A cycle followed by an engine (made of one mole of perfect gas in a cylinder with a piston) is shown in Fig. 12.10.

A to B: volume constant

B to C: adiabatic

C to D: volume constant

D to A: adiabatic

V_c = V_D = 2V_A = 2V_B

(a) In which part of the cycle heat is supplied to the engine from outside?

(b) In which part of the cycle heat is being given to the surrounding by the engine?

(c) What is the work done by the engine in one cycle? Write your answer in term of P_A, P_B, V_A.

(d) What is the efficiency of the engine?

[ γ = for the gas], ( C_v = R for one mole)

Given,

A to B: volume constant (isochoric)

B to C: adiabatic

C to D: volume constant (isochoric)

D to A: adiabatic

a.

We know that in adiabatic process there is no heat supplied or evolved from the system.

The part BC and AD should be eliminated as they are adiabatic process.

Now,

According to 1^st law of thermodynamics which states that the change in internal energy of a system is equal to the heat added to the system minus work done by the system.

For isochoric process , as = 0 Since, dv=0

.........(1)

Also we know internal energy of gas,

..........(2)

Where, n= No. of moles

= change in internal energy

= change in temperature

= specific heat of air at constant volume

Equating equation 1 and 2...

In Process AB we can clearly see, the pressure is increasing and volume is constant,

From Ideal gas eq. i.e.

We can say T

From A to B is positive

Therefore, Q is also positive. So, heat is supplied to the engine.

Heat is supplied to the engine from outside in the cycle from A to B.

b. As in part CD, the P is decreasing with V constant

is negative

Hence is negative

And as ,(when Volume is constant)

Therefore, Q is also negative.

Heat is being given to the surrounding by the engine in the cycle from C to D.

c. Work done by the system is given by

For isochoric process ,

(since from A to B and from C to D the processes are isochoric)

= 0

Similarly, .......(1)

We know that for adiabatic compression or expansion

...........(2)

Substituting equation 2 in 1 we get,

Putting C=

Similarly,

Given, V_c = V_D = 2V_A = 2V_B.........(3)

Putting the value in the above equation and solving we get,

...........(4)

Similarly,

...........(5)

Total work done=

=

=

Putting the values from 3,4 and 5 and solving we get

Total work done=

=

=

Again given, γ = for the gas

Total work done

c. We know

We know internal energy of gas,

.......(6)

Where, n = No. Of moles

C_v= Specific heat capacity at constant volume.

= Change in temperature

Also heat supplied during process A to B,

........(7)

Equating equation 7 and 8

=

Since, PV=nRT