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Finding work done on a gas?

+1 vote
air is compressed slowly and frictionlessly in a cylinder according to the relationship pV^1.4 = C from an initial volume of .8m^3 and pressure of 100kN.m^2 to a final volume of .4m^3. the symbole C represents a constant. calculate the work.

im not sure how the integrull for this would be set up.
asked Jan 21, 2013 in PHYSICS by angel12 Scholar

1 Answer

+1 vote
Initial volume ,v1=0.8 m^3

Initial pressure,p1=100 kN/m^2

Final volume,v2=0.4 m^3

Relation

pv^1.4=C

p1v1^1.4=p2v2^4

p2=100*(0.8/0.4)^1.4

     =263.9 kPa

Work done

W=p1v1^1.4(v2^(1-1.4)-v1^(1-1.4))/1-1.4

    =100*0.8^1.4(0.4^(1-1.4)-0.8^(1-1.4))/(1-1.4)

    =-63.9 kJ
answered Jan 21, 2013 by bradely Mentor

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