Question

The temperature of a sample of dilute argon gas with n = 3 moles decreases by 380 K. If 33130 J of heat are extracted from the gas, what is the work done by the gas on its surroundings? If you use the universal gas constant to solve this problem, use 8.31 J/(mol*K) rather than 8.314 J/(mol*K).

Answer 1

n = 3 moles

decreases =380 K

heat =33130 J

gas constant=8.314

Apply the Nonflow Energy Equation :
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Q = Delta U + WB

WB = Q - Delta U

Q = - 33130 J

Delta U = ( n ) ( Cvm ) ( T2 - T1 )

Cvm = ( 3/2 ) ( R ) for a monatomic gas

Cvm = ( 3/2 ) ( 8.314 J / mol - K ) = 12.511 J / mol - K

Delta U = (3 mol ) ( 12.511 J / mol - K ) ( - 380 K )

Delta U = - 14262.54 J

WB = Q - Delta U

WB = ( - 33130 J ) - ( - 114262.54 J ) = - 81133 J = - 81.13 kJ <------------------

The Surroundings do 81.13 kJ of work on the System, since WB is negative. <------------