A container contains an ideal gas with Cp = 29.1 J/mol K and Cv = 20.8 J/mol K.

Question

A container contains an ideal gas with Cp = 29.1 J/mol K and Cv = 20.8 J/mol K. It follows a thermodynamic cycle consisting of two isovolumetric and two isobaric processes. The temperature at both points 1 and 3 is 900K. a. Use the idea gas law to find the temperature at point 2. b. Use the idea gas law to find the temperature at point 4. c. Find the net work done on the gas during the cycle. d. Find the number of moles of gas in the container. e. Find the heat flow for segment 1-2. f. Find the heat flow for segment 2-3 g. Find the heat flow for segment 3-4 h. Find the heat flow for segment 4-1 i. What is the efficiency of the heat engine?

Explanation / Answer

(A) T2: 1-2 is isobaric so V1/T1 = V2/T2 --> 0.1/900 = 0.3/T2 so T2 = 0.3*900/0.1 = 2700K Ans

(B) T4: 3-4 is isobaric so V3/T3 = V4/T4 --> 0.3/900 = 0.1/T4 so T4 = 0.1*900/0.3 = 300K Ans

(C) net work = area under the curve = P1(V2-V1) - P2(V2-V1) = (P1-P2)(V2-V1) = (303900 - 101300(0.3-0.1) = (3-1)(0.3-0.1) = 0.4*105J Ans

(D) no of moles n = PV/RT. Let us use point 1 to calculate n. P1 = 303900/101300 = 3atm, V1 = 0.1*1000 = 100L, R = 0.0821 L atmK-1mol-1, T1 = 900K so n = (3*100)/(0.0821*900) = 4.06 moles Ans

(E) 1-2 heat flow: nCpdT = 4.06 * 29.1 * (2700-900) = 2.126*105J Ans

(F) 2-3 heat flow: nCvdT = 4.06 * 20.8 * (900-2700) = -1.52*105J Ans

(G) 3-4 heat flow: nCpdT = 4.06 * 29.1 * (300-900) = -0.7*105J Ans

(H) 4-1 heat flow: nCvdT = 4.06 * 20.8 * (900-300) = 0.506*105J Ans

(I) Efficiency: work done/heat absorbed * 100 = (0.4*105)/[(2.126-1.52-0.7+0.506)*105]*100 = 0.4*100/0.412 = 97.087% Ans

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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