Problem 20.6 The figure (Figure 1) is a PV diagram for a reversible heat engine

Question

Problem 20.6 The figure (Figure 1) is a PV diagram for a reversible heat engine in which 1.0 mol of argon, a nearly ideal monatomic gas, is initially at STP (point a). Points b and c are on an isotherm at T 423 K. rocess ab is at constant volume, process ac at constant pressure. Figure 1 of 1 Part A ls the path of the cycle carried out clockwise or counterclockwise? clockwise counterclockwise Submit My Answers Give Up Correct Part B What is the efficiency of this engine? e 72 Submit My Answers Give Up incorrect; Try Again; 12 attempts remaining

Explanation / Answer

A)clockwise.since integral p dv which would be negative if the the integration carried out counterclockwise.......b) 1mole of ideal gas at stp= 22.4/1000 = 0.0224meter cube..for argon.. cp= 5R/2..and cv=3R/2..cp= 20.8j/c....cv= 12.5j/c...soQ (a to b) =12.5 (423-273) =1875j..for path b to cwe use first law of thermo dynamics..dw = p dv...w= integral dvfromv=0.0224to v= 0.0352...RTintegral (dv/v)..=1590j..so total heat= 1875+1590 =3465j..note that Q btoc= area under bc curve... so w=1590-105(0.0352-0.0224)= 310j..therefore efficiency= w/Q1 =310/3465=0.09

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