550_Physics ProblemsTechnical Physics

# 550_Physics ProblemsTechnical Physics - 552 P19.2...

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552 Temperature P19.2 PV nRT 11 = and 22 = imply that P P T T 2 1 2 1 = (a) P PT T 2 12 1 0980 273 450 273 20 0 106 == + + = .. . . atm K K K atm af a f a f (b) T TP P 3 13 1 293 0 500 149 124 = = ° K a t m atm KC a f . . FIG. P19.2 P19.3 (a) TT FC =+ ° = + = ° 9 5 32 0 9 5 195 81 32 0 320 . FF (b) C = + = 273 15 195 81 273 15 77 3 . . K P19.4 (a) To convert from Fahrenheit to Celsius, we use CF =−= −=° 5 9 32 0 5 9 98 6 32 0 37 0 . . bg C and the Kelvin temperature is found as C = 273 310 K (b) In a fashion identical to that used in (a), we find T C =− ° 20 6 .C and T = 253 K P19.5 (a) T °− ° ° F H G I K J 450 450 212 32 0 000 810 CC 100 C C F . . (b) T = 450 450 C K P19.6 Require 150 °=− °+ CS ab 100 60 0 °= °+ . Subtracting, 100
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## This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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