Answer to Question #95727 in Molecular Physics | Thermodynamics for Bala Parina

Question #95727
Three moles of an ideal monatomic gas expands at a constant pressure of 3.5 atm; the volume of the gas changes from 3.0*10-2 m3 to 4.0*10-2 m3. (a) Calculate the initial and final temperatures of the gas. (a) Calculate the amount of work the gas does in expanding. (a) Calculate the change in internal energy of the gas.
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Expert's answer
2019-10-07T10:41:57-0400

Solution. (a) We use the equation


PV=νRTPV=\nu RT

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=PVνT=3.5×1.013×105×3×1023×8.31=427KT_1=\frac {PV} {\nu T}=\frac {3.5 \times 1.013 \times 10^5 \times 3 \times 10^{-2} } {3 \times 8.31}=427K

final temperature


T2=PVνT=3.5×1.013×105×4×1023×8.31=569KT_2=\frac {PV} {\nu T}=\frac {3.5 \times 1.013 \times 10^5 \times 4 \times 10^{-2} } {3 \times 8.31}=569K

(b) For an isobaric process, work is equal to


W=p(V2V1)=3.5×1.013×105×(43)×102=3545.5JW=p(V_2-V_1)=3.5 \times 1.013 \times 10^5 \times (4-3)\times 10^{-2}=3545.5J

(c) For ideal monatomic gas the change in internal energy is equal to


ΔU=32νRΔT=1.3×3×8.31×(569427)=5310.09J\Delta U= \frac {3} {2} \nu R \Delta T=1.3 \times 3 \times 8.31 \times (569-427)=5310.09J

Answer. (a) T1=427K; T2=569K; (b) 3545.5J; (c) 5310.09J


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