Answers for Exercise in Power Point (Module: Solid, Liquid & Gases)
IJSO Training: Solid, Liquid & Gases
Answers for Exercises in Power Point
p. 28:
p0 = 0.987 atm = 750 cm Hg = 75.0 cm Hg
p = 110.2 cm Hg
p. 32:
p2 = 7 L
p. 42:
T1 = 200 °C = 473 K
V2 = 100 cm3 = 0.1 L
 T2 = 200 °C = 4730 K
p.44:
p1 = 30 psi
T1 = 25 °C = 298 K
T2 = 100 °C = 373 K
V2 = 1.1 V1
 p2 = (T2 / T1 ) (V1 / V2 ) p1 = 34 psi
p.49:
(i) JK -1mol -1
(ii) JK -1
(iii) k = 1.381 x 10 -23 JK -1
p.51:
p = 700 torr = 93100 Pa
V = 1 L = 1 x 10 -3 m3
T = 25 °C = 298 K
R = 8.314 JK –1
 n = 2.6 mol
p.52:
p1 = 1700 torr
T1 = 25 °C = 298 K
T2 = 0 °C = 273 K
V2 = V1 = constant
n2 = 0.5 n1
 p2 = (n2 / n1 ) (V1 / V2 ) (T2 / T1 ) p1 = 779 torr
Answers for Exercise in Power Point (Module: Solid, Liquid & Gases)
p.58:
(i)
(ii)
(iii)
Using p1 V1 = p2 V2 separately for each gas, we have:
p2(O2) = 0.75 atm
p2(N2) = 1 atm
p2(CO2) = 0.5 atm
The partial pressures for the three gases after mixing is the same as their respective pressures at the same
volume and temperature. Therefore, their partial pressures are:
p(O2) = 0.75 atm
p(N2) = 1 atm
p(CO2) = 0.5 atm
The total pressure of the mixture is the sum of partial pressures = 2.25 atm
p.59:
p = 1 atm = 101325 Pa
T = 273.15 K
R = 8.314 J
n = 1 mol
Substituting these values into pV = nRT, we have V = 0.022414 m3 = 22.414 L
–1
 Standard molar volume of a gas = 22.414 L mol
p.99:
(1)
(2)
Many organic substances are more soluble in non-polar CO2 than in polar water
The temperature of supercritical CO2 is lower than that of boiling water and hence organic substances are less
likely to be destroyed during extraction.
(3) CO2 is easier to remove after extraction.
(4) V1 = V2 = constant
1 L = 1 x 10 -3 m3
T = 25 °C = 298 K
R = 8.314 JK –1
 n = 2.6 mol