Solution Manual for Chapter 17 (#21, 34, 39, 44, 50, 52, 60, 63)

#21. (a) If 2.0 mol of an ideal gas are initially at temperature 250 K and pressure 1.5 atm,

what’s the gas volume?

We’re dealing with an ideal gas, and we can write

with , , and the initial pressure, volume, and temperature, and the number

of moles. Now, we can find the initial volume using the ideal gas law.

(b) The pressure is now increased to 4.0 atm, and the gas volume drops to half its

initial value. What’s the new temperature?

The number of moles is fixed. From the ideal gas law, we can derive relation that

Using this, we have

#34. Suppose a single piece of welded steel railroad track stretched 5000 km across the

continental United States. If the track were free to expand, by how much would its

length change if the entire track went from a cold winter temperature of -25˚C to a hot

summer day at 40˚C?

From the definition of the coefficient of linear expansion α,

derive a differential equation that

Now, solve this differential equation,

where is the length of steel railroad. We already know the expansion coefficient of

steel from Table 17.2 on the textbook and it is . Let the length and

temperature in summer are and , in winter, and .

By simple calculation, we can find the length difference.

#39. A 3000-mL flask is initially open in a room containing air at 1.00 atm and 20˚C. The flask

is then closed and immersed in boiling water. When the air in the flask has reached

thermodynamic equilibrium, the flask is opened and air is allowed to escape. The flask is

then closed and cooled back to 20˚C. Find

(a) the maximum pressure reached in the flask.

We’re dealing with an ideal gas, and we can write

When the flask is closed and has reached thermodynamic equilibrium, the pressure has

maximum value. The flask is closed, so the number of molecules and the volume are not

changed. Let the initial pressure and temperature are and , at equilibrium, they are

and . i.e. is the maximum pressure.

Transpose the equilibrium temperature to left side,

(b) the number of moles that escape when air is released.

First, we need to find the number of moles that exist when flask is closed.

Now, we should calculate the number of moles in the flask at equilibrium temperature

and 1 atm. It is the number of moles that exist when flask is opened.

The number of escaped air molecules is the difference between and .

(c) the final pressure in the flask.

The flask is closed again and cooled back to 20˚C. So, the number of moles in the flask

is . Then, we know all value to obtain the pressure.

From the ideal gas law,

#44. At winter’s end, Lake Superior’s surface is frozen to a depth of 1.3 m; the ice density is

917 kg/m³.

(a) How much energy does it take to melt the ice?

The energy to melt the ice is obtained by this equation

The heat of fusion of water is 334 KJ/kg from Table 17.1. Let the surface area of Lake

Superior is 82100 km².

(b) If the ice disappears in 3 weeks, what’s the average power supplied to melt it?

#50. During a nuclear accident, 420 m³ of emergency cooling water at 20˚C are injected into

a reactor vessel where the reactor core is producing heat at the rate of 200 MW. If the

water is allowed to boil at normal atmospheric pressure, how long will it take to boil the

reactor dry?

The cooling water is heated until boiling point, 100˚C and then, it is evaporating. So,

total drying time is summation of heating time and evaporating time. Let heating time is

, and evaporating time is .

To boil cooling water, we need the heat Q that

Where is the mass, is the specific heat, and is the temperature difference.

Using this equation, we can derive relation that

Now, we can obtain the heating time.

To obtain the evaporating time, we need to compare the accepted heat and heat of

vaporization

Obtain the evaporating time.

The total drying time is

#52. A bowl contains 16 kg of punch (essentially water) at a warm 25˚C. What’s the minimum

amount of ice at 0˚C needed to cool the punch to 0˚C?

The heat of cooling punch is equivalent to the fusing energy of ice. Let the fusing

energy of ice is , and the heat of cooling punch is .

As I said, . So, we can obtain .

#60. A 2000-mL graduated cylinder is filled with liquid at 350 K. When the liquid is cooled to

300 K, the cylinder is full to only the 1925-mL mark. Use Table 17.2 to identify the liquid.

From the definition of the coefficient of volume expansion β,

Derive a differential equation that

Now, solve this differential equation,

where is the volume of the liquid. Let the initial volume and temperature of the

liquid is and , and the volume and temperature after the cooling is and .

Divide both sides,

Solving for β,

Therefore, this liquid is inferred to ethyl alcohol ( ).

#63. A rod of length is clamped rigidly at both ends. Its temperature increases by ΔT and

in the ensuing expansion, it cracks to form two straight pieces, as shown in Fig. 17.11.

Find an expression for the distance d shown in the figure, in terms of , ΔT, and the

linear expansion coefficient α.

We derived many times that

where is the initial length of rod, and is the temperature difference.

Using the Pythagoras’ theorem, we can derive the length of the snapped rod.

If the length of the square,

Therefore,