Physics · May / June 2020 · Paper 41 · Question 3
0625
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The question as text
Fig. 3.1 shows gas trapped in the sealed end of a tube by a dense liquid. cm3 10 20 30 40 50 60 70 trapped gas open end sealed end dense liquid Fig. 3.1 The scale marked on the sealed end of the tube is calibrated to read the volume of gas trapped above the liquid surface. Fig. 3.1 shows that initially the volume V1 of the gas is 60 cm3. The pressure of the atmosphere is 1.0 × 105 Pa.
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3(a)[1]
liquid levels in the two limbs of the tube are equal
3(b)[3]
molecules collide with the walls (of the container)
momentum of molecules changes (reverses)
this causes a force AND force spread out (over area of walls)
3(c)(i)[2]
(p2 =) p1V1 / V2 = 1.0 × 105 × 60 / 50
1.2 × 105 Pa
3(c)(ii)[2]
p2 = patm + hρg OR 1.2 × 105 – 1.0 × 105 OR 2.0 × 104 OR (ρ =) 2.0 × 104 / (0.15 × 10)
1.3 × 104 kg m–3
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