Page 44 - PRE-U STPM CHEMISTRY TERM 1

Chemistry Term 1 STPM

2
PV at P = 0, is 23.5  10 Pa m 3 (ii) Total pressure
5
5
3
Using PV = nRT = (1.30  10 ) + (3.75  10 ) + (5.60  10 ) Pa
2
23.5  10 = n  8.31  298 = 5.11  10 Pa
5
∴ n = 0.949 mol (iii) P(Ar) = 1.30  10 Pa
5
32.5
5
∴ M = 0.949 = 34.2 P(Ethane) = 3.75  10 Pa
r
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Total pressure = (1.30 + 3.75)  10 Pa
7 (a) PV = nRT = 5.05  10 Pa
5
m
PV = —–RT 9 Solid:
M r • Strong attractive forces between particles.
1
m
P = —(—–) RT • Particles have no translational motion.
V M r
m • Particles vibrate and rotate in their mean position.
But — = d (density)
V Liquid:
d • Intermediate strength attractive forces between
∴ P = —–RT
M r particles.
d M r • Particles can move freely in the body of the liquid but
or — = —–
P RT not out of it.
(b) (i) When pressure increases, the volume occupied Gas:
by the gas decreases. This causes the density • No intermolecular forces between particles.
 Volume  to increase. • Particles can move freely throughout the container
Mass
(ii) where it is placed.
10 (a) The pressure exerted by water vapour which is in
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Pressure (× 10 )/Pa 33.7 67.6 100.0
equilibrium with liquid water at a fixed temperature.
The pressure is due to the collisions of the water
Density 2.85  10 –5 2.86  10 –5 2.93  10 –5
–3
Pressure /kg m Pa –1 vapour molecules with the walls of the container.
(b)
d /kg m Pa –1
–3
P Pressure
3.0 × 10 –5
2.9 × 10 –5
Temperature
Increasing temperature increases the kinetic energy
2.8 × 10 –5 of the water molecules. More water will undergo
evaporation. At the same time, the collisions between
2.7 × 10 –5 the vapour molecules and the walls of the container is
0 10 20 30 40 50 60 70 80 90 100 more energetic.
2
P (× 10 )/Pa (c) Decreases. Fraction of the liquid surface for water to
evaporate is decreased.
d –5
P at P = 0 is 2.80  10 11 (a) Melting point: The temperature at which a solid is in
M
Using d = r equilibrium with its liquid at 101 kPa.
P RT Boiling point: The temperature at which the vapour
–5
∴ M = (2.80  10 )  (8.31  301.5)
r pressure of a liquid is equal to the external pressure
= 0.0702 kg mol –1 (101 kPa).
∴ Relative molecular mass = 70.2 (b)
8 (a) Refer to text
(b) (i) For argon: Pressure/atm Sea
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(2.0  10 )  2.60 = P  4.0 1.0
P(Ar) = 1.30  10 Pa Liquid
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For ethane: Solid
–5
(1.2  10 )  12.5 = P  4.0
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P(Ethane) = 3.75  10 Pa Vapour
For CO :
2 Temperature/°C
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(2.8  10 )  0.80 = P  4.0 0 100
3
P(CO ) = 5.60  10 Pa
2
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12 Answers.indd 367 3/26/18 4:06 PM