A gas is taken through the cyclic process described in the figure. (The x axis i

Question

A gas is taken through the cyclic process described in the figure. (The x axis is marked in increments of 1 m3.) Find the net energy transfered to the system by heat during one complete cycle. If the cycle is reversed-that is, the process follows the path ACBA-what is the net energy input per cycle by heat? kJ A sample of an ideal gas goes through the process shown in the figure below. From A to B, the process is adiabatic; from B to C, it is isobaric with 345 kJ of energy entering the system by heat; from C to D, the process is isothermal; and from D to A, it is isobaric with 371 kJ of energy leaving the system by heat. Determine the difference In internal energy Eint,B - Eint,A. kJ

Explanation / Answer

Part 1

a)

Work done in a cyclic process = Area of PV curve=0.5*2*6=6 kPa.m^3:Since the cycle is transversed in a clockwise manner work done is positive

Internal Energy Change = 0 : Since System returns to it's initial Stage

Total Heat Content(Q)=W+U=6 kPa.m^3

b)

If the cycle is reversed it becomes anticlockwise making W negative

U still remanis 0

Total Heat = -6kPa.m^3

Part 2

For process BC(isobaric process)

Q=W+U

W=P*(Vf-Vi)=3*(0.4-0.09)=93 kJ

Since Q = 345 kJ

U=345-93=252 kJ

For CD U=0:Isothermal process

For DA

Q=W+U

W=1*(0.2-1.2)=-100 kJ

Since Q=-371 kJ

U=-271 kJ

Since U for the entire cycle must be zero

U(A->B) = Energy B - Energy A = -(252-271+0)kJ=19 kJ

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.