The figure shows a reversible cycle through which 1.00 mole of a monatomic ideal
ID: 2252686 • Letter: T
Question
The figure shows a reversible cycle through which 1.00 mole of a monatomic ideal gas is taken. Process bc is an adiabatic expansion, with pb = 8.30 atm and Vb = 4.80 x 10-3 m3. For the cycle, find (a) the energy added to the gas as heat, (b) the energy leaving the gas as heat, (c) the net work done by the gas, and (d) the efficiency of the cycle.
The figure shows a reversible cycle through which 1.00 mole of a monatomic ideal gas is taken. Process bc is an adiabatic expansion, with pb = 8.30 atm and Vb = 4.80 times 10-3 m3. For the cycle, find (a) the energy added to the gas as heat, (b) the energy leaving the gas as heat, (c) the net work done by the gas, and (d) the efficiency of the cycle.Explanation / Answer
1) from b to c
it follows a adiabatic process .
SO p(V)^5/3 = constant
So P1 ( V1) 5/3 = P2 ( V2) 5/ 3
Given V2=8v1l
S0
P1/P2 = ( V2/V1) 5/3
P1/P2 = ( 8 v1/V1 ) 5/3
P1/P2 = (8) 5/3
P1/P2 = 32
So P2 =P1/32
Given P1 = 8.30 atm
So P2 = 8.30 /32
P2= 0.259375 atm
In a adiabatic process
Heat Q =0
work done = (P1V1 -P2V2 ) / (5/3 -1)
work done = ( 8.30 x 101325 x 4.80 x 10-3 - 0.259375 x 101325 x 8 x 4.80 x 10-3 ) / ( 2/3)
work done = 4541.386 J
work done = 4541.386 J
So dU = - W
Change in internal energy = -4541.386 J
2) from c to a
COnstant pressure process.
Work done = P ( V2-V1)
Work done = Pa ( vb-8b)
Work done = - 7 Pa Vb
work done = -7 x 0.259375 x 101325 x 4.8 x 10-3
work done = - 883.047 J
Heat Q = n Cp dT
Q= Cp ( ndT)
Q= Cp /R ( pdV )
Q= cp x work done / R
Q= -5 R /2 x 883.047 / R
Q= -2207.618 J
So dU = -2207.618 +883.047
dU = -1324.571 J
3) from a to b
Conatant volume process
work done = 0
change in internal energy dU = n Cv dT
dU = Cv V ( P2-p1) /R
dU = 3/2 x 4.8 x 10-3 x ( 8.30 - 0.259375) x 101325
dU = 5865.95 J
Q= dU
Q= 5865.95 J
A) Total heat added = 5865.95 J
B) total heat released = -2207.618 J
C) net work = -883.047 + 4541.386
net work done = 3658.339 J
D) efficiency = 3658.339 / 5865.95
efficiency = 62.36 %
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