Physics Solution Manual for 1100 and 2101

The water that freezes into ice also loses heat the

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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 . ______________________________________________________________________________ 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.

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