Basic Concepts: Systems, Properties, Processes, Work and Heat, and Temperature

This chapter answers Section XE4.1 “Basic Concepts” of XE4 Thermodynamics, one of the nine optional sections of the Engineering Sciences (XE) paper, any two of which a candidate chooses in the exam: “Continuum, microscopic and macroscopic approaches; Thermodynamic systems (closed and open); Thermodynamic properties, state and equilibrium; State postulate for simple compressible substances, paths and processes on property diagrams; Concepts of heat and work, different modes of work; Zeroth law of thermodynamics, concept of temperature.” It teaches the vocabulary on which every later thermodynamic calculation rests: systems and boundaries, intensive and extensive properties, equilibrium and the state postulate, quasi-static processes and cycles, heat and work with one fixed sign convention, boundary work for constant-pressure, isothermal and polytropic paths, the other modes of work, and temperature scales with the gas thermometer. The numericals are boundary work from p–V data, work in a piston–spring system, shaft and electrical work, surface-tension work, scale conversions and gas-thermometer readings.

1. Approaches, the continuum and thermodynamic systems

Thermodynamics describes matter by a few measurable bulk quantities such as p, V and T. This macroscopic (classical) approach needs no model of molecules and treats the system as a whole. The microscopic (statistical) approach follows the molecules, their positions and velocities, and obtains p and T as averages: pressure is the mean momentum transfer of molecular impacts, temperature a measure of mean translational kinetic energy. The two agree, and classical thermodynamics is used because it needs no molecular detail.

The macroscopic view is valid under the continuum assumption: matter is treated as continuous, so density and other properties are smooth functions of position. This holds when the characteristic length L of the problem is much larger than the molecular mean free path λ, that is when the Knudsen number Kn = λ/L is small. Density is defined as lim δm/δV with δV shrunk only to a volume still containing very many molecules, not to zero. For air at atmospheric pressure λ is about 0.07 μm, so a pipe or a cylinder is a continuum; a rarefied gas at very low pressure, where λ is comparable with the vessel, is not.

A system is the quantity of matter or region chosen for study; everything else is the surroundings, and the boundary (real or imaginary, fixed or moving) separates them.

SystemMass crosses boundaryHeat and work cross boundaryTypical example
Closed (control mass)NoYesGas sealed in a piston–cylinder
Open (control volume)YesYesTurbine, nozzle, compressor, a tank being filled
IsolatedNoNoIdeal sealed, insulated, rigid flask
⚠️ Closed does not mean constant volume
“Closed” refers to mass only: the gas in a piston–cylinder is a closed system although its boundary moves. An open system may carry a steady flow, with mass entering and leaving at equal rates, so that the mass inside is constant even though it is not a closed system.

2. Properties, state, equilibrium and the state postulate

A property is a measurable characteristic that has one value at each state, whatever the history. Intensive properties (p, T, density ρ, specific volume v) do not depend on the mass; extensive properties (m, V, U, H, S) are proportional to it. An extensive property divided by mass is a specific property and is intensive: v = V/m, u = U/m, h = H/m. Test: split the system by an imaginary partition; an intensive property is unchanged in each half, an extensive one halves. Because a property depends only on the state, its change depends only on the end states and ∮dZ = 0 over a cycle (a point function with an exact differential).

The state is the condition fixed by the values of the properties. It is well defined only at thermodynamic equilibrium, which needs all of: thermal equilibrium (no temperature difference), mechanical equilibrium (no unbalanced force, so uniform pressure apart from the weight of the fluid), chemical equilibrium (no reaction) and phase equilibrium (no net transfer between phases).

State postulate. The number of independent intensive properties needed to fix the state equals one (for heat transfer) plus the number of relevant reversible work modes. A simple compressible substance is one whose only reversible work mode is compression or expansion, p dV, with electrical, magnetic, surface-tension and gravitational effects absent, so two independent intensive properties fix its state, for example (T, v) or (p, v). All others follow from the equation of state f(p, v, T) = 0, which is a surface in p–v–T space.

⚠️ Two properties must be independent
In a single phase, T and p are independent. In a liquid–vapour mixture at saturation they are not, since p = p_sat(T); then T and p fix only the saturation condition and a second independent property, such as the quality x or v, is needed to fix the state.

3. Paths, processes and property diagrams

A process is a change of state and its path is the series of states passed through. In a quasi-static (quasi-equilibrium) process the system departs only infinitesimally from equilibrium at every instant, so each intermediate state has well-defined properties. This is achieved when the process is slow compared with the relaxation time, the time the system needs to restore equilibrium; gas pressure equalises at about the speed of sound, so a piston moving at a few m/s is effectively quasi-static. Only quasi-static processes can be drawn as continuous curves on a property diagram such as p–v. A free expansion, a rapid compression or a stirring process passes through non-equilibrium states and only its end points can be marked.

Quasi-static is not the same as reversible: a slow process with friction is quasi-static but irreversible (the full treatment is in the Second Law section). A cycle returns the system to its initial state, so every property change over the cycle is zero. A process is steady when properties at a point do not change with time, and uniform when they do not vary with position. Named processes are listed below.

ProcessConstraintPolytropic pVⁿ = C
Isobaricp constantn = 0
IsothermalT constant (ideal gas: pV constant)n = 1
Reversible adiabatic, ideal gasQ = 0 and no dissipationn = γ
IsochoricV constantn → ∞

4. Heat, work, sign convention and boundary work

Heat Q is energy crossing the boundary because of a temperature difference. Work W is energy crossing the boundary in any other way, such that the sole effect on the surroundings could be the raising of a weight. Both are transient boundary phenomena, not properties: a system has energy, never “heat” or “work”. They are path functions with inexact differentials, written δQ and δW, and their values depend on the process, not only on the end states. A process with Q = 0 is adiabatic; a gas heated by compression has received work, not heat.

ℹ️ Sign convention used throughout this chapter set
Q is positive when heat flows INTO the system. W is positive when the system does work ON the surroundings (expansion, shaft output); work done on the system (compression, stirring, electrical input) is negative. The first law is then Q − W = ΔE. Some texts take W as work done on the system and write Q + W = ΔE; always read the stem for the convention.

For a quasi-static expansion the boundary work is δW = p dV and W₁₂ = ∫p dV, the area under the curve on a p–V diagram: positive for expansion, negative for compression, and dependent on the path. The net work of a cycle is the enclosed area, positive if the cycle runs clockwise. Standard results for the work done BY the system:

  • Isobaric: W = p(V₂ − V₁); isochoric: W = 0.
  • Isothermal ideal gas (pV = C): W = p₁V₁ ln(V₂/V₁) = mRT ln(V₂/V₁).
  • Polytropic pVⁿ = C, n ≠ 1: W = ∫C V⁻ⁿ dV = (p₁V₁ − p₂V₂)/(n − 1), with p₂ = p₁(V₁/V₂)ⁿ.

Worked example. A gas at 500 kPa and 0.1 m³ expands to 0.3 m³. Polytropic with n = 1.2: p₂ = 500 × (1/3)^1.2 = 133.8 kPa, so W = (500 × 0.1 − 133.8 × 0.3)/0.2 = (50 − 40.14)/0.2 = 49.3 kJ. Isothermal (n = 1): W = 500 × 0.1 × ln 3 = 54.9 kJ. Isobaric at 500 kPa: W = 500 × 0.2 = 100 kJ. On the same expansion the work falls as n rises, because the p–V curve falls more steeply and the area under it shrinks.

⚠️ When ∫p dV does not apply
W = ∫p dV needs a quasi-static process with p the system pressure. In a free expansion into vacuum the volume rises but W = 0, since there is no resisting force. In a rigid tank ∫p dV = 0, yet a paddle can still do work on the gas: stirring is a non-boundary work mode.

5. Other modes of work

ModeExpressionRemark
Shaft / stirringW = Tω = 2πNT/60 (N in rpm, T in N·m, result in W)Work on the system is negative in our convention
ElectricalW = VIt (energy input)Supply to a resistor inside the system: work on the system
Spring½k(x₂² − x₁²) stored in the springForce F = kx, k in N/m
Surface tensionδW = −σ dA by the systemA soap film has two surfaces: A = 2 × film area
Stretched wireδW = −F dL by the systemExtension needs work input

Stirring and electrical heating of a closed system are always work into the system and can never be reversed to give work out, which is why they matter in the second law. A film of surface tension σ resists growth, so enlarging its area needs work input σ ΔA, which is negative as work by the system.

Worked example (piston with a linear spring). Gas in a cylinder pushes a piston that bears on a linear spring, so the gas pressure rises linearly with volume: from 200 kPa at 0.04 m³ to 400 kPa at 0.10 m³. The p–V line is straight, so the area is a trapezium: W = ½(200 + 400) × (0.10 − 0.04) = 18 kJ done by the gas. Of this, 200 × 0.06 = 12 kJ is work against the constant 200 kPa and the triangle ½ × 200 × 0.06 = 6 kJ is stored in the spring, taken as unstretched at the start.

6. Zeroth law and the concept of temperature

Zeroth law: if two bodies are each in thermal equilibrium with a third body, they are in thermal equilibrium with each other. It makes thermometry meaningful: the third body is the thermometer, and temperature is the property that has the same value for all bodies in thermal equilibrium. A thermometer uses a thermometric property that varies with temperature: the length of a liquid column, the resistance of a metal, the emf of a thermocouple, or the pressure of a gas at constant volume.

Scales. Celsius and Fahrenheit are relative scales, Kelvin and Rankine are absolute. T(K) = t(°C) + 273.15; t(°F) = 1.8 t(°C) + 32; T(R) = 1.8 T(K) = t(°F) + 459.67. A temperature difference of 1 K equals 1 °C, 1.8 °F and 1.8 R, so a rise of 50 °C is a rise of 90 °F, not 122. The classical absolute scale assigns 273.16 K to the triple point of water; since 2019 the kelvin is defined by fixing the Boltzmann constant, k = 1.380649 × 10⁻²³ J/K, and the triple point stays 273.16 K within measurement uncertainty.

Constant-volume gas thermometer. A bulb of gas at fixed volume is connected to a manometer. Its pressure p is the thermometric property, and T = 273.16 × (p/p_tp) in the limit p_tp → 0, where p_tp is the pressure when the bulb is at the triple point of water. At finite gas density different gases read slightly different values; extrapolating the reading to zero density removes the difference, giving the ideal-gas temperature scale. Example: p_tp = 80.0 kPa and p = 99.6 kPa at the unknown temperature give T = 273.16 × 99.6/80.0 = 340.1 K.

⚠️ Differences versus readings
Convert a temperature reading with the offset (+273.15, or × 1.8 + 32) but a temperature difference only with the factor (× 1.8). Properties such as p V = mRT and efficiencies need absolute temperature; never insert °C into a gas law.

Key takeaways

  • A closed system exchanges energy but not mass, an open system (control volume) exchanges both, an isolated system neither. The continuum model holds when the mean free path is much smaller than the length scale (small Knudsen number).
  • Intensive properties do not scale with mass (p, T, v), extensive ones do (m, V, U, H); specific values are intensive. A property is a point function, ∮dZ = 0; heat and work are path functions.
  • State postulate: independent intensive properties = 1 + number of reversible work modes, so a simple compressible substance needs two. In a saturated mixture p and T are not independent; add the quality.
  • Only quasi-static processes are curves on property diagrams; free expansion and stirring show end states only. Quasi-static is not reversible: friction can be present.
  • Sign convention: Q positive into the system, W positive by the system, so Q − W = ΔE. Boundary work W = ∫p dV is the area under the p–V curve; polytropic W = (p₁V₁ − p₂V₂)/(n − 1), isothermal W = p₁V₁ ln(V₂/V₁).
  • Shaft work 2πNT/60, electrical work VIt, spring work ½k(x₂² − x₁²), surface-tension work σ ΔA (two surfaces for a film). The zeroth law justifies the thermometer; T(K) = t(°C) + 273.15, differences scale by 1.8 for °F; a gas thermometer gives T = 273.16 p/p_tp.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. Water flows steadily through a heat exchanger. For analysis, a fixed region of space enclosing the heat exchanger between its inlet and outlet pipe sections is chosen. How is this system best classified?

    1. Closed system, because the volume of the chosen region does not change
    2. Open system (control volume), because mass crosses the boundary of the chosen region
    3. Isolated system, because the flow is steady and properties at each point do not change with time
    4. Closed system, because the mass contained inside the region stays constant in steady flow
    Show answer

    Answer: B — Open system (control volume), because mass crosses the boundary of the chosen region

    A system is open if mass crosses its boundary, whatever happens to the mass inside. Water enters and leaves the region, so it is a control volume. The tempting reasoning is that steady flow keeps the contained mass constant; that is true, but constancy of the contained mass is not the same as no mass crossing the boundary. A fixed volume does not make a system closed, and an isolated system exchanges neither mass nor energy, whereas here heat is exchanged with the second stream.

  2. A closed system of gas executes a cycle: it expands, and is then compressed back to its initial state. Which statement about the cycle is correct?

    1. The net boundary work is zero because the final state coincides with the initial state
    2. The net heat transfer is zero because the internal energy returns to its initial value
    3. The cyclic integral of δW is zero because work is a property fixed by the state
    4. The cyclic integrals of dp, dV and dT vanish, but that of δW need not vanish
    Show answer

    Answer: D — The cyclic integrals of dp, dV and dT vanish, but that of δW need not vanish

    Properties return to their initial values, so ∮dp = ∮dV = ∮dT = 0. Work is a path function, and the net work of a cycle equals the area enclosed on the p–V diagram, which is non-zero unless the expansion and compression curves coincide. The net heat is also non-zero in general: by the first law it equals the net work. Saying the net work is zero because the end state is the same confuses a path function with a property.

  3. A gas in a uniform equilibrium state is divided into two equal halves by an imaginary partition. Which properties of one half have the same value as the corresponding property of the whole gas?

    1. Pressure
    2. Internal energy
    3. Specific volume
    4. Volume
    Show answer

    Answer: A — Pressure; C — Specific volume

    Pressure and specific volume are intensive properties, so they do not depend on the amount of gas and are unchanged in each half. Internal energy and volume are extensive: each is proportional to mass, so each half has half the value of the whole.

  4. Take the state postulate as: the number of independent intensive properties that fix the equilibrium state equals 1 plus the number of relevant reversible work modes. A pure substance in a vessel can exchange reversible compression (p dV) work and also reversible surface-tension work with its surroundings, and no other work. How many independent intensive properties fix its state?

    1. 1
    2. 2
    3. 3
    4. 4
    Show answer

    Answer: C — 3

    There are two reversible work modes (compression and surface tension), so the count is 1 + 2 = 3. A simple compressible substance has only one work mode and needs 2. The tempting answer 2 forgets the surface-tension mode; the extra mode adds one more independent property, such as the surface area per unit mass.

  5. For which of the following is the continuum assumption least justified?

    1. Helium at a pressure so low that its mean free path is comparable with the vessel size
    2. Air at atmospheric pressure and room temperature flowing through a pipe of 5 cm diameter
    3. Water standing in a laboratory beaker a few centimetres across, at room temperature
    4. Steam at several megapascals passing through turbine passages a few centimetres wide
    Show answer

    Answer: A — Helium at a pressure so low that its mean free path is comparable with the vessel size

    The continuum model needs a small Knudsen number, Kn = λ/L, the mean free path over the characteristic length. In the low-pressure vessel λ is comparable with L, so density and pressure cannot be defined as smooth point values. For air at atmospheric pressure λ is about 0.07 μm, thousands of times smaller than a 5 cm pipe, and in liquids and dense steam it is smaller still, so the other situations are all safely continuum.

  6. A rigid, insulated vessel is divided by a thin diaphragm: one side holds an ideal gas and the other side is evacuated. The diaphragm is then ruptured and the gas fills the vessel. Which statement is correct?

    1. The process is a continuous curve on the p–V diagram, since the initial and final states are both equilibrium states
    2. Only the end states can be marked on the p–V diagram, and the gas does no work since nothing resists its expansion
    3. The work done by the gas is p ΔV, because the volume available to the gas has increased
    4. The gas does work equal to its fall in internal energy, so its temperature must fall as it expands
    Show answer

    Answer: B — Only the end states can be marked on the p–V diagram, and the gas does no work since nothing resists its expansion

    Rupture is a non-quasi-static process: the gas passes through non-equilibrium states, so only the end states can be plotted. The gas pushes against no moving boundary and no resisting pressure, and the vessel is rigid, so W = 0; with Q = 0 as well, the internal energy is unchanged and for an ideal gas the temperature stays the same. W = p ΔV is the tempting error: it needs a quasi-static expansion against the system pressure.

  7. A gas executes the cycle 1→2→3→4→1 on a p–V diagram. Process 1→2: isobaric expansion at 300 kPa from 0.1 m³ to 0.3 m³. Process 2→3: isochoric fall of pressure to 100 kPa. Process 3→4: isobaric compression at 100 kPa from 0.3 m³ to 0.1 m³. Process 4→1: isochoric rise of pressure to 300 kPa. Taking work done by the system as positive, the net work of the cycle is

    1. −40 kJ
    2. +60 kJ
    3. +40 kJ
    4. +80 kJ
    Show answer

    Answer: C — +40 kJ

    Work occurs only on the isobaric legs. 1→2: 300 × (0.3 − 0.1) = +60 kJ. 3→4: 100 × (0.1 − 0.3) = −20 kJ. The isochoric legs give zero. Net = 60 − 20 = +40 kJ, the enclosed rectangle (300 − 100) × 0.2. The sign is positive because the expansion occurs at the higher pressure, a clockwise cycle. The value +60 kJ counts the expansion only, and +80 kJ adds the magnitudes instead of subtracting.

  8. The same ideal gas expands quasi-statically from the same initial state to twice its initial volume along three paths: an isothermal path (n = 1), a polytropic path pV^1.3 = constant, and a reversible adiabatic path pV^1.4 = constant. Which gives the correct decreasing order of the boundary work done by the gas?

    1. Isothermal, then polytropic (n = 1.3), then adiabatic
    2. Adiabatic, then polytropic (n = 1.3), then isothermal
    3. Polytropic (n = 1.3), then isothermal, then adiabatic
    4. All three equal, because the initial and final volumes are the same
    Show answer

    Answer: A — Isothermal, then polytropic (n = 1.3), then adiabatic

    Work is the area under the p–V curve. For pVⁿ = C the pressure falls faster with volume as n increases, so the area shrinks. With W₁₂ in units of p₁V₁ for a doubling of volume: isothermal ln 2 = 0.693, n = 1.3 gives (1 − 2^(−0.3))/0.3 = 0.626, n = 1.4 gives (1 − 2^(−0.4))/0.4 = 0.605. The end volumes alone do not fix the work, because work is a path function.

  9. Use the convention that Q is positive for heat flowing into the system and W is positive for work done by the system on its surroundings. A gas in a piston–cylinder is compressed, with 200 kJ of work done on the gas, while 50 kJ of heat flows from the gas to the surroundings. The values of Q and W for the gas are

    1. Q = +50 kJ and W = +200 kJ
    2. Q = +50 kJ and W = −200 kJ
    3. Q = −50 kJ and W = +200 kJ
    4. Q = −50 kJ and W = −200 kJ
    Show answer

    Answer: D — Q = −50 kJ and W = −200 kJ

    Heat leaves the gas, so Q = −50 kJ. Work is done on the gas, so by the stated convention W = −200 kJ. The internal energy then changes by Q − W = −50 − (−200) = +150 kJ, which is why a compressed gas warms up even while it loses heat. Pairing a negative Q with a positive W, or the reverse, applies the convention to the wrong quantity.

  10. A soap film has two free surfaces, each with surface tension σ = 0.03 N/m. The film is stretched slowly so that its area (one face) increases by 40 cm². The minimum work that must be supplied to the film is

    1. 0.12 mJ
    2. 0.24 mJ
    3. 2.4 mJ
    4. 1.2 mJ
    Show answer

    Answer: B — 0.24 mJ

    The work input is σ ΔA_total with both surfaces counted: 2 × 0.03 N/m × 40 × 10⁻⁴ m² = 2.4 × 10⁻⁴ J = 0.24 mJ. Using a single surface gives 0.12 mJ, the usual slip. The value 2.4 mJ comes from slipping a factor of 10 in the unit conversion (cm² to m² is a factor 10⁻⁴), and 1.2 mJ is the same slip applied to a single surface.

  11. A gas expands quasi-statically from p₁ = 800 kPa and V₁ = 0.05 m³ to V₂ = 0.20 m³ along a polytropic path pV^1.25 = constant. The boundary work done by the gas, in kJ, to one decimal place, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 46.9

    p₂ = p₁ (V₁/V₂)^1.25 = 800 × (0.25)^1.25 = 800 × 0.17678 = 141.42 kPa. W = (p₁V₁ − p₂V₂)/(n − 1) = (800 × 0.05 − 141.42 × 0.20)/0.25 = (40 − 28.28)/0.25 = 46.86 kJ ≈ 46.9 kJ. Using n = 1 (isothermal) would give 55.5 kJ and using the final pressure only would be wrong.

  12. A gas in a cylinder initially at 100 kPa and 0.02 m³ pushes a frictionless, massless piston of area 0.1 m². The atmosphere at 100 kPa acts on the outer face of the piston, which is also attached to a linear spring of stiffness 40 kN/m that is unstretched in the initial state. The gas is heated until its volume doubles. The work done by the gas, in kJ, to one decimal place, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 2.8

    Piston displacement x = ΔV/A = 0.02/0.1 = 0.2 m. Final gas pressure p₂ = p_atm + kx/A = 100 + 40 × 0.2/0.1 = 180 kPa. The pressure varies linearly with volume, so W = ½(100 + 180) × 0.02 = 2.8 kJ. Check: 100 × 0.02 = 2.0 kJ against the atmosphere plus ½kx² = ½ × 40 × 0.2² = 0.8 kJ stored in the spring. Counting only the atmospheric part (2.0 kJ) forgets the spring.

  13. A rigid, well-insulated tank contains a gas. A paddle wheel driven by a motor delivers a torque of 20 N·m at 300 rpm for 10 minutes, and an electrical resistance heater inside the tank draws 5 A at 12 V for the same 10 minutes. The total work input to the gas, in kJ, to the nearest whole number, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 413

    Shaft power = 2πNT/60 = 2π × 300 × 20/60 = 628.3 W. Over 600 s: 628.3 × 600 = 376.99 kJ. Electrical power = VI = 12 × 5 = 60 W, over 600 s = 36 kJ. Total = 376.99 + 36 = 413.0 kJ. Both are work done on the gas (W negative as work by the system), and ∫p dV = 0 for the rigid tank, so none of this is boundary work.

  14. Use the convention that W is positive for work done by the system. Which of the following statements are correct?

    1. The boundary work of a quasi-static process between two given end states depends on the path followed.
    2. A gas that warms up when rapidly compressed in a perfectly insulated cylinder has received heat.
    3. The net work of a cycle run clockwise on a p–V diagram equals the enclosed area and is positive.
    4. Stirring a fluid in a rigid insulated tank is work done on the fluid although ∫p dV = 0.
    Show answer

    Answer: A — The boundary work of a quasi-static process between two given end states depends on the path followed.; C — The net work of a cycle run clockwise on a p–V diagram equals the enclosed area and is positive.; D — Stirring a fluid in a rigid insulated tank is work done on the fluid although ∫p dV = 0.

    Work is a path function, so the boundary work depends on the path; the cyclic area rule gives a positive net work for a clockwise cycle; and stirring is a non-boundary work mode that enters the system even with no volume change. The compressed gas warms because work, not heat, is supplied: the insulation means Q = 0, and heat requires a temperature difference across the boundary.

  15. Which statement is the zeroth law of thermodynamics, and what does it justify?

    1. Energy can be neither created nor destroyed, which justifies the energy balance of a closed system
    2. Heat flows unaided only from a hotter to a colder body, which justifies the analysis of reversible engines
    3. Bodies each in thermal equilibrium with a third are in equilibrium with one another, which justifies the thermometer
    4. The entropy of a perfect crystal tends to zero at absolute zero, which justifies defining an absolute temperature scale
    Show answer

    Answer: C — Bodies each in thermal equilibrium with a third are in equilibrium with one another, which justifies the thermometer

    The zeroth law is transitivity of thermal equilibrium: the thermometer is the third body, and equal readings mean thermal equilibrium without bringing the two bodies together. The energy-conservation statement is the first law, the direction-of-heat-flow statement belongs to the second law, and the zero-entropy statement is the third law.

  16. Which of the following statements about temperature are correct?

    1. A temperature rise of 40 °C is a rise of 72 °F.
    2. A gas at 27 °C has an absolute temperature of 300.15 K.
    3. In pV = mRT the temperature may be inserted in °C, provided the pressure is absolute.
    4. Different gases in a constant-volume thermometer read exactly the same temperature at any finite gas density.
    Show answer

    Answer: A — A temperature rise of 40 °C is a rise of 72 °F.; B — A gas at 27 °C has an absolute temperature of 300.15 K.

    A temperature difference converts by the factor only: 40 × 1.8 = 72 °F. A reading converts with the offset: 27 + 273.15 = 300.15 K. The gas law needs absolute temperature, so °C gives a wrong answer. Different gases agree only in the limit of zero density, which is why the reading is extrapolated to zero gas pressure.

  17. A constant-volume gas thermometer is used in the low-density limit, where the ideal-gas relation applies. Its gas pressure is 50.0 kPa when the bulb is at the triple point of water (273.16 K) and 68.2 kPa when the bulb is in contact with a hot body. The temperature of the hot body, in K, to one decimal place, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 372.6

    At constant volume, T ∝ p, so T = 273.16 × (p/p_tp) = 273.16 × 68.2/50.0 = 273.16 × 1.364 = 372.59 K ≈ 372.6 K. The result is an absolute temperature already; no further 273.15 is added.

  18. The Fahrenheit and Celsius readings are related by t_F = 1.8 t_C + 32. At what temperature, in °C, is the Fahrenheit reading numerically exactly twice the Celsius reading? The answer in °C is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 160

    Set t_F = 2 t_C: 2 t_C = 1.8 t_C + 32, so 0.2 t_C = 32 and t_C = 160 °C. Check: 1.8 × 160 + 32 = 320 °F, which is twice 160.