WJEC Physics for A2: Study and Rev Guide

223 Questions and answers Q & A 5 The product of the pressure and volume of an ideal gas may be expressed as pV = nRT . The product may also be written in terms of the mean square speed of the molecules as pV = 1 3 Nmc 2 (a) Derive in clear steps a formula that shows how the internal energy of the ideal gas depends on the temperature of the gas. [4] (b) A canister of volume 0.025 m 3 contains helium gas at a pressure of 305kPa and a temperature of 18°C . Calculate: (i) the internal energy of the gas [2] (ii) the number of molecules of helium in the canister. [2] Tom’s answer (a) Combining pV = nRT and pV = 1 3 Nmc 2 we get nRT = 1 3 mc 2 ✓ The energy of the gas is the kinetic energy of the molecules ✓ = 1 2 mc 2 ✗ So the internal energy is 3 2 nRT ✗ (b) (i) From part (ii), the number of moles is 0.0510 So U = 3 2 nRT = 3 2 × 0.0510 × 8.31 × 18 = 11.4 J ✗✓ ecf (ii) n = pV RT = 305 × 0.025 8.31 × 18 = 0.0510 ✗ \ Number of molecules = 0.0510 × 6.02 × 10 23 = 3.07 × 10 22 ✓ ecf Examiner commentary (a) Tom starts well by combining the two given equations and identifying the internal energy of the gas with the molecular kinetic energy, which he unfortunately gives as the mean energy of an individual molecule. He then mysteriously writes the equation which is given in the Data booklet. (b) Tom answers this in an unusual way by first answering part (ii) and using it to answer part (i). This is legitimate and the examiner has applied the e.c.f. rules accordingly, though the mark allocation for part (i) becomes rather generous. (ii) Tom uses the correct equation for calculating the number of moles but unfortunately he has fallen into both unit traps: he should have converted to Pa and K. However, he then goes on to use n to calculate the number of molecules and gains the second mark. (i) The ecf value of n is accepted but he again loses out on the unit trap. Tom scores 4 out of 8 marks. Seren’s answer (a) Ideal gas molecules don’t have forces between them so the internal energy is just their kinetic energy. ✓ \ Internal energy, U = 1 2 Nmc 2 ✓ From equation 2, Nmc 2 = 3 pV , so U = 1 2 × 3 pV = 3 2 pV ✓ So, from equation 1, U = 3 2 nRT ✓ (b) (i) U = 3 2 pV = 3 2 × 305 × 10 3 Pa × 0.025 m 3 ✓ = 11 4000 J (3 s.f.) ✓ (ii) n = pV RT = 305 × 10 3 Pa × 0.025 m 3 8.31 J K –1 × 291 K = 3.15 mol ✓ Examiner commentary (a) A good answer by Seren. She makes the important point about the lack of intermolecular forces and correctly states the internal energy as U = 1 2 Nmc 2 . Following this she logically and clearly uses the given equation to derive the Data booklet formula. Note that the mark is for the working towards the equation and not for the equation itself. (b) (i) Seren has seen the easiest way of tackling this using the U = 3 2 pV , which is not in the Data booklet, but arose from her part (a) working. She skilfully avoids the unit trap. (ii) Again, Seren converts both units and correctly calculates the number of moles of the gas. Unfortunately she fails (forgets?) to go on to calculate the number of molecules. Seren scores 7 out of 8 marks.