THE PROPERTIES OF GASES
7
(b) The total pressure is determined from the total amount of gas, n = nCH4 + nAr + nNe . nCH4 =
0.320 g 16.04 g mol−1
= 1.995 × 10−2 mol
nAr =
0.175 g 39.95 g mol−1
= 4.38 × 10−3 mol
n = (1.995 + 0.438 + 1.115) × 10−2 mol = 3.548 × 10−2 mol p=
nRT (3.548 × 10−2 mol) × (62.36 L Torr K−1 mol−1 ) × (300 K) [1] = V 3.137 L = 212 Torr
E1.14(b)
This is similar to Exercise 1.14(a) with the exception that the density is first calculated. RT [Exercise 1.11(a)] p 33.5 mg ρ= = 0.1340 g L−1 , 250 mL
M=ρ
M= E1.15(b)
p = 152 Torr,
T = 298 K
(0.1340 g L−1 ) × (62.36 L Torr K−1 mol−1 ) × (298 K) = 16.4 g mol−1 152 Torr
This exercise is similar to Exercise 1.15(a) in that it uses the definition of absolute zero as that temperature at which the volume of a sample of gas would become zero if the substance remained a gas at low temperatures. The solution uses the experimental fact that the volume is a linear function of the Celsius temperature. Thus V = V0 + αV0 θ = V0 + bθ, b = αV0 At absolute zero, V = 0, or 0 = 20.00 L + 0.0741 L◦ C−1 × θ(abs. zero) θ (abs. zero) = −
E1.16(b)
20.00 L 0.0741 L◦ C−1
= −270◦ C
which is close to the accepted value of −273◦ C. nRT (a) p= V n = 1.0 mol T = (i) 273.15 K; (ii) 500 K V = (i) 22.414 L; (ii) 150 cm3 (1.0 mol) × (8.206 × 10−2 L atm K−1 mol−1 ) × (273.15 K) 22.414 L = 1.0 atm
(i) p =
(1.0 mol) × (8.206 × 10−2 L atm K−1 mol−1 ) × (500 K) 0.150 L = 270 atm (2 significant figures)
(ii) p =
(b) From Table (1.6) for H2 S a = 4.484 L2 atm mol−1 nRT an2 p= − 2 V − nb V
b = 4.34 × 10−2 L mol−1
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