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# ch14 - CHAPTER 14 THE IDEAL GAS LAW AND KINETIC THEORY...

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CHAPTER 14 THE IDEAL GAS LAW AND KINETIC THEORY ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1. (d) This statement is not true, because the number of molecules that have a mass of ten grams depends on the molecular mass of the substance, which is different for different substances. 2. 10.2 g 3. (b) For an ideal gas, we know that nRT P V = . Increasing n , T , and V together may or may not cause P to increase. Increasing n and T certainly causes P to increase. However, increasing V causes P to decrease. Therefore, the result of increasing all three variables together depends on the relative amounts by which they are increased. 4. 2.25 mol 5. 1.25 kg/m 3 6. (a) If the speed of every atom in a monatomic ideal gas were doubled, the root-mean-square speed of the atoms would be doubled. According to the kinetic theory of gases, the Kelvin temperature is proportional to the square of the root-mean-square speed (see Equation 14.6). Therefore, the Kelvin temperature would be multiplied by a factor of 2 2 = 4. 7. (d) According to the kinetic theory of gases, the average translational kinetic energy per molecule is proportional to the Kelvin temperature (see Equation 14.6). Since each gas has the same temperature, each has the same average translational kinetic energy. However, this kinetic energy is 2 1 rms 2 mv , and depends on the mass m . Since each type of molecule has the same kinetic energy, the molecules with the larger masses have the smaller translational rms speeds v rms . 8. (c) According to the kinetic theory of gases, the internal energy U of a monatomic ideal gas is 3 2 U nRT = (Equation 14.7). However, the ideal gas law indicates that PV = nRT , so that 3 2 U PV = . Since P is doubled and V is reduced by one-half, the product PV and the internal energy are unchanged.

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209 THE IDEAL GAS LAW AND KINETIC THEORY 9. (e) Increasing the cross-sectional area of the diffusion channel and increasing the difference in solute concentrations between its ends certainly would increase the diffusion rate. However, increasing its length would decrease the rate (see Equation 14.8). Therefore, increasing all three factors simultaneously could lead to a decrease in the diffusion rate, depending on the relative amounts of the change in the factors. 10. 6.0 × 10 11 kg/s
210 THE IDEAL GAS LAW AND KINETIC THEORY CHAPTER 14 THE IDEAL GAS LAW AND KINETIC THEORY PROBLEMS _____________________________________________________________________________ _ 1. REASONING AND SOLUTION Since hemoglobin has a molecular mass of 64 500 u, the mass per mole of hemoglobin is 64 500 g/mol. The number of hemoglobin molecules per mol is Avogadro's number, or 6.022 × 10 23 mol –1 . Therefore, one molecule of hemoglobin has a mass (in kg) of 22 23 1 64 500 g/mol 1 kg = 1.07×10 kg 1000 g 6.022×10 mol - - ÷ ÷ _____________________________________________________________________________ _ 2. REASONING The number of molecules in a known mass of material is the number n of moles of the material times the number N A of molecules per mole (Avogadro's number).

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