PHYSICS - OBJECTIVE -6

 13. The shown p-V diagram represents the thermodynamic cycle of an engine,operating with an ideal monoatomic gas. The amount of heat, extracted from the source in a single cycle is 

In the above thermodynamic cycle, input energy is at constant pressure (isentropic) .

So Q1=Heat input ; Q1 = nCpΔT

Cp =5/2 R for monoatomic gas (R is the Avagadro constant)

So, Q1 =5/2nRΔT

Similarly ,Q2 = Heat rejected ; Q2=nCvΔT0

Cp - Cv = R ; So, Cv=5/2R - R =3/2R

Hence, Q2 =3/2nRΔT0

Now , from the graph ΔT = 2ΔT0

Q1 + Q2 = 5/2nR x2ΔT + 3 /2nRΔT = 13/2RΔT

Now , from ideal gas equation PV =nRT

So, RΔT =PΔV

Hence we can deduce , 

Q1 + Q2 = 13/2PV0

So, the answer is (b)

 EXPLANATION: Here, the engine cycle can be approximated to Otto cycle .Here the input is at constant pressure (hence it is isentropic) The output is at constant volume (which is adiabatic).Here the automobile engine undergoes huge pressure difference ,since it is controlled combustion. 

So, my point is the same amount of engine volume has been displaced at the input and output strokes.This shows that the automobile is having fairly low efficiency ( efficiency is low). This problem results in huge pollution since output gases are heavily unfinished(unburned) while undergoing combustion in the piston combustion chamber.This process is completed by addition of muffler to exhaust which reduces the sound of expulsion of output gases ( gases which has particles suspended in it )