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YF_ISM_18

# YF_ISM_18 - THERMAL PROPERTIES OF MATTER 18 18.1(a IDENTIFY...

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18-1 T HERMAL P ROPERTIES OF M ATTER 18.1. (a) I DENTIFY : We are asked about a single state of the system. S ET U P : Use Eq.(18.2) to calculate the number of moles and then apply the ideal-gas equation. E XECUTE : tot 3 0.225 kg 56.2 mol 4.00 10 kg/mol m n M = = = × (b) pV nRT = implies / p nRT V = T must be in kelvins; (18 273) K 291 K T = + = 6 3 3 (56.2 mol)(8.3145 J/mol K)(291 K) 6.80 10 Pa 20.0 10 m p = = × × 6 5 (6.80 10 Pa)(1.00 atm/1.013 10 Pa) 67.1 atm p = × × = E VALUATE : Example 18.1 shows that 1.0 mol of an ideal gas is about this volume at STP. Since there are 56.2 moles the pressure is about 60 times greater than 1 atm. 18.2. I DENTIFY : pV nRT = . S ET U P : 1 41.0 C 314 K T = = ° . 0.08206 L atm/mol K R = . E XECUTE : n , R constant so pV nR T = is constant. 1 1 2 2 1 2 pV p V T T = . 3 2 2 2 1 1 1 (314 K)(2)(2) 1.256 10 K 983 C p V T T p V ⎞⎛ = = = × = ⎟⎜ ⎠⎝ ° . (b) (1.30 atm)(2.60 L) 0.131 mol (0.08206 L atm/mol K)(314 K) pV n RT = = = . tot (0.131 mol)(4.00 g/mol) 0.524 g m nM = = = . E VALUATE : T is directly proportional to p and to V , so when p and V are each doubled the Kelvin temperature increases by a factor of 4. 18.3. I DENTIFY : pV nRT = . S ET U P : T is constant. E XECUTE : nRT is constant so 1 1 2 2 pV p V = . 3 1 2 1 3 2 0.110 m (3.40 atm) 0.959 atm 0.390 m V p p V = = = . E VALUATE : For T constant, p decreases. 18.4. I DENTIFY : . pV nRT = S ET U P : 1 20.0 C 293 K T = = ° . E XECUTE : (a) n , R , and V are constant. constant p nR T V = = . 1 2 1 2 p p T T = . 2 2 1 1 1.00 atm (293 K) 97.7 K 175 C 3.00 atm p T T p = = = = − ° . (b) 2 1.00 atm p = , 2 3.00 L V = . 3 3.00 atm p = . n , R , and T are constant so constant pV nRT = = . 2 2 3 3 p V p V = . 2 3 2 3 1.00 atm (3.00 L) 1.00 L 3.00 atm p V V p = = = . E VALUATE : The final volume is one-third the initial volume. The initial and final pressures are the same, but the final temperature is one-third the initial temperature. 18

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18-2 Chapter 18 18.5. I DENTIFY : pV nRT = S ET U P : Assume a room size of 20 ft 20 ft 10 ft × × . 3 3 4000 ft 113 m V = = . Assume a temperature of 20 C 293 K T = = ° and a pressure of 5 1.01 10 Pa p = × . 3 6 3 1 m 10 cm = . E XECUTE : (a) 5 3 3 (1.01 10 Pa)(113 m ) 4.68 10 mol (8.315 J/mol K)(293 K) pV n RT × = = = × . 3 23 27 A (4.68 10 mol)(6.022 10 molecules/mol) 3 10 molecules N nN = = × × = × . (b) 27 25 3 19 3 3 3 10 molecules 3 10 molecules/m 3 10 molecules/cm 113 m N V × = = × = × E VALUATE : The solution doesn't rely on the assumption that air is all 2 N . 18.6. I DENTIFY : pV nRT = and the mass of the gas is tot m nM = . S ET U P : The temperature is 22.0 C 295.15K. T = ° = The average molar mass of air is 3 28.8 10 kg mol M = × . For helium 3 4.00 10 kg mol M = × . E XECUTE : (a) 3 3 tot (1.00 atm)(0.900 L)(28.8 10 kg/mol) 1.07 10 kg. (0.08206 L atm/mol K)(295.15 K) pV m nM M RT × = = = = × (b) 3 4 tot (1.00 atm)(0.900 L)(4.00 10 kg/mol) 1.49 10 kg. (0.08206 L atm/mol K)(295.15 K) pV m nM M RT × = = = = × E VALUATE : A N pV n N RT = = says that in each case the balloon contains the same number of molecules. The mass is greater for air since the mass of one molecule is greater than for helium.
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