Theory, worked examples and 15 previous-year questions with full solutions — IIT-JAM, GATE, JEST and CSIR NET-JRF, 2020 onwards. Click Show solution under any question.
Theory
3.1 Statements of the second law
Kelvin–Planck statement. It is impossible to construct a device that, operating in a cycle, takes heat from a single reservoir and converts it completely into work, with no other effect.
Clausius statement. It is impossible for heat to flow, in a cyclic process, from a colder body to a hotter body with no other change taking place.
Equivalence. The two statements are equivalent, as each fails if the other does. Suppose first that the Clausius statement were false. Let an engine take \(Q_1\) from the hot reservoir, do work \(W\), and reject \(Q_2\) to the cold one. A “Clausius violator” could then carry \(Q_2\) back to the hot reservoir for free, and the combination would take \(Q_1-Q_2\) from the hot reservoir and convert it entirely into \(W\), violating Kelvin–Planck. Conversely, if the Kelvin–Planck statement were false, a perfect engine could turn heat \(Q\) from the cold body into work, and that work could drive heat from the cold body into the hot body (for example by friction), violating Clausius.
3.2 Heat engines and refrigerators
A heat reservoir is a body of effectively infinite heat capacity, whose temperature does not change when heat is exchanged with it. A heat engine has three parts: a source, a hot reservoir at \(T_1\) supplying heat \(Q_1\) (e.g. a furnace or nuclear reactor); a working substance, which goes through a cycle (e.g. steam/water); and a sink, a cold reservoir at \(T_2\) receiving heat \(Q_2\) (e.g. a condenser or the atmosphere). The scheme is shown in Figure 3.1.
Figure 3.1 Schematic of a heat engine working between a hot and a cold reservoir.
Over one cycle \(\Delta U=0\), so the work done is \(W=Q_1-Q_2\), and the efficiency is
A refrigerator is an engine run backwards: work \(W\) removes heat \(Q_2\) from the cold body and delivers \(Q_1=Q_2+W\) to the hot body. Its coefficient of performance and that of a heat pump are
For Carnot devices, \(\text{COP}_{\rm ref}=\dfrac{T_2}{T_1-T_2}\).
3.3 The Carnot cycle
A Carnot engine uses an ideal gas taken through four reversible steps (Figure 3.2). In the first, an isothermal expansion at \(T_1\) (1 → 2), the gas absorbs heat \(Q_1\) from the source and does work. Since \(\Delta U=0\),
\[Q_1=W_1=nRT_1\ln\frac{V_2}{V_1}.\]
In the second, an adiabatic expansion (2 → 3), the gas cools from \(T_1\) to \(T_2\), doing work \(W_2=\dfrac{nR(T_1-T_2)}{\gamma-1}\). In the third, an isothermal compression at \(T_2\) (3 → 4), heat \(Q_2\) is rejected:
Finally, in an adiabatic compression (4 → 1), the gas warms back to \(T_1\), with \(W_4=\dfrac{nR(T_2-T_1)}{\gamma-1}=-W_2\).
Figure 3.2 The Carnot cycle on a \(P\)–\(V\) diagram: isotherms at \(T_1\) and \(T_2\) joined by two adiabats.
Efficiency. The two adiabatic works cancel, so \(W=nRT_1\ln\dfrac{V_2}{V_1}+nRT_2\ln\dfrac{V_4}{V_3}\). On the adiabats, \(T_1V_2^{\gamma-1}=T_2V_3^{\gamma-1}\) and \(T_1V_1^{\gamma-1}=T_2V_4^{\gamma-1}\); dividing, \(\dfrac{V_2}{V_1}=\dfrac{V_3}{V_4}\). Hence \(W=nR(T_1-T_2)\ln\dfrac{V_2}{V_1}\), and
No engine working between two reservoirs is more efficient than a reversible engine working between the same two reservoirs.
All reversible engines between the same two reservoirs have the same efficiency, whatever the working substance (ideal gas, photon gas, …).
Proof of (1). Suppose an engine X had \(\eta_X>\eta_C\). Run the Carnot engine backwards as a refrigerator, driven by X, with both taking the same \(Q_1\) at the hot reservoir. X produces more work than the refrigerator needs. The surplus is net work obtained from the cold reservoir alone, which violates Kelvin–Planck.
Part (2) allows a thermodynamic temperature scale to be defined by \(Q_2/Q_1=T_2/T_1\); it coincides with the ideal-gas scale.
\(T\)–\(S\) diagram. The Carnot cycle is a rectangle: the isotherms are lines of constant \(T\) and the adiabats lines of constant \(S\). For an engine it is traversed clockwise. The area of the rectangle is \(W\), and the area under the upper isotherm is \(Q_1\).
Example 3.1 An ideal gas engine operates in a cycle which, when represented on a \(P\)–\(V\) diagram, is a rectangle. Call \(P_1\) and \(P_2\) the lower and higher pressures, and \(V_1\) and \(V_2\) the lower and higher volumes.
(a) Calculate the work done in one complete cycle.
(b) Indicate in which parts of the cycle heat is absorbed and in which it is liberated.
(c) Calculate the quantity of heat flowing into the gas in one cycle.
(d) Show that the efficiency of the engine is
\[\eta=\frac{\gamma-1}{\dfrac{\gamma P_2}{P_2-P_1}+\dfrac{V_1}{V_2-V_1}} .\]
Solution. Label the corners A\((V_1,P_2)\), B\((V_2,P_2)\), C\((V_2,P_1)\), D\((V_1,P_1)\), traversed A → B → C → D → A (clockwise), as in Figure 3.3. Use \(nC_V=\dfrac{nR}{\gamma-1}\), \(nC_P=\dfrac{\gamma nR}{\gamma-1}\) and \(nR\,\Delta T=\Delta(PV)\).
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