Saved Bookmarks
| 1. |
One mole of an idealgas iscompressedfrom 500cm^(3) againsta constantpressureof 1.216 xx 10^(5)Pa. Theworkinvolved in theprocessis 35.50 J. calculatethe finalvolume . |
|
Answer» Constantexternalpressure `= P_(EX)` `= 1.216 xx10^(5)Pa = 1.216 xx 10^(5)NM^(-2)` Forcompressionworkobtained= W=+36 .5 J Initialvolume= `V_(1)= 500cm^(3)= 500xx 10^( -6)m^(3)( :'1 m^(3)=10^(6) cm^(3))` Finalvolume `=V_(2)= ?` Sincethe COMPRESSION thetakesplaceagainsta constantpressureit isan irreversibleprocess andworkobtained is `W= -P_(ex) (V_(2)-V_(1))` `36.5 =-1.216xx 10^(5)xx (V_(2) - 500xx 10^(-6))` `:.(36.5)/(1.216 xx 10^(5))= - (V_(2) -50 xx 10^(5))` `30 xx 10^(-5)=-(V_(2) -50 xx 10^(-5))` `:. 3010^(5) = - V_(2) + 50 xx 10^(-5)` `:. V_(2)= (50 -300) xx 10^(-5)=20 xx 10^(-5)m^(3)` `= 20 xx 10^(-5) xx 10^(6) cm^(3) = 200 cm^(3)` |
|