Three moles of gas initially at a pressure of 2.00 atm and a volume of 0.300 L h
ID: 2244887 • Letter: T
Question
Three moles of gas initially at a pressure of 2.00 atm and a volume of 0.300 L has internal energy equal to 93.0 J. In its final state, the gas is at a pressure of 1.50 atm and a volume of 0.800 L, and its internal energy equals 180 J.
WIAF = J WIBF = J WIF = J Three moles of gas initially at a pressure of 2.00 atm and a volume of 0.300 L has internal energy equal to 93.0 J. In its final state, the gas is at a pressure of 1.50 atm and a volume of 0.800 L, and its internal energy equals 180 J. For the paths IAF, IBF, and IF in the figure above, calculate the work done on the gas. For the paths IAF, IBF, and IF in the figure above, calculate the net energy transferred to the gas by heat in the process.Explanation / Answer
Work done can be calculated by are ubder the curve
W(IAF)= W(IA)+ W(AF)
W(IA)= 0 (dV=0)
W(AF)= PdV= 1.5*101325 Pa *(0.8-0.3)/1000 L= 75.99 J
W(IBF)= W(IB)+ W(BF)
W(BF)= 0, (dV=0)
W(IB)= 2*101325 Pa (0.8-0.3)/1000 L= 101.325 J
W(IF)= area under the curve
=> area of triangle IAF+ area of rectangle= (0.5*0.5*1/2)*101325/1000 +(1.5*0.5)*101325/1000= 88.65J
U initial= 93 J
U final= 180 J
Q=U+W (From 1st law of thermodynamics)
Q(IAF) = dU+ 0.75= (180-93)+75.99 J= 162.99 J
Q(IBF)= dU+1= (180-93)+101.325 J= 188.325 J
Q(IF)= dU+0.875 J= (180-93)+ 88.65= 175.65 J
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.