Exercise Solution

7. We derive the coefficient of expansion from a ratio of lengths

(DL/L₀)(1/DT).

Since the lengths cancel, the coefficient does not depend on the specific unit of length used.

We just need to use the same units for delta L and the initial length.

 

2. We find the number of atoms in a pure substance by dividing the mass of the substance by the mass of a single atom. The atomic mass of gold is 197 and that of silver is 108 (can be obtained from the inside back cover of your texts)

 

NAu  = (2.65 x 10^-2 kg/(197)(1.66x10^-27 kg/atom) = 108

 

NAg = (2.65 x 10-2 kg)/(108)(1.66x10-27kg/atom) = 197

 

NAu/NAg = 0.548     ð    NAu = 0.548NAg

Because a gold atom is heavier than a silver atom, there are fewer gold atoms in the given mass.

 

3. Temperature Conversions

 

a. 20 C

b. 3300 F

 

12. We use the volume expansion formula

 

DV = bV₀DT = (1x10^-6/C)[4/3p(8.75 cm/2)³](200 C - 30 C) = 6.0 x 10^-2 cm³     1 m³ = 1000 L  10⁶ cm³ = 1000 L   

 

26. Oops, more conversions

 

We use the relationships provided in the text

 

a. 359 K

b. 299 K

c. 173 K

d. 5773 K

e. 0.37 K

 

29. We use the ideal gas law. We know that the amount of gas is constant, so the value PV/T is constant.

 

P₁V₁/T1 = P₂V₂/T₂    ð   V₂ = V₁(P₁/P₂)(T₂/T₁)

= 3.00 m³)(1.00 atm/3.20 atm)(273 + 38 K/273 K = 1.07 m³   or 1070 liters   since 1 m³ = 1000 L

 

42.  NOT GRADED

 

49. The equation is   v_rms = (3kT/m)^1/2

 

We can see that the speed is proportional to the square root of the absolute temperature.

 

Therefore, doubling the rms speed with no change to the mass, we must multiply the absolute temperature by a factor of 4

 

T_fast = 4T_slow = 4(273 + 20) K = 1172 K = 899 C    or 1172 K             since K = C + 273.15   

                       

Completion of Problem I Started in the Examples Section of Chapter 13

 

A cubic box with a volume of 5.1 x 10^-2 m³ is filled with air at atmospheric pressure at 20 degrees C. The box is closed and heated to 180 degrees C. Assume that there is no leakage from the box and that the box is rigid - volume does not change. What is the net force on each side of the box?

 

Force is pressure times area and, from above, we know the area (one side of the cube)

The net force on each side is the pressure difference between the inside and the outside times the area of a side.

 

We are given the outside pressure (1 atmosphere)

 

The net force on each side of the box will be the pressure difference between the inside and outside of the box, times the area of a side of the box.  The outside pressure is 1 atmosphere.  The ideal gas law is used to find the pressure inside the box, assuming that the mass of gas and the volume are constant.

 

The mass of the gas will be constant (nothing added to or removed from the box). Also since the box is rigid, the volume remains constant.

 

Use the ideal gas law

PV = nRT   ð P/T = nR/V = constant             

P₂/T₂ = P₁/T₁   ð   P₂ = P₁T₂/T₁ = (1.00 atm)(273 + 180)K/(273 + 20)K = 1.55 atm  

 

F = (Delta Pressure)(Area)

 

What is delta pressure?   Delta pressure = 1.55 atm - 1 atm = 0.55 atm

How do you find the area?

The area of a side of the box is given by

 

A = L² = [(volume of box)^1/3]² = (5.1x10^-2 m²)^2/3 = 1.4 x 10^-1 m²

 

       The net force on a side of the box is the pressure difference times the area.

 

       1 atm = 1.01x105 N/m²       

 

F = (Delta Pressure)(Area)

 = (0.55 atm)(1.01 x10⁵ N/m²/atm)(1.4 x 10^-1 m²) = 7.6 x 10³ N