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Definemolarspecificheatcapacities. |
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Answer» Solution : (a)Linearmolecule : Energyofonemole ` = ( 7)/(2 )kTxx N_(A)= (7)/(2) RT ` ` C_V= ( DU )/(dT )= ( d )/(dT )[ 7/2 RT ]` `C_v =7/2 R ` `C_p= C_v+ R= 7/2R+R =(9R )/(2)` ` thereforegamma =( C_p )/( C_(V )) = ((9)/(2)R )/((7)/(2)R) = 9/7 = 1.28 ` (b )Non linear molecule : Energyof amole`=6/2kTxx N_(A)= 6/2RT= 3RT ` `C_V= (dU )/(dT ) = 3R ` `C_p = C_v+R = 3R+R = 4R ` ` thereforegamma= (c_p )/(C_(v)) = ( 4R) /( 3R ) = (4)/(3 ) = 1.33` NOTE that according to kin e tic theory model of gases the SPECIFIC heat capacity at constant volume and constant pressure are independent of TEMPERATURE. But in reality it is not sure. The specific heat capacity varies with the temperature. |
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