557_Physics ProblemsTechnical Physics

557_Physics - Chapter 19 P19.39 mf PV = nRT mi = nf m f = mi so Pf V f RTi Pf = RT f PVi Pi i = ni FP I GH P JK f i m = m i m f = m i P19.40 559 F

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 19 559 P19.39 PV nRT = : m m n n PV RT RT P P f i f i ff f i ii f i == = so mm P P fi f i = F H G I K J mmm m PP P ifi if i =− = F H G I K J = F H G I K J = 12 0 26 0 439 . . . kg 41.0 atm atm 41.0 atm kg P19.40 My bedroom is 4 m long, 4 m wide, and 2.4 m high, enclosing air at 100 kPa and 20 293 °= C K. Think of the air as 80.0% N 2 and 20.0% O 2 . Avogadro’s number of molecules has mass 0 800 28 0 0 200 32 0 0 028 8 .. . af bg g mol g mol kg mol += Then PV nRT m M RT F H G I K J gives m PVM RT × = 100 10 384 00288 8314 293 45 4 5 . . . Nm m kgmo l Jmo lK K kg ~10 kg 23 2 ej e j *P19.41 The CO 2 is far from liquefaction, so after it comes out of solution it behaves as an ideal gas. Its molar mass is M =+ = 12 0 2 16 0 44 0 . g mol g mol g mol . The quantity of gas in the cylinder is
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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 .

Ask a homework question - tutors are online