Page 244 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 19] EXPANSION OF SOLIDS, LIQUIDS, AND GASES 229

SOLVED PROBLEM 19.15
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A tank whose capacity is 0.1 m contains helium at an absolute pressure of 10 bar and a temperature of
20 C. A rubber weather balloon is inflated with this helium. (a) The gas cools as it expands, and when the
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pressure of the helium in the balloon is 1 bar, its temperature is −40 C. Find the volume of the balloon.
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(b) Eventually the helium in the balloon absorbs heat from the air around it and returns to 20 C. Find the
volume of the balloon at this time.
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(a) Here T 1 = 20 C = 293 K, T 2 =−40 C = 233 K, V 1 = 0.1m , p 1 = 10 bar, p 2 = 1 bar. From the ideal gas
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law,
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(233 K)(10 bar)(0.1m )
T 2 p 1 V 1 3
V 2 = = = 0.8m
T 1 p 2 (293 K)(1 bar)
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The volume of the balloon is therefore 0.7 m at this time, since the tank retains 0.1 m of the helium after the
expansion.
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(b) Here p 1 = 10 bar, p 3 = 1 bar, V 1 = 0.1m , and, since T 1 = T 3 , Boyle’s law can be used. We have
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p 1 V 1 (10 bar)(0.1m ) 3
V 3 = = = 1m
p 2 1 bar
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The volume of the balloon is therefore 0.9 m , assuming it is still attached to the tank.
SOLVED PROBLEM 19.16
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A gas sample occupies 5 ft at 60 F and atmospheric pressure. (a) Find its volume at 200 F and a gauge
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pressure of 50 lb/in. . (b) Find its gauge pressure when it has been compressed to 1 ft and the temperature
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has been reduced to 0 F.
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(a) Here T 1 = 60 + 460 = 520 R, V 1 = 5ft , and p 1 = 15 lb/in. . When
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2
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T 2 = 200 + 460 = 660 R and p 2 = 50 lb/in. + 15 lb/in. = 65 lb/in. 2
the volume V 2 is
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2
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(660 R)(15 lb/in. )(5ft )
T 2 p 1 V 1 3
V 2 = = 2 = 1.46 ft
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T 1 p 2 (520 R)(65 lb/in. )
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(b)Now T 2 = 0 + 460 = 460 R and V 2 = 1ft . Hence the new absolute pressure is
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3
2
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T 2 p 1 V 1 (460 R)(15 lb/in. )(5ft ) 2
p 2 = = 3 = 66 lb/in.
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T 1 V 2 (520 R)(1ft )
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The new gauge pressure is 66 − 15 = 51 lb/in. .
SOLVED PROBLEM 19.17
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The density of air at 0 C and 1 bar of pressure is 1.293 kg/m . Find its density at 100 C and 2-bar
pressure.
At a pressure of p 1 = 1 bar and an absolute temperature of T 1 = 0 + 273 = 273 K, the mass of a volume of
air V 1 of density d 1 is m = d 1 V 1 . At p 2 = 2 bar and T 2 = 100 + 273 = 373 K, the volume V 2 is
m T 2 p 1 V 1
V 2 = =
d 2 T 1 p 2
Since m = d 1 V 1 ,
d 1 V 1 T 2 p 1 V 1
=
d 2 T 1 p 2