Solution. (a) We use the equation
"PV=\\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
"T_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
"T_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(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
"\\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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