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# ch20 - CHAPTER 20 THE THERMAL BEHAVIOR OF MATTER Problem 4...

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CHAPTER 20 THE THERMAL BEHAVIOR OF MATTER ActivPhysics can help with these problems: Activities 8.1–8.4 Section 20-1: Gases Problem 1. At Mars’s surface, the planet’s atmosphere has a pressure only 0.0070 times that of Earth, and an average temperature of 218 K. What is the volume of 1 mole of the Martian atmosphere? Solution The molar volume of an ideal gas at STP for the surface of Mars can be calculated as in Example 20-1. However, expressing the ideal gas law for 1 mole of gas at the surfaces of Mars and Earth as a ratio, P M V M /T M = P E V E /T E , and using the previous numerical result, we find V M = ( P E /P M )( T M /T E ) × V E = (1 / 0 . 0070)(218 / 273)(22 . 4 × 10 - 3 m 3 ) = 2 . 56 m 3 . Problem 2. How many molecules are in an ideal gas sample at 350 K that occupies 8.5 L when the pressure is 180 kPa? Solution From Equation 20-1, N = PV/kT = (180 kPa)(8 . 5 × 10 - 3 m 3 ) / (1 . 38 × 10 - 23 J / K)(350 K) = 3 . 17 × 10 23 . Problem 3. What is the pressure of an ideal gas if 3.5 moles occupy 2.0 L at a temperature of - 150 C? Solution The ideal gas law in terms of the gas constant per mole, Equation 20-2, gives P = nRT/V = (3 . 5 mol)(8 . 314 J / K · mol)(123 K) / (0 . 002 m 3 ) = 1 . 79 × 10 6 Pa . (The absolute temperature must be used, but any convenient units for the gas constant can be used, e.g., R 0 . 0821 L · atm / K · mol . Then P = (3 . 5 mol)(0 . 0821 L · atm / K · mol)(123 K) / (2 L) = 17 . 7 atm . ) Problem 4. An ideal gas occupies a volume V at 100 C . If the gas pressure is held constant, by what factor does the volume change (a) if the Celsius temperature is doubled and (b) if the kelvin temperature is doubled? Solution To compare different states of an ideal gas, it is often convenient to express Equation 20-1 as a ratio: ( P 1 V 1 /P 2 V 2 ) = ( N 1 T 1 /N 2 T 2 ) . In this problem, the pressure is constant, and if no gas escapes or enters, so is N , therefore V 1 /V 2 = T 1 /T 2 . (a) T is the absolute temperature. If T 2 = 100 C = 373 K , and T 1 = 200 C = 473 K , then V 1 = (473 / 373) V 2 = 1 . 27 V 2 . (b) If T 1 = 2 T 2 , then V 1 = 2 V 2 . (The fact that V T for a given mass of ideal gas at constant pressure is known as the law of Charles and Gay-Lussac.) Problem 5. If 2.0 mol of an ideal gas are at an initial temperature of 250 K and pressure of 1.5 atm, (a) what is the gas volume? (b) The pressure is now increased to 4.0 atm, and the gas volume drops to half its initial value. What is the new temperature? Solution (a) From Equation 20-2: V = nRT P = (2 mol)(8 . 314 J / mol · K)(250 K) (1 . 5 atm)(1 . 013 × 10 5 Pa / atm) = 2 . 74 × 10 - 2 m 3 = 27 . 4 L . (b) The ideal gas law in ratio form (for a fixed quantity of gas, N 1 = N 2 ) gives: T 2 T 1 = P 2 V 2 P 1 V 1 , or T 2 = 4 . 0 atm 1 . 5 atm 0 . 5 V 1 V 1 (250 K) = 333 K . Problem 6. The solar corona is an extended atmosphere of hot (2 × 10 6 K) gas surrounding the cooler visible

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2 CHAPTER 20 surface of the Sun. The gas pressure in the solar corona is about 0.03 Pa. What is the coronal density in particles per cubic meter? Compare with Earth’s atmosphere.
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ch20 - CHAPTER 20 THE THERMAL BEHAVIOR OF MATTER Problem 4...

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