# Chapter 11 - Worked Solutions

### to selected extension questions from the OUP text Senior Physics - Concepts in Context by Walding, Rapkins and Rossiter

Q45.
25 % of energy goes into rotational energy etc., therefore 75% of the energy put in goes into Kinetic energy.
The average kinetic energy of a molecule :Ek = 3/2.k.T
At T1, Ek = 3/2.k.T1
total Ek = 3/2 kT1 x N
At T2 the total Ek = 3/2 k T2 x N
Change in total Ek = 3/2 k T2 x N - 3/2 k T1 x N = 3/2 k x N (T2 - T1) = 1.38 x 10-23 x 2 x 6.02 x 1023 x (35 - 10)
= 6.23 x 102 J
This is ¾ of the total energy put in.
Total Energy put into the system = 4/3 x 6.23 x 102 J = 8.31 x 102 J
Q46.
Since Helium and Argon are ideal gases, the energy put into them changes the kinetic energy of the molecules.
For Helium: total Ek = 3/2 k T N
= 3/2 k (273 + 50) x 2 x NA
= 3/2 k x 323 x 2 x NA
For Argon: total Ek = 3/2 kT N
= 3/2 k (273 + 20) x 4 x NA
= 3/2 k x 293 x 4 x NA
Total energy = kinetic energy for Helium + kinetic energy for Argon.
= 3/2 k x 323 x 2 x NA + 3/2 k x 293 x 4 x NA
= 3/2 k NA (646 + 1172)
= 3/2 k NA x 1818 J
For the whole system (consisting of 6 moles):total Ek = 3/2 kT N
3/2 k NA x 1818 J = 3/2 kT x 6 x NA
1818 = 6 T
T = 303 K
T = 30 oC
Q47. (a)
2 moles of X at 70 oC mixed with 1 mole of Y at 25 oC
All of the energy of X is Ek = 3/2 k T N
= 3/2 1.38 x 10-23 x 2 x 6.02 x 1023 x (273+ 70) = 8549 J
The total Ek in Y = 3/2 k T N = 1.38 x 10-23 x 1 x 6.02 x 1023 x (273 + 25) = 3713 J
This is only = ¾ of the total energy in Y
Therefore the total energy in Y = 4/3 x 3713 J = 4951 J
Total energy of the system = 8549 + 4951 J = 13500 J
Since, on average, the kinetic energy of each mole of the combined gas is the same (since the temperature is constant.) The kinetic energy of the system is 9/10 of the total energy of the system.
( For Y the Ek : other energies = 3 : 1. Let the Ek of a mole of Y be 3, then the Ek of each mole of X has to be 3. Then the Ek of the system = 9/10 of the total energy)
Ek = 9/10 x 13500 = 12150 = 3/2 k T N
Hence T = Ek/3/2kN = 12150/(3/2 x 1.38 x 10-23 x 3 x 6.02 x 1023) = 325 K = 52 oC

Q47. (b)
2 moles of Y at 80 oC mixed with 2 mole of X at 35 oC
Kinetic Energy of Y = 3/2 k T N =3/2 k (273+ 80) 2 NA = 3/2 k NA706
Total energy in Y = 4/3.3/2. k NA 706 = 3/2 k NA 941
Kinetic Energy of X = 3/2k T N = 3/2 k (273+ 35) 2 NA = 3/2 k NA616
Energy of the system = energy in Y + energy in X
= 3/2 k NA (941 + 616)
= 3/2 k NA 1557
Ek of system = 12/14 of the total energy = 12/14 x 3/2 k NA 1557
that is 3/2 k N T = 12/14 x 3/2 k NA 1557. (See question 47a)
T = 3/2 k NA 1335/(3/2 k NA4) = 333.6 K = 60.6 oC

Q48.
(a)
The pressure at the surface equals one atmospheric pressure = 1 Atmos.
The volume at the surface V2 = 3 V1
P1 V1 = P2 V2
P1 V1 = 1 Atmos x 3 V1
P1 = 3 Atmos
but 2 Atmos of this pressure is due to the depth of water
P =r g h.
2 x 1.013 x 105= 1000 x 9.8 x h
h = 20.6 m
(b)
P2 = 1 Atmos, T1 = 7 + 273 K, P1 = 3 Atmos, T2 = 27 + 273 K
P1.V1/T1 = P2.V2/T2
V2 = 3V1 x 300/280 = 3.2 V1.

Q49. T = 20 oC = 293 K
P = 76 cm of Hg = 1.013 x 105 Pa
The total number of molecules remains the same:
Originally the number of molecules = N
P V = N k T or N = P V/kT
Originally N = (1.013 x 105 x3 V)/293 k
Finally the number of molecules are N1 and N2
Hence N = N1 + N2
Finally N1 = P V/293 k and N2 = P 2 V/373 k
therefore equating and solving for P
P = 1.15 x 105 Pa or 86 cm of Hg

Q50.
Acetone expands by:DV = b V1 DT
= 14.87 x 10-4 x .5 x10-3 x 12
= 89.2 x 10-7 m3
= 8.9 mL
therefore the volume of air in the top = 10 - 8.9 mL = 1.1 mL
P1 = 1 Atmos
V1 = 10 mL
T1 = 20 + 273 = 293 K
P2 = ?
V2 = 1.1 mL
T2 = 32 + 273 = 305K
P1.V1/T1 = P2.V2/T2
P2 = 1 x 10 x 305/(293 x 1.1) = 9.46 Atmos

Q51.
Since it is at constant pressure V1/ T1 = V2 /T2
at 90o M V = 30 L
at 120oM V = 45 L
In K
V1/ T1 = V2 /T2
30/x = 45/(x + 30)
(x + 30 since a change of 1 oM is the same as a change of 1K)
therefore 30x + 900 = 45x
900 = 15x or x = 60
check: 30/60 = 45/90
therefore x = 60K
therefore 90oM = 60K = -213 oC
and 120oM = 90 K = -303oC
therefore oM = oC - 303 Return to Physics Text Home Page