The Workbook

If two bodies are each in thermal equilibrium with a third body, they are in thermal equilibrium with each other.

The law defines temperature: bodies in thermal equilibrium have the same temperature. A thermometer is the third body. When it reaches equilibrium with a patient, its reading is the patient's temperature.

\(Q = \Delta U + W\): the heat given to a system raises its internal energy and does work.

\(Q\) is positive when heat enters the system. \(W\) is positive when the system does work on its surroundings, as in an expansion. Heat leaving and work done on the gas are negative.

1. \(\Delta U = Q - W = 500 - 200 = 300\) J.

2. \(Q = -300\) J and \(W = -100\) J (work done on the gas). \(\Delta U = -300 - (-100) = -200\) J.

A state function has a definite value in each state, so its change depends only on the initial and final states. \(U\) depends only on the state of the gas.

Heat and work depend on the path: they are energy in transit, not properties of a state. For two paths from \(A\) to \(B\), \(\Delta U\) is the same but \(Q\) and \(W\) can differ. Only their difference \(Q - W\) is fixed.

\(W = P(V_2 - V_1)\), the area of a rectangle under the horizontal line on the \(P\)-\(V\) diagram.

\(W = 10^5 \times (3 - 2) \times 10^{-3} = 100\) J.

1. The area under the curve on a \(P\)-\(V\) diagram is \(\int P\,dV\), the work done by the gas.

2. Trapezium: \(W = \dfrac{200 + 100}{2} \times (6 - 2) = 150 \times 4 = 600\) kPa L \(= 600\) J.

3. Positive. The volume increases, so the gas pushes its surroundings outward and does work on them.

1. \(AB\): \(V\) constant, \(W = 0\). \(BC\): \(W = 300 \times 4 = 1200\) J. \(CD\): \(W = 0\). \(DA\): \(W = 100 \times (2 - 6) = -400\) J.

2. \(W_{net} = 0 + 1200 + 0 - 400 = 800\) J, which is the area of the rectangle, \(200\) kPa \(\times 4\) L.

3. In a cycle \(\Delta U = 0\), so \(Q_{net} = W_{net} = 800\) J. Run anticlockwise, every \(W\) changes sign: \(W_{net} = -800\) J, and the gas gives out \(800\) J of heat.

1. \(AB\): \(V\) constant, \(W = 0\). \(BC\): area of the trapezium under \(BC\), \(\dfrac{300 + 100}{2} \times 4 = 800\) J. \(CA\): \(W = 100 \times (2 - 6) = -400\) J.

2. \(W_{net} = 0 + 800 - 400 = 400\) J, the area of the triangle: \(\dfrac{1}{2} \times 4 \times 200 = 400\) J.

1. \(V \propto T\) along a line through the origin. From \(PV = nRT\), \(\dfrac{V}{T} = \dfrac{nR}{P}\) is constant, so \(P\) is constant. The process is isobaric.

2. \(W = P\Delta V = 100 \times (4 - 2) = 200\) kPa L \(= 200\) J.

3. A horizontal line at \(P = 100\) kPa from \(V = 2\) L to \(V = 4\) L, with the arrow towards larger \(V\).

1. \(AB\): \(P \propto T\) through the origin, so \(\dfrac{P}{T} = \dfrac{nR}{V}\) is constant and \(V\) is constant. \(AB\) is isochoric. \(BC\): \(P\) is constant, so \(BC\) is isobaric.

2. \(AB\): the volume does not change, so \(W = 0\).

3. \(BC\): \(W = P\Delta V = nR\Delta T = 1 \times 8.3 \times (900 - 600) = 2490\) J.

\(W = nRT\ln\dfrac{V_2}{V_1} = 1 \times 8.3 \times 300 \times \ln 2 = 2490 \times 0.693 = 1726\) J, about \(1.73\) kJ.

\(W = P\Delta V = 2 \times 10^5 \times (2 - 5) \times 10^{-3} = -600\) J. So \(600\) J of work is done on the gas.

\(Q = -400\) J. \(\Delta U = Q - W = -400 - (-600) = +200\) J. The internal energy rises, so the gas becomes hotter.

At constant volume \(W = 0\). \(\Delta U = nC_V\Delta T = 1 \times 20.75 \times 100 = 2075\) J. \(Q = \Delta U + W = 2075\) J.

\(W_{net} = Q_{in} - Q_{out} = 1000 - 600 = 400\) J. In a full cycle the gas returns to its starting state, so \(\Delta U = 0\).

Fraction \(= \dfrac{400}{1000} = 0.4\), that is \(40\%\). This is the efficiency of the engine.

1. Expansion: \(W\) positive. Adiabatic: \(Q = 0\). Then \(\Delta U = -W\) is negative: the gas cools.

2. Compression: \(W\) negative. Isothermal: \(T\) constant, so \(\Delta U = 0\). Then \(Q = W\) is negative: the gas gives out heat.

Internal energy is a state function. After a complete cycle the gas is back in its initial state, so \(\Delta U = 0\).

The first law, \(Q = \Delta U + W\), then gives \(Q_{net} = W_{net}\).

The \(P\)-\(V\) graph is a straight line, so the area under it is a trapezium: \(W = \dfrac{150 + 50}{2} \times (3 - 1) = 100 \times 2 = 200\) kPa L \(= 200\) J.

Force on the piston \(= PA = 10^5 \times 0.01 = 1000\) N. Work \(= F \times d = 1000 \times 0.1 = 100\) J.

\(\Delta V = A \times d = 0.01 \times 0.1 = 10^{-3}\) m\(^3\), and \(P\Delta V = 10^5 \times 10^{-3} = 100\) J. The two agree.

Path 1: \(\Delta U = Q - W = 800 - 500 = 300\) J. \(\Delta U\) is the same for every path from \(A\) to \(B\).

Path 2: \(Q = \Delta U + W = 300 + 300 = 600\) J.

The internal energy of an ideal gas depends only on its temperature, \(U = \dfrac{f}{2}nRT\). So \(\Delta U\) is fixed by \(\Delta T\) alone. At constant volume \(Q = \Delta U = nC_V\Delta T\), and the same \(\Delta U\) goes with the same \(\Delta T\) in any other process.

\(U = \dfrac{3}{2} \times 2 \times 8.3 \times 300 = 7470\) J.

Isothermal: \(T\) does not change, so \(\Delta U = 0\). The work done on the gas leaves as heat.

The air is compressed quickly, so there is no time for heat to leave: \(Q \approx 0\). Work is done on the air, so \(W\) is negative and \(\Delta U = -W\) is positive. The internal energy and temperature of the air rise, and the pump becomes warm.

1. Isothermal: \(T\) constant, \(PV = \) constant. \(\Delta U = 0\), so \(Q = W = nRT\ln\dfrac{V_2}{V_1}\).

2. Adiabatic: no heat exchange, \(Q = 0\), \(PV^\gamma = \) constant. \(W = -\Delta U = \dfrac{P_1V_1 - P_2V_2}{\gamma - 1}\).

3. Isobaric: \(P\) constant, \(\dfrac{V}{T} = \) constant. \(W = P\Delta V = nR\Delta T\), \(Q = nC_P\Delta T\).

4. Isochoric: \(V\) constant, \(\dfrac{P}{T} = \) constant. \(W = 0\), \(Q = \Delta U = nC_V\Delta T\).

\(W = nRT\ln\dfrac{V_2}{V_1} = 2 \times 8.3 \times 300 \times 0.693 = 3452\) J.

\(T\) is constant, so \(\Delta U = 0\) and \(Q = W = 3452\) J. All the heat absorbed becomes work.

At constant \(T\), \(\dfrac{V_2}{V_1} = \dfrac{P_1}{P_2} = \dfrac{1}{4}\). \(W = nRT\ln\dfrac{1}{4} = -2490 \times 1.386 = -3452\) J. So \(3452\) J of work is done on the gas.

\(\Delta U = 0\), so \(Q = W = -3452\) J: the gas gives out \(3452\) J of heat.

\(PV^\gamma = \) constant, \(TV^{\gamma - 1} = \) constant, \(P^{1 - \gamma}T^\gamma = \) constant.

From \(PV = nRT\), \(P = \dfrac{nRT}{V}\). Put this in \(PV^\gamma = \) constant: \(\dfrac{nRT}{V}V^\gamma = nRTV^{\gamma - 1} = \) constant. Since \(nR\) is fixed, \(TV^{\gamma - 1} = \) constant.

\(T_1V_1^{\gamma - 1} = T_2V_2^{\gamma - 1}\), so \(T_2 = T_1\left(\dfrac{V_1}{V_2}\right)^{\gamma - 1} = 300 \times 8^{2/3} = 300 \times 4 = 1200\) K.

\(P_2 = P_1\left(\dfrac{V_1}{V_2}\right)^\gamma = P_1 \times 8^{5/3} = 32P_1\).

\(W = \dfrac{P_1V_1 - P_2V_2}{\gamma - 1} = \dfrac{nR(T_1 - T_2)}{\gamma - 1}\).

\(W = \dfrac{1 \times 8.3 \times (300 - 1200)}{2/3} = -11205\) J. So \(11205\) J of work is done on the gas.

\(Q = 0\), so \(\Delta U = -W = +11205\) J. Check: \(nC_V\Delta T = 12.45 \times 900 = 11205\) J.

1. Curve \(2\) is the adiabat: it is steeper. In an adiabatic expansion the gas cools as well as spreads out, so its pressure falls faster than along the isotherm.

2. Slope of adiabat \(= \gamma \times\) slope of isotherm, so the ratio is \(\gamma = 1.4\).

Isotherm: \(PV = c\), so \(P\,dV + V\,dP = 0\) and \(\dfrac{dP}{dV} = -\dfrac{P}{V}\).

Adiabat: \(PV^\gamma = c\), so \(V^\gamma dP + \gamma PV^{\gamma - 1}dV = 0\) and \(\dfrac{dP}{dV} = -\gamma\dfrac{P}{V}\).

The adiabatic slope is \(\gamma\) times the isothermal slope, and \(\gamma > 1\).

\(W = nR\Delta T = 1 \times 8.3 \times 100 = 830\) J. \(\Delta U = nC_V\Delta T = 20.75 \times 100 = 2075\) J.

\(Q = \Delta U + W = 2905\) J, which is \(nC_P\Delta T\) with \(C_P = 29.05\) J/mol K.

Rigid vessel: \(W = 0\). \(\Delta U = nC_V\Delta T = 2 \times 12.45 \times 150 = 3735\) J. \(Q = \Delta U = 3735\) J.

1. \(1\), horizontal: isobaric. \(2\), the gentler curve: isothermal. \(3\), the steeper curve: adiabatic. \(4\), vertical: isochoric.

2. Path \(1\). The work is the area under the curve, and the isobaric line stays highest over the whole expansion. The adiabat does the least.

1. The isothermal path \(AB\): the area under it is larger. The shaded gap is the extra work done along \(AB\) compared with \(AC\).

2. \(C\). Along \(AC\) no heat enters, so the work is paid for by the internal energy and the gas cools. Along \(AB\) the temperature stays at its original value.

3. \(AB\) absorbs heat equal to the work done. \(AC\) absorbs none.

1. \(X\): heat flows freely to the bath, so the gas stays at the bath temperature. Isothermal. \(Y\): no heat can leave. Adiabatic.

2. Heat flow takes time. In \(X\) the piston must move slowly so that the heat produced by compression has time to leave and \(T\) stays constant. In \(Y\) no heat is meant to leave, so moving fast does no harm and in practice helps.

3. \(Y\). The work done on the gas cannot escape as heat, so it raises the internal energy and the temperature.

1. \(A\): \(V = 8\) L, \(P = 100\) kPa. \(B\): \(V = 1\) L, \(P = 3200\) kPa.

2. \(A\): \(100 \times 8^{5/3} = 100 \times 32 = 3200\). \(B\): \(3200 \times 1^{5/3} = 3200\). The products agree, in kPa L\(^{5/3}\).

3. \(T_B = T_A\left(\dfrac{V_A}{V_B}\right)^{\gamma - 1} = 300 \times 8^{2/3} = 1200\) K. Check with \(PV = nRT\): \(\dfrac{P_BV_B}{P_AV_A} = \dfrac{3200}{800} = 4\), so \(T_B = 4T_A\).

\(T_2 = T_1\left(\dfrac{V_1}{V_2}\right)^{\gamma - 1} = 400 \times \left(\dfrac{1}{8}\right)^{2/3} = \dfrac{400}{4} = 100\) K.

Isothermal: \(\dfrac{dP}{dV} = -\dfrac{P}{V}\), so \(B = -V\left(-\dfrac{P}{V}\right) = P\). Adiabatic: \(\dfrac{dP}{dV} = -\gamma\dfrac{P}{V}\), so \(B = \gamma P\).

Air: \(B_{iso} = 10^5\) Pa and \(B_{adi} = 1.4 \times 10^5\) Pa. Sound compresses the air too fast for heat to flow, so the adiabatic modulus governs the speed of sound (Laplace's correction).

Isothermal: the temperature stays constant only if the heat produced or needed has time to flow in or out. So the change must be slow. Example: the slow leak of air from a balloon over hours.

Adiabatic: no heat must flow, so the change must be over before heat can move. Example: the compression stroke of a diesel engine, or a bicycle pump.

\(W = P(V_2 - V_1)\). With \(PV = nRT\) at constant \(P\), \(PV_2 - PV_1 = nRT_2 - nRT_1\), so \(W = nR\Delta T\).

\(W = 3 \times 8.3 \times 50 = 1245\) J.

\(\gamma = 1.4 = 1 + \dfrac{2}{f}\), so \(f = 5\): a diatomic gas.

\(C_V = \dfrac{5}{2}R = 20.75\) J/mol K and \(C_P = C_V + R = 29.05\) J/mol K.

Isothermal: \(\Delta U = 0\), so \(W = Q = 1000\) J.

The adiabatic compression does work on the gas with no heat leaving, so \(\Delta U\) is positive and the gas warms above its original temperature. The isothermal expansion had left \(T\) unchanged, so the final temperature is higher.

1. \(W = 0\), so \(\Delta U = Q = 500\) J.

2. \(\Delta U = Q - W = 700 - 200 = 500\) J.

3. \(T\) constant, so \(\Delta U = 0\). The \(400\) J leaves as work.

4. \(Q = 0\), so \(\Delta U = -W = -300\) J.

The air expands quickly as it escapes, with no time to take in heat: an adiabatic expansion. It does work pushing the outside air aside, and this work comes from its internal energy. Its temperature falls, so the jet feels cold.

Isobaric: \(P = \) constant, so \(x = 0\). Isothermal: \(PV = \) constant, \(x = 1\). Adiabatic: \(PV^\gamma = \) constant, \(x = \gamma\). Isochoric: \(V = \) constant, which is \(P^{1/x}V = \) constant with \(x \to \infty\).

\(P = \dfrac{k}{V^x}\) with \(k = P_1V_1^x = P_2V_2^x\). \(W = \int_{V_1}^{V_2} \dfrac{k}{V^x}dV = k\dfrac{V_2^{1 - x} - V_1^{1 - x}}{1 - x}\).

\(kV_2^{1 - x} = P_2V_2\) and \(kV_1^{1 - x} = P_1V_1\), so \(W = \dfrac{P_2V_2 - P_1V_1}{1 - x} = \dfrac{P_1V_1 - P_2V_2}{x - 1}\). With \(PV = nRT\) this is \(\dfrac{nR(T_1 - T_2)}{x - 1}\).

For one mole, \(Q = C\,\Delta T\), \(\Delta U = C_V\Delta T\) and \(W = \dfrac{R(T_1 - T_2)}{x - 1} = \dfrac{R\,\Delta T}{1 - x}\).

\(C\,\Delta T = C_V\Delta T + \dfrac{R\,\Delta T}{1 - x}\), so \(C = C_V + \dfrac{R}{1 - x}\).

\(x = 0\): \(C = C_V + R = C_P\), correct for constant pressure. \(x = \gamma\): \(C = C_V + \dfrac{R}{1 - \gamma} = C_V - C_V = 0\), correct for an adiabatic process. \(x \to \infty\): \(C = C_V\), correct for constant volume.

\(x = 1\) gives \(C \to \infty\). In an isothermal process heat is absorbed but \(\Delta T = 0\), so the heat per degree is infinite.

\(P \propto V\) means \(PV^{-1} = \) constant, so \(x = -1\). \(C = C_V + \dfrac{R}{1 - (-1)} = C_V + \dfrac{R}{2}\).

Monatomic: \(C = \dfrac{3}{2}R + \dfrac{1}{2}R = 2R\). Diatomic: \(C = \dfrac{5}{2}R + \dfrac{1}{2}R = 3R\).

1. \(PV^2 = (PV)V = nRTV = \) constant, so \(TV = \) constant. Doubling \(V\) halves \(T\), as given.

2. \(W = \dfrac{nR(T_1 - T_2)}{x - 1} = \dfrac{8.3 \times 150}{1} = 1245\) J. \(\Delta U = nC_V\Delta T = 12.45 \times (-150) = -1867.5\) J.

3. \(Q = \Delta U + W = -1867.5 + 1245 = -622.5\) J: the gas gives out heat. Check: \(C = \dfrac{3}{2}R + \dfrac{R}{1 - 2} = \dfrac{1}{2}R\), and \(Q = \dfrac{1}{2} \times 8.3 \times (-150) = -622.5\) J.

1. Area under \(AB\), a trapezium: \(W = \dfrac{100 + 300}{2} \times 2 = 400\) kPa L \(= 400\) J.

2. \(T \propto PV\): \(\dfrac{T_B}{T_A} = \dfrac{300 \times 3}{100 \times 1} = 9\).

3. Absorbs. The gas does positive work and its temperature rises, so both \(W\) and \(\Delta U\) are positive and \(Q = \Delta U + W > 0\).

\(C = \dfrac{3}{2}R + \dfrac{R}{1 - 1.5} = \dfrac{3}{2}R - 2R = -\dfrac{1}{2}R\).

A negative \(C\) means \(Q\) and \(\Delta T\) have opposite signs. When the gas is compressed along this path, the work done on it is more than enough to raise its temperature. The extra energy leaves as heat: the gas gives out heat while it becomes hotter.

1. The gas expands into a vacuum. There is nothing to push against, so no force acts through a distance and \(W = 0\).

2. The walls are insulated, so no heat can enter or leave: \(Q = 0\).

3. \(\Delta U = Q - W = 0\). For an ideal gas \(U\) depends only on \(T\), so the temperature does not change.

The temperature stays at \(300\) K. With \(T\) fixed and the volume four times larger, \(P_2 = P_1\dfrac{V_1}{V_2} = 4 \times \dfrac{1}{4} = 1\) atm.

An isothermal process passes slowly through equilibrium states, each with a definite \(P\), \(V\) and \(T\). In a free expansion the gas rushes out and is in turmoil. It has no single pressure or temperature until it settles. Only the end states are equilibrium states, so the process cannot be drawn as a curve on a \(P\)-\(V\) diagram.

1. \(AB\): \(W = P_AV_A\ln\dfrac{V_B}{V_A} = 400 \times 1 \times \ln 4 = 400 \times 1.386 = 554\) J. \(BC\): \(W = 100 \times (1 - 4) = -300\) J. \(CA\): \(V\) constant, \(W = 0\).

2. \(W_{net} = 554 - 300 + 0 = 254\) J. The shaded area enclosed by the cycle is this net work.

1. \(AB\) (constant \(V\)): \(Q = C_V\Delta T = 12.45 \times 300 = 3735\) J. \(BC\) (constant \(P\)): \(Q = C_P\Delta T = 20.75 \times 600 = 12450\) J. \(CD\): \(Q = 12.45 \times (-600) = -7470\) J. \(DA\): \(Q = 20.75 \times (-300) = -6225\) J.

2. \(Q_{net} = 3735 + 12450 - 7470 - 6225 = 2490\) J. Area \(= 100\) kPa \(\times 24.9\) L \(= 2490\) J. They agree, as they must, since \(\Delta U = 0\) in a cycle.

3. Heat absorbed \(= 3735 + 12450 = 16185\) J. Efficiency \(= \dfrac{2490}{16185} = 0.154\), about \(15\%\).

1. Area of the triangle \(= \dfrac{1}{2} \times 4 \times 200 = 400\) J. Anticlockwise, so \(W_{net} = -400\) J: net work is done on the gas.

2. \(\Delta U = 0\) in a cycle, so \(Q_{net} = W_{net} = -400\) J. The gas gives out \(400\) J of heat.

3. A refrigerator or heat pump: work is put in and heat is moved from a cold body to a hot one.

From Question 16, \(TP^{(1 - x)/x} = \) constant, so \(T_2 = T_1\left(\dfrac{P_2}{P_1}\right)^{(x - 1)/x} = 300 \times 8^{0.5/1.5} = 300 \times 8^{1/3} = 600\) K.

\(V = \dfrac{nRT}{P}\). Put this in \(PV^x = \) constant: \(P\left(\dfrac{nRT}{P}\right)^x = (nR)^x\,T^xP^{1 - x} = \) constant.

Take the \(x\)-th root: \(TP^{(1 - x)/x} = \) constant.

\(C_V + \dfrac{R}{1 - x} = 3R\) with \(C_V = \dfrac{3}{2}R\): \(\dfrac{R}{1 - x} = \dfrac{3}{2}R\), so \(1 - x = \dfrac{2}{3}\) and \(x = \dfrac{1}{3}\).

\(W_{net} = \) enclosed area \(= 800\) J. \(Q_{out} = Q_{in} - W = 2000 - 800 = 1200\) J. Efficiency \(= \dfrac{800}{2000} = 40\%\).

Clockwise: the expansion happens along the upper curve, where \(P\) is higher, so the positive work of expansion exceeds the negative work of compression. \(W_{net}\) is positive. Anticlockwise the compression is along the upper curve, so \(W_{net}\) is negative.

The molecules of a real gas attract each other. As the gas spreads out the molecules move apart against this attraction, so their potential energy rises. With \(\Delta U = 0\) the kinetic energy must fall, and the temperature drops a little. An ideal gas has no such attraction.

Both start and end at \(300\) K, so \(\Delta U = 0\) in each.

Isothermal: \(W = nRT\ln 4 = 8.3 \times 300 \times 1.386 = 3452\) J, and \(Q = W = 3452\) J. Free: \(W = 0\) and \(Q = 0\). The same change of state, but the isothermal path turns heat into work and the free expansion does nothing useful.

The gas spreads into the vacuum by itself, but it never gathers back into one half of the vessel by itself. To restore the initial state the surroundings must do work on the gas and take heat from it. This leaves a permanent change in the surroundings. A process that cannot be undone without such a change is irreversible.