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Unformatted text preview: C. Therefore,
liquid carbon dioxide exists in equilibrium with its vapor phase at 0 °C when the vapor
pressure is 3.5 × 106 Pa .
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75. SSM REASONING The definition of percent relative humidity is given by Equation 12.6
as follows:
Partial pressure of water vapor
Percent relative humidity =
× 100
Equilibrium vapor pressure of
water at the existing temperature
Using R to denote the percent relative humidity, P to denote the partial pressure of water
vapor, and PV to denote the equilibrium vapor pressure of water at the existing temperature,
we can write Equation 12.6 as
P
R=
×100
PV
The partial pressure of water vapor P is the same at the two given temperatures. The relative
humidity is not the same at the two temperatures, however, because the equilibrium vapor
pressure PV is different at each temperature, with values that are available from the vapor
pressure curve given with the problem statement. To determine the ratio R10/R40, we will
apply Equation 12.6 at each temperature.
SOLUTION Using Equation 12.6 and reading the values of PV, 10 and PV, 40 from the vapor
pressure curve given with the problem statement, we find
R10
R40 = P / PV, 10
P / PV, 40 = PV, 40
PV, 10 = 7200 Pa
= 5.5
1300 Pa 672 TEMPERATURE AND HEAT 76. REASONING In order to find the change in the temperature, we must first determine the
higher temperature T1 and the lower temperature T2. Each temperature corresponds to a
different equilibrium vapor pressure PV of water on the curve given in problem 75. We will
employ the definition of percent relative humidity (denoted by R) in order to find the
equilibrium vapor pressures that will allow us to locate the temperatures T1 and T2 on this
curve:
P
Percent relative humidity = R =
×100
(12.6)
PV In Equation 12.6, the quantity P denotes the actual partial pressure of water vapor in the
atmosphere. The dew point is the temperature at which the equilibrium water vapor pressure
equals P. The dew point does not change, so the actual partial pressure P of water vapor in
the atmosphere also remains unchanged. The vapor pressure curve of problem 75 shows that
the dew point is 14 °C when P = 1800 Pa. Solving Equation 12.6 for the equilibrium vapor
pressure PV of water at the existing temperature, we obtain
PV = P
× 100
R (1) SOLUTION When the relative humidity is R1 = 50.0%, then, Equation (1) gives the
equilibrium vapor pressure of water PV1: PV1 = P
1800 Pa
×100 =
×100 = 3600 Pa
R1
50.0 On the vapor pressure curve of problem 75, this corresponds to a temperature of
approximately T1 = 26 °C. Similarly, when the percent relative humidity is 69%, the
equilibrium vapor pressure of water is
PV2 = P
1800 Pa
× 100 =
×100 = 2600 Pa
R2
69 Again making use of the vapor pressure curve of problem 75, we find that the equilibrium
vapor pressure of water is 2600 Pa at a temperature of approximately T2 = 21 °C. Therefore,
the relative humidity increases from 50.0% to 69% when the temperature drops by T1 − T2 = 26 o C − 21 o C = 5 C o Chapter 12 Problems 673 77. REASONING The definition of relative humidity is given by Equation 12.6 as: Percent relative humidity = Partial pressure of water vapor
×100
Equilibrium vapor pressure of
water at the existing temperature The partial pressure of water vapor is given in the problem statement. The equilibrium vapor
pressure can be found by consulting the vapor pressure curve for water that accompanies
problem 75.
SOLUTION By using the vapor pressure curve for water given in problem 75, we estimate
that at a temperature of 37 °C the water vapor in the lungs has an equilibrium vapor pressure
of 6.3 × 103 Pa. The relative humidity is, then, 5.5 ×103 Pa Percent relative humidity = ×100 = 87% 6.3 × 103 Pa ______________________________________________________________________________ 78. REASONING We will rely on Equation 12.6, which defines what is meant by “percent
relative humidity:” Partial pressure
of water vapor
Percent relative humidity =
× 100
Equilibrium vapor pressure of
water at the existing temperature (12.6) Note especially that the denominator on the right in this equation is not the total atmospheric
air pressure. Thus, if the partial pressure of water vapor in the air remains the same and an
increase in temperature causes the equilibrium vapor pressure of water to increase, the
percent relative humidity will decrease.
SOLUTION
a. The percentage of atmospheric pressure is 1300 Pa × 100% = 1.3%
5 1.013 × 10 Pa b. The percentage is not 100%. The percentage is that determined in part a, namely,
1.3% .
c. The relative humidity at 35 °C is 674 TEMPERATURE AND HEAT 1300 Pa × 100% = 24% 5500 Pa (12.6) As expected, this value is less than the 100% value at 10 °C. 79. REASONING To bring the water to the point where it just begins to boil, its temperature
must be increased to the boiling point. The heat Q that must be added to raise the
temperature of a substance of mass m by an amount ∆T is given by Equati...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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