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 :
---------------------------------------...
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. <---------**---

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