Solution. (a) We use the equation
PV=νRT where P=3.5 atm is pressure; V is volume; R=8.31 J/(mol K) is gas constant; T is temperature. Therefore initial temperature
T1=νTPV=3×8.313.5×1.013×105×3×10−2=427Kfinal temperature
T2=νTPV=3×8.313.5×1.013×105×4×10−2=569K(b) For an isobaric process, work is equal to
W=p(V2−V1)=3.5×1.013×105×(4−3)×10−2=3545.5J(c) For ideal monatomic gas the change in internal energy is equal to
ΔU=23νRΔT=1.3×3×8.31×(569−427)=5310.09J
Answer. (a) T1=427K; T2=569K; (b) 3545.5J; (c) 5310.09J
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