YF_ISM_18

YF_ISM_18 - THERMAL PROPERTIES OF MATTER 18 18.1. (a)...

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18-1 T HERMAL P ROPERTIES OF M ATTER 18.1. (a) IDENTIFY: We are asked about a single state of the system. SET UP: Use Eq.(18.2) to calculate the number of moles and then apply the ideal-gas equation. EXECUTE: tot 3 0.225 kg 56.2 mol 4.00 10 kg/mol m n M == = × (b) pV nRT = implies / pn R TV = T must be in kelvins; (18 273) K 291 K T =+ = 6 33 (56.2 mol)(8.3145 J/mol K)(291 K) 6.80 10 Pa 20.0 10 m p × × 65 (6.80 10 Pa)(1.00 atm/1.013 10 Pa) 67.1 atm p × = EVALUATE: 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. IDENTIFY: pV nRT = . SET UP: 1 41.0 C 314 K T ° . 0.08206 L atm/mol K R =⋅ . EXECUTE: n , R constant so pV nR T = is constant. 11 2 2 12 pV TT = . 3 22 21 (314 K)(2)(2) 1.256 10 K 983 C ⎛⎞ = × = ⎜⎟ ⎝⎠ ° . (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 mn M = . EVALUATE: 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. IDENTIFY: pV nRT = . SET UP: T is constant. EXECUTE: nRT is constant so = . 3 1 3 2 0.110 m (3.40 atm) 0.959 atm 0.390 m V pp V = . EVALUATE: For T constant, p decreases. 18.4. IDENTIFY: . pV nRT = SET UP: 1 20.0 C 293 K T ° . EXECUTE: (a) n , R , and V are constant. constant R . = . 2 1 1.00 atm (293 K) 97.7 K 175 C 3.00 atm p 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 32 3 1.00 atm (3.00 L) 1.00 L 3.00 atm p VV p = . EVALUATE: 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. IDENTIFY: pV nRT = SET UP: Assume a room size of 20 ft 20 ft 10 ft ×× . 33 4000 ft 113 m V == . Assume a temperature of 20 C 293 K T ° and a pressure of 5 1.01 10 Pa p . 36 3 1 m 10 cm = . EXECUTE: (a) 53 3 (1.01 10 Pa)(113 m ) 4.68 10 mol (8.315 J/mol K)(293 K) pV n RT × . 32 3 2 7 A (4.68 10 mol)(6.022 10 molecules/mol) 3 10 molecules Nn N × × = × . (b) 27 25 3 19 3 3 3 10 molecules/m 3 10 molecules/cm 113 m N V × ×= × EVALUATE: The solution doesn't rely on the assumption that air is all 2 N. 18.6. IDENTIFY: pV nRT = and the mass of the gas is tot mn M = . SET UP: 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 . EXECUTE: (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 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 M RT × = EVALUATE: A Np V n NR T 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. 18.7. IDENTIFY: We are asked to compare two states. Use the ideal gas law to obtain 2 T in terms of 1 T and ratios of pressures and volumes of the gas in the two states. SET UP: pV nRT = and n , R constant implies / constant pV T nR and 11 1 2 2 2 / / pV T = EXECUTE: 1 (27 273) K 300 K T =+ = 5 1 p 65 6 2 2.72 10 Pa 1.01 10 Pa 2.82 10 Pa p (in the ideal gas equation the pressures must be absolute, not gauge, pressures) 63 22 21 46.2 cm 300 K 776 K 499 cm pV TT ⎛⎞ × = ⎜⎟ × ⎝⎠ 2 (776 273) C 503 C T =− ° = ° EVALUATE: The units cancel in the / VV volume ratio, so it was not necessary to convert the volumes in 3 cm
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This note was uploaded on 03/14/2012 for the course MAE 162D taught by Professor Shaefer during the Spring '11 term at UCLA.

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YF_ISM_18 - THERMAL PROPERTIES OF MATTER 18 18.1. (a)...

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