5) The figure shows a reversible cycle through which 1.93 mol of a monatomic ide

Question

5) The figure shows a reversible cycle through which 1.93 mol of a monatomic ideal gas is taken. Assume that p = 2p0,V = 2V0,p0 = 4.05 105 Pa, and V0 = 0.0227 m3. 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)?

5) The figure shows a reversible cycle through which 1.93 mol of a monatomic ideal gas is taken. Assume that p = 2p0,V = 2V0,p0 = 4.05 ½ 105 Pa, and V0 = 0.0227 m3. 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)? r now contains modified data Materials inspection

Explanation / Answer

Given data

Numberof moles n = 1.93

p0 = 4.05 × 105 Pa

   p = 2 p0

      = 2( 4.05 × 105 Pa)

      = 8.10× 105 Pa

v0 = 0.0227 m3

    v = 2v0

       =2(0.0227 m3)

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a)

The work done during the cycle = Rectangulaer area enclosed in the pV diagram

                                              W = (v-v0 ) (p-p0 )

                                                   = p0 v0

                                                                         =(4.05 × 105 Pa)(0.0227 m3)

                                                 = 9.1935 kJ

b)

The amount of energy added during the stroke abc is,

   Q( abc) = (13/2) p0 v0

                          =(13/2) (9.1935 kJ)

                = 59.76 kJ

c)

The effiiceiency of the cycle is,

                       = W/Q (H)

                           = ( p0 v0)/((13/2) p0 v0)

                            = 15.4 %

d)

The efficiency of a Carnot engine operating between the highest and lowest temperatures is,

                       = 1- Ta / TC

                          = 1- (p0 v0)/ (2p0 2v0)

                         = 1- 1/4

                         = 75%

e)

The efficiency of carnot engine in part (d) is greater than the part (c).

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