The figure shows a reversible cycle through which 2.44 mol of a monatomic ideal

Question

The figure shows a reversible cycle through which 2.44 mol of a monatomic ideal gas is taken. Assume that p = 2p_0, V = 2V_0, P_0 = 3.91 Times 10^5 Pa, and V_0 = 0.0201 m^3. Calculate (a) the work done during the cycle, (b) the energy added as heat during stroke abc, and (c) the efficiency of the cycle, (d) What is the efficiency of a Carnot engine operating between the highest and lowest temperatures that occur in the cycle? (e) Is this greater than or less than the efficiency calculated in (c)?

Explanation / Answer

(a) Work done in the cycle
We will calcualte the work done for each cycle then add all of them
For a- b is a constant volume
Wa-b = 0
For b -c
W = 2Po(V-Vo) = 2Po(2V0 -Vo) = 2PoVo = 2*3.91*105*0.0201 = 15.7182 kJ
For c-d
again constant volume
So Wc-d = 0
For d-a
Wd-a = Po(Vo -2Vo) = -PoVo = -7.8591 kJ
total work done
15.7182 -7.8591 = 7.8591 KJ = 7859.1 J
(c)
Efficiency = Workdone / Heat given
Heat is given during process a-b and b-c
Temp at A
PV = nRT
TA = 387.41 K
temp at C
TC = 1549.64 K
temp at B
PV = nRT
TB = 2*3.91*105*(0.0201) / (2.44*8.314) = 775 K
Now heat during a-b = nCV (TB - TA) = 2.44*(8.314/0.67)(775 - 387) = 13160.8 J
Now heat during b-c = nCP(TC - TB) = 2.44*((1.67*8.314)/0.67)(1549 - 775) = 39136.5 J
total heat given = 13160.8 + 39136.5 = 52297.344 J
Efficiecny = workdone /heat = 7859.1 / 52297.344 = 0.15027
Efficiency is 15 %

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