3 .0L of an ideal monatomic gas at 1.00 atm and 300 K are contained in a cylinde
ID: 2106632 • Letter: 3
Question
3 .0L of an ideal monatomic gas at 1.00 atm and 300 K are contained in a cylinder with a
piston. The gas first heats up isochorically to twice its original pressure (step 1). It then expands
isothermally to twice its original volume (step 2), cools down isochorically back to its original
temperature (step 3), and finally compresses isothermally back to its original pressure (step 4).
Explanation / Answer
process 1: isochronic dW=0 dU=Q Q=m*Cv*(Tf-Ti) Ti=300k Cv=R*3/2 P=nRT/V V=nRT/P V is const =>Ti/Pi=Tf/Pf =>Tf=Ti*Pf/Pi=2*Ti=600k Q=m*Cv*300=m*450R Process 2:isothermal W=-Q W=-nRTlnVf/Vi=-nR*600*ln2 Pf=P/2=1 Process 3:Pi=1 Pf=Tf*Pi/Ti=600*1/300=2 Q=m*Cv*Tf-Ti=-m450R Process 4=Pi=2;Pf=1 =>Vf/Vi=Pi/Pf=2 W=-Q=-nRTln2=-nR300ln2 Toatal w=-900nRln2 Q=900nRln2
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