Question

Physical Chemistry I

Calculate the values q, w, deltaU, and deltaH for the reversible adiabatic expansion of 1 mole of an ideal monatomic gas from 5.00m³ to 8.00m³. The temperature of the gas is initially 298K with C_v = 3/2nR.

Answer 1

For a monatomic ideal gas :
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Cpm = ( 5/2 ) ( R )

Cpm = ( 5/2 ) ( 8.314 kJ / kgmol - K ) = 20.785 kJ / kgmol - K

Cvm = ( 3/2 ) ( R )

Cvm = ( 3/2 ) ( 8.314 ) = 12.471 kJ / kgmol - K

k = Cpm / Cvm = [ ( 5/2 ) ( R ) ] / [ ( 3/2 ) ( R ) ] = 1.667

Since expansion is adiabatic :
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q = 0.0 kJ / kgmol <------------------------

Q = 0.0 kJ <---------------------------------------...

For ideal gas with adiabatic and reversible process and constant specific heats,
you have the following :
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( P ) ( V )^k = Constant = ( P1 ) ( V1 )^1.667 = ( P2 ) ( V2 )^1.667

( T ) ( V )^k - 1 = Constant = ( T1 ) ( V1 )^0.667 = ( T2 ) ( V2 )^0.667

T2 = ( T1 ) ( V1 / V2 )^0.667

T2 = ( 298 ) ( 5.0 / 8 )^0.667 = 217.805 K

Delta u = ( Cvm ) ( T2 - T1 )

Delta u = ( 12.471 ) ( 217.805 - 298 ) = - 1000 kJ / kgmol <-----------------------

Delta h = ( Cpm ) ( T2 - T1 )

Delta h = ( 20.785 ) ( 217.805 - 298 ) = - 1666.85 kJ / kgmol <---------------------

Delta U = ( n ) ( Delta u )

Delta U = ( 1.000 ) ( - 1000 ) = - 1000 kJ <-----------------------------

Delta H = ( n ) ( Delta h )

Delta H = ( 1.000 ) ( - 1666.85 ) = - 1666.85 kJ <-----------------------------------

q = Delta u + wB

wB = q - Delta u

wB = ( 0.0 ) - ( - 1000 ) = 1000 kJ / kgmol <-------------------------

WB = ( n ) ( wB )

WB = ( 1.000 ) ( 1000 ) = 1000 kJ <-------------------------