Page 241 - Schaum's Outline of Theory and Problems of Applied Physics

226 EXPANSION OF SOLIDS, LIQUIDS, AND GASES [CHAP. 19

Since b = 3a, the coefficient of volume expansion of lead is
−5 ◦
−5 ◦
b = (3)(3 × 10 / C) = 9 × 10 / C
At 20 C the volume of a mass m of lead is V 0 = m/d 0 , and at 200 Citis
◦
◦
m
V = = V 0 + V = V 0 + bV 0 T
d
Substituting V 0 = m/d 0 yields
m m bm T m
= + = (1 + b T )
d d 0 d 0 d 0
d 0
d =
1 + b T
◦
◦
◦
Here T = 200 C − 20 C = 180 C, and so
11 g/cm 3
d = = 10.8 g/cm 3
1 + (9 × 10 / C)(180 C)
−5 ◦
◦
BOYLE’S LAW
At constant temperature, the volume of a sample of gas is inversely proportional to the absolute pressure applied
to the gas. The greater the pressure, the smaller the volume. This relationship is known as Boyle’s law. If p 1 is
the gas pressure when its volume is V 1 and p 2 is its pressure when its volume is V 2 , then Boyle’s law states that
T = constant
p 1 V 1 = p 2 V 2

SOLVED PROBLEM 19.6
A 1-L sample of nitrogen at 0 C and 1-atm pressure is compressed to 0.5 L. If the temperature is
◦
unchanged, what happens to the pressure of the sample?
Since p 2 = p 1 V 1 /V 2 and here V 1 /V 2 = 2, the pressure doubles to 2 atm.

SOLVED PROBLEM 19.7
A scuba diver’s 12-L tank is filled with air at an absolute pressure of 150 bar. If the diver uses 30 L of
air per minute at the same 2.5-bar absolute pressure as the water pressure at her depth of 15 m below the
surface, how long can she stay at that depth?
From Boyle’s law the volume of air available at a pressure of 2.5 bar is

p 1 V 1 (150 bar)(12 L)
V 2 = = = 720 L
p 2 2.5 bar
However, 12 L of air remains in the tank, so she can use only 708 L. Hence
708 L
t = = 23.6 min
30 L/min
SOLVED PROBLEM 19.8
3
A steel cylinder contains 3 m of air at a gauge pressure of 15 bar. What volume would this amount of
air occupy at sea-level atmospheric pressure of about 1 bar?
The absolute pressure of the air in the tank is

p 1 = gauge pressure + atmospheric pressure
= 15 bar + 1 bar = 16 bar