Physics (AS & A Level) · Oct / Nov 2016 · Paper 42 · Question 2
9702
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2(a)(i)[1]
number of moles/amount of substance [1]
2(a)(ii)[1]
kelvin temperature/absolute temperature/thermodynamic temperature [1]
2(b)[3]
pV = nRT 4.9 × 105 × 2.4 × 103 × 10–6 = n × 8.31 × 373 n = 0.38 (mol) number of molecules or N = 0.38 × 6.02 × 1023 = 2.3 × 1023 [3]
or pV = NkT (C1) 4.9 × 105 × 2.4 × 103 × 10–6 = N × 1.38 × 10–23 × 373 (M1) number of molecules or N = 2.3 × 1023 (A1)
2(c)[3]
volume occupied by one molecule = (2.4 × 103) / (2.3 × 1023) = 1.04 × 10–20 cm3 mean spacing = (1.04 × 10–20) 1/3 = 2.2 × 10–7 cm (allow 1 s.f.) [3]
(allow other dimensionally correct methods e.g. V = (4/3)πr3)
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