Question Solved1 Answer 2.25 Five moles of nitrogen are expanded from an initial state of 3 bar and 88°C to a final state of 1 bar and 88°C. You may consider N, to behave as an ideal gas. Answer the following questions for each of the following reversible processes: (a) The first process is isothermal expansion. (i) Draw the path on a Pu diagram and label it path A. (ii) Calculate the following: w,q, Au, Ah. (b) The second process is heating at constant pressure followed by cooling at constant volume. (i) Draw the path on the same Pu diagram, label it path B. (ii) Calculate the following: w,q, Au, Ah.

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Transcribed Image Text: 2.25 Five moles of nitrogen are expanded from an initial state of 3 bar and 88°C to a final state of 1 bar and 88°C. You may consider N, to behave as an ideal gas. Answer the following questions for each of the following reversible processes: (a) The first process is isothermal expansion. (i) Draw the path on a Pu diagram and label it path A. (ii) Calculate the following: w,q, Au, Ah. (b) The second process is heating at constant pressure followed by cooling at constant volume. (i) Draw the path on the same Pu diagram, label it path B. (ii) Calculate the following: w,q, Au, Ah.
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Transcribed Image Text: 2.25 Five moles of nitrogen are expanded from an initial state of 3 bar and 88°C to a final state of 1 bar and 88°C. You may consider N, to behave as an ideal gas. Answer the following questions for each of the following reversible processes: (a) The first process is isothermal expansion. (i) Draw the path on a Pu diagram and label it path A. (ii) Calculate the following: w,q, Au, Ah. (b) The second process is heating at constant pressure followed by cooling at constant volume. (i) Draw the path on the same Pu diagram, label it path B. (ii) Calculate the following: w,q, Au, Ah.
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  NitrogenInitial no. of motes =n_(1)=5Initeal prestuie =P_(1)=3 bar =303.9kPaInitial temperatixe =T_(1)=88^(@)C+361KN_(2)rarr Ideal gasIt is revensible process.{:[" Thitial volume "=v_(1)=(n_(1)Rsigma_(1))/(P_(1))=(5xx5.314 xx361)/(303.9)],[(P_(N))/(1pi)=R=8.314(J)/(molk)=(101.3kpaxx22.4(p^(3))/(man)xx)/(270.15k)],[V_(1)=40.38m^(3)","V_(1)=(40.38)/(5)=8.076(m^(3))/(mal)]:}a) Isothermal process P_(2)=1 bar T_(2)=$8^(@)c=36 ik v_(2)= ? (P_(1)v_(1))/(Tk)=(P_(2)v_(2))/(T_(2)) (n_(1)=(n_(2))/()=:} cast) =>v_(2)=(P_(1)V_(1))/(P_(2))=(3)/(1)xx8.076=24.228m//ma cos v_(2)=24.228 xx5=121.14m.since it is isothermat process, DeltaU_(0)=0,Delta h=0omega=8314 xx361 xx ln((8.076)/(24.228))=-3297.32J//" mol "(-ve workdone by the system )for Tsothermal systam Delta hat(P)=d omega+dQ{:[=>d omega=-dQ],[:.Q=+3297.32J//mol.]:}(b) heating at cost pressuse {:CP_(1)=P_(1)^(')=3bal){: ... See the full answer