Thermodynamics

The zeroth law says: if bodies A and B are each in thermal equilibrium with a third body C, then A and B are in thermal equilibrium with each other. This is why a thermometer works, and it lets us define temperature as the property that is equal for bodies in thermal equilibrium.

A gas in equilibrium has a well-defined pressure \(P\), volume \(V\) and temperature \(T\), linked by an equation of state such as \(PV=nRT\). Any quantity fixed by the state alone is a state function. Quantities that depend on how the gas got there are path functions.

Intensive variables (\(P, T\), density) do not depend on the amount of gas. Extensive variables (\(V, U\), mass) double when you double the system. During a fast, turbulent process the gas has no single \(P\) or \(T\), so only the start and end states can be plotted on a \(P\)–\(V\) diagram.

NCERT physics takes \(W\) as work done by the gas. Chemistry books (and some physics books) take \(W\) as work done on the gas. Both describe the same physics.

A safe habit: read every statement as "energy into the gas" or "energy out of the gas". Heat in and work done on the gas both raise \(U\). Heat out and work done by the gas both lower \(U\).

For an ideal gas \(U\) depends only on \(T\), so \(\Delta U = nC_V\Delta T\) for every process, not only at constant volume. The other two terms change with the process. Work is by the gas.

Let the gas push a piston of area \(A\) through a small distance \(dx\). The force is \(PA\), so the work is \(dW = PA\,dx = P\,dV\). Adding up the strips from \(V_1\) to \(V_2\):

• Moving right (expansion): \(dV > 0\), so \(W > 0\).
• Moving left (compression): \(W < 0\). The area is the same size but counted negative.
• Vertical line (isochoric): no area, \(W = 0\).
• Two paths between the same end points enclose different areas, so \(W\) (and \(Q\)) differ while \(\Delta U\) is the same.

For a sudden, irreversible change, the gas pressure is not defined during the process. Then use the external pressure: \(W = P_{\rm ext}\Delta V\) when \(P_{\rm ext}\) is constant.

Going round a closed loop, the expansion part adds area and the compression part subtracts area. What is left is the area enclosed by the loop.

• Clockwise loop on a \(P\)–\(V\) diagram (P up, V across): net work by the gas is positive. The gas expands at high pressure and is compressed at low pressure. This is a heat engine.
• Anticlockwise loop: net work is negative. Work is done on the gas. This is a refrigerator.
• Because \(\Delta U = 0\) over a cycle, net heat absorbed equals the net work: \(Q_{\rm net} = W_{\rm net}\).

The rule "clockwise means positive" holds only with \(P\) on the vertical axis and \(V\) on the horizontal axis. If the axes are swapped (V vertical), clockwise means negative work. On \(P\)–\(T\) or \(V\)–\(T\) graphs the loop area is not work at all. Convert to \(P\)–\(V\) first.

• On a \(V\)–\(T\) graph, a straight line through the origin means \(V \propto T\): isobaric. Its work is \(nR\Delta T\).
• On a \(P\)–\(T\) graph, a straight line through the origin means \(P \propto T\): isochoric. Its work is zero.
• A straight line on a \(P\)–\(V\) graph that is not horizontal: the area is a trapezium, \(W = \tfrac12(P_1+P_2)(V_2-V_1)\).
• Unit check: 1 kPa × 1 L = 10³ Pa × 10⁻³ m³ = 1 J. But 1 atm × 1 L ≈ 101.3 J.

1. Constant volume: \(W = 0\), so \(Q = \Delta U = nC_V\Delta T\). This defines \(C_V\).
2. Constant pressure: \(Q = \Delta U + P\Delta V\). For an ideal gas \(P\Delta V = nR\Delta T\), so \(Q = nC_V\Delta T + nR\Delta T\).
3. By definition this heat is \(nC_P\Delta T\). Hence \(C_P = C_V + R\), i.e. \(C_P - C_V = R\).

Per unit mass the relation reads \(c_P - c_V = R/M\), where \(M\) is the molar mass in kg/mol. So it differs from gas to gas.

At constant pressure, \(Q = nC_P\Delta T\), \(\Delta U = nC_V\Delta T\) and \(W = nR\Delta T\). Dividing:

Slopes give ranking questions. On a \(V\)–\(T\) graph, a steeper isobar means a lower pressure. On a \(P\)–\(T\) graph, a steeper isochore means a smaller volume. The lines pass through the origin only when \(T\) is in kelvin.

1. Along an isotherm \(P = nRT/V\) with \(T\) constant.
2. \(W = \int_{V_1}^{V_2} P\,dV = nRT\int_{V_1}^{V_2}\frac{dV}{V} = nRT\ln\frac{V_2}{V_1}\).
3. Since \(P_1V_1 = P_2V_2\), \(V_2/V_1 = P_1/P_2\). So \(W = nRT\ln\frac{P_1}{P_2}\).
4. With common logarithms: \(W = 2.303\,nRT\log_{10}\frac{V_2}{V_1}\).

\(\Delta U = 0\), so \(Q = W\). In expansion the gas absorbs heat equal to the work it does. In compression \(W < 0\) and the gas gives out the same amount of heat.

• Thin conducting (diathermic) walls, good contact with a large reservoir, and a slow process so heat has time to flow.
• Heat flows but \(T\) does not change, so the molar heat capacity \(C = Q/(n\Delta T)\) is infinite.
• Slope on \(P\)–\(V\): from \(PV = \) const, \(\dfrac{dP}{dV} = -\dfrac{P}{V}\).
• Isothermal bulk modulus: \(B_T = -V\dfrac{dP}{dV} = P\).

Link to the speed of sound. Newton assumed sound compressions are isothermal and used \(v = \sqrt{P/\rho}\). For air at 1.013 × 10⁵ Pa and 1.293 kg/m³ this gives about 280 m/s, well below the measured 332 m/s. Laplace noted that the compressions are too fast for heat to flow, so they are adiabatic: \(v = \sqrt{\gamma P/\rho} \approx 331\) m/s.

In Joule's free expansion, a gas in an insulated container rushes into an evacuated part. It pushes nothing, so \(W = 0\). The walls are insulated, so \(Q = 0\). Hence \(\Delta U = 0\).

• For an ideal gas the final temperature equals the initial temperature, and \(P_fV_f = P_iV_i\).
• For a real gas with attractive forces, the temperature usually drops a little.
• The process is irreversible and fast. The intermediate states are not equilibrium states, so it has no \(P\)–\(V\) curve. Neither \(nRT\ln(V_2/V_1)\) nor \(PV^\gamma = \) const applies.

Isothermal expansion and free expansion share the same end states for an ideal gas. They are different paths with different \(Q\) and \(W\).

1. Adiabatic: \(dQ = 0\), so \(dU = -dW\), that is \(nC_V\,dT = -P\,dV\).
2. Differentiate \(PV = nRT\): \(P\,dV + V\,dP = nR\,dT\).
3. Substitute \(n\,dT = -P\,dV/C_V\): \(P\,dV + V\,dP = -\dfrac{R}{C_V}P\,dV = -(\gamma-1)P\,dV\), using \(R/C_V = \gamma - 1\).
4. Rearrange: \(\gamma P\,dV + V\,dP = 0\), so \(\dfrac{dP}{P} = -\gamma\dfrac{dV}{V}\).
5. Integrate: \(\ln P = -\gamma\ln V + \) const, so \(PV^\gamma = \) constant.

Using \(P = nRT/V\) gives the other forms: \(TV^{\gamma-1} = \) const and \(P^{1-\gamma}T^{\gamma} = \) const (equivalently \(T \propto P^{(\gamma-1)/\gamma}\)).

Write \(P = K/V^\gamma\) where \(K = P_1V_1^\gamma = P_2V_2^\gamma\). Then

Check with the first law: \(W = -\Delta U = -nC_V(T_2 - T_1) = \dfrac{nR(T_1 - T_2)}{\gamma - 1}\), since \(C_V = R/(\gamma-1)\). In expansion \(T_2 < T_1\) and \(W > 0\); in compression \(W < 0\) and the gas heats up.

Slope of an isotherm: \(dP/dV = -P/V\). Slope of an adiabat: \(dP/dV = -\gamma P/V\). Through the same point the adiabat is steeper by a factor \(\gamma\). For the same reason the adiabatic bulk modulus is \(B_S = \gamma P\), while the isothermal one is \(B_T = P\).

"Slow" does not mean isothermal. A slow process in an insulated cylinder is a reversible adiabatic process. A fast process (a bursting tyre, a sound wave) is nearly adiabatic because heat has no time to flow.

Two more useful facts. Using \(P = \mu RT/V\), the process \(PV^n = \) const is the same as \(TV^{n-1} = \) const. And the slope of the curve on a \(P\)–\(V\) diagram is \(dP/dV = -nP/V\), so a larger n gives a steeper curve through the same point.

1. Write \(P = K/V^n\) with \(K = P_1V_1^n = P_2V_2^n\).
2. \(W = \int_{V_1}^{V_2} KV^{-n}\,dV = \dfrac{K\left(V_2^{1-n} - V_1^{1-n}\right)}{1-n} = \dfrac{P_1V_1 - P_2V_2}{n-1} = \dfrac{\mu R(T_1 - T_2)}{n-1}\) for \(n \ne 1\).
3. \(\Delta U = \mu C_V(T_2 - T_1)\) for any ideal-gas process.
4. \(Q = \Delta U + W = \mu\left(C_V - \dfrac{R}{n-1}\right)(T_2 - T_1)\).
5. Defining \(Q = \mu C\Delta T\): \(C = C_V + \dfrac{R}{1-n} = \dfrac{R}{\gamma - 1} - \dfrac{R}{n-1}\).

Fractions of the heat: \(\dfrac{\Delta U}{Q} = \dfrac{C_V}{C}\) and \(\dfrac{W}{Q} = \dfrac{R/(1-n)}{C}\). For \(P \propto V\) with a monatomic gas, \(C = 2R\), so 3/4 of the heat goes to U and 1/4 to work.

A general shortcut for any process law: \(C = C_V + \dfrac{P}{\mu}\dfrac{dV}{dT}\). It comes from \(dQ = \mu C_V\,dT + P\,dV\).

For \(1 < n < \gamma\), \(R/(1-n)\) is negative and larger in size than \(C_V\), so \(C < 0\). Then Q and ΔT have opposite signs. In such an expansion, the gas absorbs heat but does even more work than the heat it gets. The extra comes from U, so the gas cools while heat flows in.

To find C for a law written in T, turn it into \(PV^n = \) const using \(PV = \mu RT\):

• \(V \propto T^2\): \(T \propto V^{1/2}\), so \(P \propto T/V \propto V^{-1/2}\). Thus \(PV^{1/2} = \) const, n = 1/2, \(C = C_V + 2R\).
• \(PT = \) const: \(P \propto 1/T\) and \(T \propto PV\), so \(P^2V = \) const, that is \(PV^{1/2} = \) const. Again n = 1/2.
• \(P = kV^2\): \(PV^{-2} = \) const, n = −2, \(C = C_V + R/3\).
• \(V \propto T^k\) in general: \(TV^{n-1} = \) const gives \(n = 1 - 1/k\).

A heat engine has three parts: a hot source at \(T_1\), a working substance (the gas) and a cold sink at \(T_2\). Per cycle it takes \(Q_1\), gives out work \(W\) and rejects \(Q_2\). (The lesson above writes these as \(Q_h\) and \(Q_c\).)

Method for a cycle drawn on a graph:

1. Find \(W\) as the enclosed area, with its sign from the direction.
2. Find \(Q\) for each step: \(nC_V\Delta T\) (isochoric), \(nC_P\Delta T\) (isobaric), \(nRT\ln(V_2/V_1)\) (isothermal), 0 (adiabatic).
3. Add only the positive \(Q\) values. That sum is \(Q_1\).
4. \(\eta = W/Q_1\). Dividing by the net heat would always give 100%, which is wrong.

A Carnot engine uses an ideal gas through four reversible steps:

1. A→B: isothermal expansion at \(T_1\). The gas absorbs \(Q_1 = nRT_1\ln(V_B/V_A)\).
2. B→C: adiabatic expansion. T falls from \(T_1\) to \(T_2\).
3. C→D: isothermal compression at \(T_2\). The gas rejects \(Q_2 = nRT_2\ln(V_C/V_D)\).
4. D→A: adiabatic compression. T rises from \(T_2\) back to \(T_1\).

The two adiabats give \(T_1V_B^{\gamma-1} = T_2V_C^{\gamma-1}\) and \(T_1V_A^{\gamma-1} = T_2V_D^{\gamma-1}\). Dividing, \(V_B/V_A = V_C/V_D\). So the logarithms cancel and

The adiabatic works B→C and D→A are equal and opposite (each is \(nR(T_1-T_2)/(\gamma-1)\)), so they cancel. The net work is \(Q_1 - Q_2\).

To convert a cycle: label each corner with its (P, V, T), identify each side from the table, then redraw on \(P\)–\(V\). Decide the sign of the work only from the \(P\)–\(V\) picture.

• Kelvin–Planck statement: No process is possible whose sole result is the absorption of heat from a reservoir and its complete conversion into work. So no engine has 100% efficiency; some heat must go to a sink.
• Clausius statement: No process is possible whose sole result is the transfer of heat from a colder body to a hotter body. So a refrigerator needs work input; its COP is finite.
• The two statements are equivalent: breaking one would let you break the other.

A reversible process is quasistatic (a chain of equilibrium states) and free of dissipation. It can be run backwards with both system and surroundings restored. Friction, free expansion, heat flow across a finite temperature difference and mixing of gases are irreversible.

Carnot's theorem: no engine working between two given temperatures is more efficient than a reversible engine, and all reversible engines between the same two temperatures have the same efficiency, \(1 - T_2/T_1\). \(\eta = 1\) would need \(T_2 = 0\) K, which cannot be reached.

Run an engine backwards. Work \(W\) is put in, heat \(Q_2\) is taken from the cold body and \(Q_1 = Q_2 + W\) is given to the hot body.

Engine reversed: for a reversible machine between the same reservoirs, the same \(Q_1, Q_2, W\) appear with arrows reversed. Then \(\beta = Q_2/W = (Q_1 - W)/W = 1/\eta - 1\), so

\(\eta = 1 - T_2/T_1\). Change each temperature by a small amount:

Since \(T_2 < T_1\), \(T_2/T_1^2 < 1/T_1\). So lowering the sink by \(\Delta T\) raises the efficiency more than raising the source by the same \(\Delta T\).

Example: between 500 K and 300 K, \(\eta = 40\%\). Raising the source by 100 K gives \(1 - 300/600 = 50\%\). Lowering the sink by 100 K gives \(1 - 200/500 = 60\%\).

For a refrigerator, a colder inside (smaller \(T_2\)) or a hotter room makes \(T_1 - T_2\) larger, so \(\beta = T_2/(T_1 - T_2)\) gets smaller. Cooling to very low temperatures costs a lot of work.