An engine made of one mole of a perfect gas in a cylinder with a piston is taken once around the cycle $A\to B\to C\to D\to A$ on a $P$-$V$ diagram, where $A\to B$ is at constant volume, $B\to C$ is adiabatic, $C\to D$ is at constant volume and $D\to A$ is adiabatic. The volumes satisfy $V_C=V_D=2V_A=2V_B$, so states $A$ and $B$ share the smaller volume $V_A$ and states $C$ and $D$ share the larger volume $2V_A$. On the diagram $A$ is the lower-left corner with $B$ directly above it (same volume $V_A$, higher pressure $P_B>P_A$); $C$ is at the upper-right and $D$ directly below it (both at volume $2V_A$), with the adiabatic $B\to C$ expanding the gas and the adiabatic $D\to A$ compressing it. Take $\gamma=\tfrac{5}{3}$ and $C_v=\tfrac{3}{2}R$ for the one mole of gas. (a) In which part of the cycle is heat supplied to the engine from outside? (b) In which part of the cycle is heat given to the surroundings by the engine? (c) Find the work done by the engine in one cycle, in terms of $P_A$, $P_B$ and $V_A$. (d) Find the efficiency of the engine.

An engine made of one mole of a perfect gas in a cylinder with a piston is taken once around the cycle A→ B→ C

NCERT Exemplar · Q24

A\to B\to C\to D\to A

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

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

P_A

P_B

and

V_A

(d) Find the efficiency of the engine.

Punjab PsebLong· 5mImportance★★★★★est

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Concept understanding — Heat Engine Efficiency

A heat engine is any device that takes in heat, does some useful work, and then dumps the leftover heat somewhere else. Your car engine, a steam turbine, even your own body — all are heat engines.

The Intuition: You Can't Get Something for Nothing

Imagine you buy a bag of coal. You burn it to get heat. You want that heat to push a piston and turn a wheel. But you quickly notice: the engine gets hot, the exhaust pipe gets hot, the radiator gets hot. Some of that coal's heat is always wasted. You never turn 100% of the heat into motion.

Why? Because heat naturally flows from hot to cold. To make it do work, you have to let some of it "fall" to a lower temperature. The work you get is the difference between what went in and what came out. That difference is the efficiency.

Note

Think of a water wheel. Water falls from a high reservoir to a low one. The work you get is from the drop in height. You never get back the energy of the whole river — only the part that fell. Heat engines work the same way: heat "falls" from a hot source to a cold sink, and you harvest the drop.

The Precise Statement

Let’s label the quantities:

$Q_h$ = heat absorbed from the hot reservoir (the fuel, the furnace, the sun)
$Q_c$ = heat rejected to the cold reservoir (the exhaust, the cooling water, the atmosphere)
$W$ = useful work output

By energy conservation (the First Law of Thermodynamics):

W = Q_h - Q_c

The efficiency $\eta$ is simply the fraction of the input heat that becomes work:

\eta = \frac{W}{Q_h} = \frac{Q_h - Q_c}{Q_h} = 1 - \frac{Q_c}{Q_h}

\boxed{\eta = 1 - \frac{Q_c}{Q_h}}

This is the definition. It is always less than 1 because $Q_c$ is always greater than zero — you must reject some heat.

What This Tells You

To increase efficiency, you want $Q_c$ as small as possible — dump as little heat as possible. But you can never make $Q_c = 0$ ; that would violate the Second Law of Thermodynamics.
The ratio $Q_c/Q_h$ is the fraction of input heat that is wasted. If you waste half, $\eta = 0.5$ (50%).
Real engines (car engines, power plants) have efficiencies around 25–40%. The rest is lost as heat to the environment. …

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Only the two constant-volume legs exchange heat: $A\to B$ absorbs heat (pressure and temperature rise) and $C\to D$ rejects heat. The adiabats $B\to C$ and $D\to A$ carry no heat but do the work. Adding the two adiabatic works gives $W=\tfrac32(P_B-P_A)V_A[1-2^{-2/3}]\approx0.55(P_B-P_A)V_A$ , and dividing by the absorbed heat gives $\eta=1-2^{-2/3}\approx37\%$ .

Setup

Using the adiabatic relation $PV^{\gamma}=$ const on the two adiabats (with volume ratio 2):

B\to C:\ P_C=\frac{P_B}{2^{\gamma}},\qquad D\to A:\ P_D=\frac{P_A}{2^{\gamma}}.

(a) Heat supplied from outside

$A\to B$ is at constant volume with $P_B>P_A$ , so temperature rises and $Q_{AB}=C_v(T_B-T_A)>0$ : heat is supplied during $A\to B$ .

(b) Heat rejected to surroundings

$C\to D$ is at constant volume with $P_C>P_D$ , so temperature falls and $Q_{CD}=C_v(T_D-T_C)<0$ : heat is given to the surroundings during $C\to D$ . (The adiabats $B\to C$ and $D\to A$ exchange no heat.)

(c) Work done in one cycle

Constant-volume legs do no work, so $W=W_{BC}+W_{DA}$ , with adiabatic work $W=\dfrac{P_iV_i-P_fV_f}{\gamma-1}$ :

W_{BC}=\frac{P_BV_A-P_C(2V_A)}{\gamma-1},\qquad W_{DA}=\frac{P_D(2V_A)-P_AV_A}{\gamma-1}.

Adding and substituting $P_C=P_B2^{-\gamma}$ , $P_D=P_A2^{-\gamma}$ :

$W=\frac{V_A}{\gamma-1}\Big[(P_B-P_A)-2\cdot2^{-\gamma}(P_B-P_A)\Big]=\frac{V_A(P_B-P_A)}{\gamma-1}\big(1-2^{1-\gamma}\big).$ …

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Setup

(a) Heat supplied from outside

(b) Heat rejected to surroundings

(c) Work done in one cycle

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