Worksheet markscheme
Cambridge 9702_s11_qp_42 Q2
1. (a) amount of substance M1
containing same number of particles as in 0.012 kg of carbon-12 A1 [2]
(b) pV = nRT C1
amount = (2.3 × 105 × 3.1 × 10–3) / (8.31 × 290)
+ (2.3 × 105 × 4.6 × 10–3) / (8.31 × 303) C1
= 0.296 + 0.420 C1
= 0.716 mol A1 [4]
(give full credit for starting equation pV = NkT and N = nNA)
Cambridge 9702_s12_qp_41 Q2
2. (a) (i) either random motion
or constant velocity until hits wall/other molecule B1 [1]
(ii) (total) volume of molecules is negligible M1
compared to volume of containing vessel A1
or
radius/diameter of a molecule is negligible (M1)
compared to the average intermolecular distance (A1) [2]
(b) either molecule has component of velocity in three directions
or c2 = cX2 + cY2 + cZ2 M1
random motion and averaging, so <cX2> = <cY2> = <cZ2> M1
<c2> = 3<cX2> A1
so, pV = ⅓Nm<c2> A0 [3]
(c)
<c2> ∝
T or crms ∝
C1
temperatures are 300 K and 373 K C1
crms = 580 m s–1 A1 [3]
(Do not allow any marks for use of temperature in units of oC instead of K)
Cambridge 9702_s04_qp_4 Q2
3. (a) mean (value of the) square M1
of the speeds (velocities) of the atoms/particles/molecules A1 [2]
(b) (i) p = 1/3ρ<c2> C1
<c2> = 3 x 2 x 105/2.4 = 2.5 x 105 C1
r.m.s speed = 500 m s-1 A1 [3]
(ii) new <c2> = 1.0 x 106 or <c2> increases by factor of 4 C1
<c2> ∝ T or 3/2 kT = 1/2 m<c2> C1
T = {(1.0 x 106) / (2.5 x 105)} x 300
= 1200 K A1 [3]
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