Question

8. A container filled with air (0.0177kg), held at constant pressure (152.OkPa), is compressed from 10.0L to 5.0L. Note: The average mass of an air molecule is 4.84x10-26kg. a. What is the initial and final temperature of the gas, before/after the compression? b. Calculate the work output by the gas. C. Calculate Q for this process.

Answer 1

Answer Given that A container filled with art 0.0177 at constant Preassure 152.0k Pg → The Compressed from 10:02 to S.OL. we know that PV = nRT where up = pressure volume n = mole number R = Real gas constant 8.314 J.mo moln T = Temparature ink. → of NA in the A vogurdo number je NA = 6.023x10 23 then Since given that total mass content m = 0.0177 kg and air molecule has mass m= 4.48 x 10 -26 lcg.

so mole number m 0.012 NAM 6.0 22x10 3 * 4.84 x 1026 MR Pe = 152 kPa given 152 X lo -33 loxom = lol So Initial Temperature 152 x 103 x 10 x 10 -3 0.0177 8.131 x 2x) 23 6.023X10 x 4.94X10 Ti 301. 1053875k T; 301.1k of the final volume in VR final Temparture is Ty 7 vi ut Ti -) T,

T, - 150 552 6 939 By gas POV wood done PC V - 3 ) - 15 8 x 103 x ( 10-5) x 1039 w = -7607 we know the ratio y molor nect capacities 121.4 = 7/5 of air is - cp als 5/cp =) cp / cu also lenow that cp-cuk we have 7R cp = 7 feet entered sat the system hepat = ) 0.0177 47/2 x 8.3) 4X (100.5526938 301. 1083875) 6.023x10 23 x 4.8 4 x 10-20

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