Answers to Chapter 13 Homework
1. The contraction of the glass causes the enclosed volume to decrease as if it were glass.
The volume of water that can be added is
Dv = Dvglass - D>vwater = V0bglassDT - V0>bwaterDT = V0(bglass - bwater)DT
= (350mL)[27 x 10-6(C°)-1 - 210 x 10(C°)-1](20°C - 100°C) = 5.1mL.
2. (a) T1(K) = T1(°C) + 273 = 4000°C + 273°C = 4273K;
T2(K) = T1(°C) + 273 = 15 x 106°C + 273 = 15 x 106K.
(b) The difference in each case is 273, so we have
Earth: (273)(100)/(4273) = 6.4%;
Sun: (273)(100)/(15 x 106) = 0.0018%.
3. If we assume oxygen is an ideal gas, we have
PV = nRT = (m/M)RT;
(1.013 x 105Pa)V = [m/(32g/mol)(103kg/g)](8.315J/mol.K)(273K), which gives
m/V = 1.43kg/m3.
4. (a) For the ideal gas we have
PV = nRT;
(1.000atm + 0.350atm)(1.013 x 105Pa/atm)V = (25.50mol)(8.314J/mol.K)(283K),
which gives, V = 0.439m3.
(b) For the two states of the gas we can write
P1V1 = nRT1 and P2V2 = nRT2, which can be combined to give
(P2/P1)(V2/V1) = T2/T1;
[(1.00atm + 1.00atm)/(1.00atm + 0.350atm)](1/2) = (T2/283K),
which gives T2 = 210K = - 63°C.
5. (a) We find the number of moles from
n = r(4/3)V/M = r(3/4)4pR2d/M
= (1000kg/m3)3p(6.4 x 106m)2(3 x 103m)(103g/kg)/(18g/mol) = 6 x 1022mol.
(b) For the number of molecules we have
N = nN = ( 6 x 1022mol.)(6.02 x 1023molecules/mol) = 4 x 1046 molecules.
6. The average kinetic energy depends on the temperature:
1/2mvrms2 = 3/2KT.
We form the ratio at the two temperatures, and use the ideal gas law:
(m2/m1)(vrms2/vrms1)2 = T2/T1 = 1, so we have
(vrms2/vrms1)2 = (m2/m1), or vrms2/vrms1 = (m2/m1)1/2.