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Unformatted text preview: THE THERMAL BEHAVIOR OF MATTER E XERCISES Section 17.1 Gases 18. The molar volume of an ideal gas at STP for the surface of Mars can be calculated as in Example 17.1. However, expressing the ideal gas law for 1 mole of gas at the surfaces of Mars and Earth as a ratio, / / , M M M E E E P V T P V T = and using the previous numerical result, we find ( / )( / ) (1/0.0070)(218/273)(22.4 M E M M E E V P P T T V = = 3 3 3 10 m ) 2.56 m . = 19. I NTERPRET We are dealing with an ideal gas. We are given the pressure, temperature, and volume, and want to find the number of gas molecules. D EVELOP We shall use the idealgas law, , pV NkT = given in Equation 17.1, to find the number of molecules. E VALUATE From Equation 17.1, we have 3 3 23 23 (180 kPa)(8.5 10 m ) 3.17 10 (1.38 10 J/K)(350 K) PV N kT = = = A SSESS One mole has 23 6.02 10 A N = molecules. Thus, we have about 0.53 mole of molecules in the system. 20. The ideal gas law in terms of the gas constant per mole, Equation 17.2, gives / (3.5 mol) P nRT V = = 3 6 (8.314 J/K mol)(123 K)/(0.002 m ) 1.79 10 Pa. = (The absolute temperature must be used, but any convenient units for the gas constant can be used, e.g., 0.0821 L atm/K mol. R . Then (3.5 mol)(0.0821 L atm/K mol) P = (123 K)/(2 L) 17.7 atm.) = 21. I NTERPRET We are dealing with an ideal gas. We want to know how volume changes with temperature. D EVELOP To compare different states of an ideal gas, it is often convenient to express Equation 17.1 as a ratio: 1 1 1 1 2 2 2 2 PV N T PV N T = In this problem, the pressure is constant, and if no gas escapes or enters, then N is also constant. Therefore, the above equation becomes 1 1 2 2 V T V T = where T is in the Kelvin scale. E VALUATE (a) If 2 100 C 373 K T = = and 1 200 C 473 K T = = then 1 2 2 473 K 1.27 373 K V V V = = (b) If 1 2 2 , T T = then 1 2 2 . V V = A SSESS The fact that ~ V T for a given mass of ideal gas at constant pressure is known as the law of Charles and GayLussac. 22. (a) From Equation 17.2: 17.1 17 17.2 Chapter 17 5 2 3 (2 mol)(8.314 J/mol K)(250 K) (1.5 atm)(1.013 10 Pa/atm) 2.74 10 m 27.4 L nRT V P = = = = (b) The ideal gas law in ratio form (for a fixed quantity of gas, 1 2 ) N N = gives: 2 2 2 1 2 1 1 1 1 0.5 4.0 atm , or (250 K) 333 K 1.5 atm T PV V T T PV V = = = 23. I NTERPRET We treat air molecules as ideal gas. Given the pressure, temperature, and volume, we want to find the number of air molecules. D EVELOP We shall use the idealgas law, , pV NkT = given in Equation 17.1, to find the number of molecules. E VALUATE The number of air molecules is 10 3 3 7 23 (10 Pa)(10 m ) 2.65 10 (1.38 10 J/K)(273 K) PV N kT = = = A SSESS One mole has 23 6.02 10 A N = molecules. Thus, we have about 17 4.4 10 mole of molecules in the system....
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 Spring '09
 Prof.Kim
 Physics

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