  The O 2 molecules are so close together that the repulsive forces between their electron clouds become significant.  Thus, the volume of the gas particles is not negligible as compared to the volume of the container. pV/nT (J K–1mol–1) p/Pa 8.31 For part 13(d) O2 Ideal gas

5 Notes for consideration In explaining deviation of a particular real gas versus ideal gas, it is still useful to look back at the two assumptions of ideal gas behavior (i.e. negligible IMF and negligible particle size) in (a) and how these two assu mptions are violated when we look at a real gas. Here, O2 gas deviates from ideal gas behavior be cause 1) it experiences significant intermolecular attractive forces (id-id) and 2) molecular size of O 2 are not negligible – both points violate the ideal gas assumptions. The second point on repulsive forces will help to bring out the trend in positive deviation. 14 (a) (i) At the intercept of the pV axis (low pressure), both gases approach ideal gas behaviour.  pV = nRT = (1)(8.31)(298) = 2.48 x 103 J (or Pa m3) (to 3 sf) (ii) CO2 shows greater negative deviation at moderately high pressure and greater positive deviation at very high pressure.  For greater negative deviation: CO2 has a greater electron cloud/ number of electrons compared to H2. Thus, CO2 forms stronger instantaneous dipole – induced dipole interactions as its electron cloud is more polarisable.  For greater positive deviation: The molecular size of CO2 molecules is significantly larger than that of H2. The size of electron cloud for CO2 is bigger than that of H2. This results in stronger repulsive forces between CO2 molecules compared to H2 molecules. (b) At moderately high pressures, the pV of HF has a greater negative deviation from ideality than CO 2 due to stronger hydrogen bonds between HF molecules as compared to the weaker id-id interactions between CO2 molecules.

6 Comments for tutors: You may wish to discuss positive deviation of HF. Usually assessme nt questions will have clearer mark allocation and specify what needs to be discussed. In the case of HF, likely negative deviation is the focus due to strong hydrogen bonds. At very high pressure, the pV of HF has a positive deviation which is lesser than CO 2 but greater than H 2. This is because HF has a smaller molecular size than CO 2 but larger molecular size than H2. 15 (a) Amount of oxygen = 120/(16.0  2) = 3.75 mol Amount of hydrogen = 5.00/(1.0  2) = 2.50 mol Mole fraction of oxygen, xO2 = 3.75/(3.75 + 2.50) = 0.600 Let the final total pressure be pT. pO2 = xO2 pT , 3.00 = 0.600 x p T  pT = 5.00 atm (b) On exploding the mixture: 2H 2 (g) + O2 (g)  2H2O (l) 2.50 3.75 H2 is limiting and all H2 will be used up. Amount of oxygen that remains after the reaction = 3.75 - (2.50/2) = 2.50 mol Assume that the pressure and volume of water is negligible (water formed condenses upon cooling)  the final pressure is only due to 2.50 mol of O2. Under the same conditions, 3.75 mol of oxygen has a pressure of 3.00 atm  2.50 mol of oxygen would have a pressure = 2.50/3.75 x 3.00 = 2.00 atm 16 Since Xe and F2 are mixed in equal volumes, the partial pressure of both gases is the same. Let the partial pressure of Xe (or F2) be p. Initial total pressure = p + p = 2p (Dalton’s Law of partial pressure) Final total pressure = 0.70 x 2p = 1.4p Final partial pressure of Xe = (1.4p) = 0.8p Final partial pressure of F 2 = 1.4p – 0.8p = 0.6p Since V is constant (sealed vessel) and assuming constant T and ideal gas behaviour: pV = nRT  n  p Hence, amount of Xe reacted  (p – 0.8p) = 0.2p amount of F 2 reacted  (p – 0.6p) = 0.4p i . e . 22 0.2 1 0.4 2 Xe Xe FF np p np p  or 1 mol Xe reacts with 2 mol F2.  Xe (g) + 2F 2 (g)  XeF4 (s) 7 4