Physics / Pressure Coefficient of a Gas
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EXPERIMENT 33 · PHYSICS LABORATORY

Pressure Coefficient of a Gas theory

Understand pressure coefficient of a gas: the equations, assumptions, supporting concepts and measurement method.

THE WORKING RELATIONSHIP
P_t = P_0 (1 + α_v t) = P_0 + (P_0 α_v) · t ⇔ Y = m X + b ⇔ α_v = m / P_0 = m / b

Keep the physical quantities, graph variables and units consistent.

01Kinetic Theory of Gases and the Origin of Thermal Gas Pressure

According to the classical kinetic theory of gases formulated by Clausius, Maxwell, and Boltzmann, a macroscopic gas consists of an enormous ensemble of point-like molecules of mass m moving with randomly oriented velocities in continuous chaotic motion. The absolute pressure P exerted by the gas against the interior walls of its container represents the macroscopic time-averaged rate of normal momentum transfer imparted by molecular collisions per unit surface area: P = (1/3) n m <v²> = (2/3) n <E_k>, where n = N/V is the molecular number density and <E_k> = (1/2) m <v²> is the mean translational kinetic energy of a molecule. In classical statistical thermodynamics, temperature is fundamentally defined through the equipartition theorem as a measure of this average translational kinetic energy: <E_k> = (3/2) k_B T, where k_B = 1.380649 × 10⁻²³ J K⁻¹ is the Boltzmann constant and T is the absolute thermodynamic temperature in kelvins. Substituting this kinetic definition into the pressure equation yields P = n k_B T = (N/V) k_B T, which is the microscopic statement of the ideal gas law. When the gas is confined to a rigid container of invariant volume (V = constant), the number density n remains strictly fixed. Consequently, raising the temperature T increases the root-mean-square molecular velocity according to v_rms = √(3 k_B T / m). This elevated velocity has a twofold effect on wall collisions: first, each individual collision transfers greater momentum (Δp = 2 m v_x); second, the frequency of collisions per second against each unit area of the wall increases proportionally to v_rms. Because both the momentum transfer per impact and the impact frequency scale with molecular speed, the macroscopic pressure scales directly with <v²> and therefore directly with absolute temperature T. This kinetic derivation provides the theoretical foundation for Charles’ law at constant volume: pressure is directly proportional to thermodynamic temperature because temperature is the direct macroscopic reflection of molecular kinetic energy.

P = \frac{1}{3} n m \langle v^2 \rangle = \frac{2}{3} n \langle E_k \rangle = n k_B T = \frac{N k_B T}{V} \implies \left(\frac{\partial P}{\partial T}\right)_V = \frac{N k_B}{V} = \text{constant}
How does the microscopic kinetic theory explain why gas pressure increases when a sealed rigid bulb is heated?

Heating the gas increases the mean translational kinetic energy of the molecules (<E_k> = (3/2) k_B T), which raises their average speed. Because the volume is constant, faster molecules collide with the container walls more frequently and deliver greater momentum per collision, resulting in a higher time-averaged normal force per unit area, which macroscopically constitutes increased pressure.

02Charles’ Law at Constant Volume and Gay-Lussac’s Isochoric Relation

Charles’ law at constant volume (often historically designated Gay-Lussac’s second law or Amontons’ law) states that for an ideal gas held at constant volume, the pressure P varies linearly with temperature. When referenced to the Celsius temperature scale where 0 °C corresponds to the freezing point of water, the mathematical relationship is written as P_t = P_0 (1 + α_v t), where P_0 is the absolute pressure at 0.00 °C, t is the Celsius temperature, and α_v is the thermal pressure coefficient at constant volume. The coefficient α_v is formally defined as the fractional rate of pressure increase per degree Celsius relative to the ice-point pressure: α_v = (1 / P_0) (dP / dt)_V. For an ideal gas adhering to the equation of state PV = nRT, taking the partial derivative of pressure with respect to Celsius temperature t (where T = t + T_0 and T_0 is the absolute temperature of the ice point) yields (dP / dt)_V = nR / V. At t = 0 °C, the ice-point pressure is P_0 = nR T_0 / V. Dividing the derivative by P_0 gives α_v = (nR / V) / (nR T_0 / V) = 1 / T_0. Because the ice point on the thermodynamic Kelvin scale is defined as T_0 = 273.15 K, the theoretical thermal pressure coefficient for an ideal gas is α_v = 1 / 273.15 K⁻¹ ≈ 0.00366099 K⁻¹ (or approximately 3.661 × 10⁻³ K⁻¹). Remarkably, this pressure coefficient α_v is identically equal to the thermal volume expansion coefficient β obtained under isobaric conditions (β = (1 / V_0) (dV / dt)_P = 1 / T_0). The identity α_v = β holds strictly for any substance that obeys the ideal gas equation of state, reflecting the underlying symmetry of the thermodynamic surface P(V, T). In an experimental plot of P against t, the data form a straight line with slope m = P_0 α_v and vertical intercept b = P_0. From these two regression parameters, the experimental pressure coefficient is computed directly as α_v = m / b, completely independent of any external temperature calibration beyond the Celsius scale definition.

P_t = P_0 (1 + \alpha_v t) \iff \alpha_v = \frac{1}{P_0} \left(\frac{\partial P}{\partial t}\right)_V = \frac{m}{b} = \frac{1}{T_0} \approx \frac{1}{273.15\text{ K}} \approx 0.003661\text{ K}^{-1}
Why is the thermal pressure coefficient α_v at constant volume numerically equal to the thermal volume expansion coefficient β at constant pressure for an ideal gas?

Both coefficients represent the reciprocal of the absolute ice-point temperature (1 / T_0 = 1 / 273.15 K). From the ideal gas law PV = nRT, (∂P/∂T)_V = nR/V and P_0 = nRT_0/V, so α_v = 1/T_0. Similarly, (∂V/∂T)_P = nR/P and V_0 = nRT_0/P, giving β = 1/T_0. Therefore, α_v = β = 1/T_0.

03The Jolly Constant-Volume Air Thermometer: Design and Manometry

The constant-volume air thermometer, perfected by Philipp von Jolly in 1874, remains the premier primary instrument for thermometric metrology across the cryogenic and moderate temperature domains. The core apparatus comprises a spherical borosilicate glass bulb (capacity ~100 cm³) submerged in a stirred heating bath and fused to an unheated glass capillary tube of narrow capillary bore (diameter ~0.5 mm). The capillary bends vertically downward into a closed manometer tube that terminates in a pointed glass or ivory index pin (the fiducial pointer). This closed limb is joined via heavy-duty vulcanized rubber or flexible fluoropolymer tubing to an open manometer limb mounted on a vertical sliding carriage controlled by a rack-and-pinion screw drive. Liquid mercury fills the flexible tube and both manometer limbs. When the air inside the bulb is heated, its expansion tends to push the mercury column downward. To maintain a strictly constant volume, the experimenter raises the open limb until the apex of the convex mercury meniscus in the closed capillary exactly touches the sharp tip of the fiducial index pin. Because the volume enclosed between the bulb and the pointer tip is geometrically invariant, the condition of constant volume is rigorously fulfilled. The total absolute pressure P of the enclosed gas is then obtained from the fundamental hydrostatic balance equation: P = B + ρ_Hg g (h_open − h_closed) = B + Δh, where B is the prevailing barometric pressure of the surrounding atmosphere and Δh = h_open − h_closed is the vertical height difference between the mercury menisci in the open and closed tubes. If the open tube meniscus stands higher than the closed tube meniscus, Δh is positive (gauge overpressure); if lower (as occurs below 0 °C), Δh is negative. By referencing all readings to the sharp index pointer, the Jolly air thermometer eliminates errors associated with scale calibration variations along the closed tube and ensures that the volume of the gas remains constant within less than 0.01% across the entire experimental range.

P = B + \rho_{\text{Hg}} g (h_{\text{open}} - h_{\text{closed}}) = B + \Delta h_{\text{mmHg}} = B + \frac{\Delta h}{7.5006} \text{ (in kPa)}
Why is it critical to adjust the mercury meniscus to touch the fiducial pointer tip before taking every pressure reading?

The fiducial pointer defines the boundary of the constant volume V_0. If the meniscus were allowed to move away from the pointer, the gas volume would vary with temperature, violating the isochoric condition (V = const) and invalidating the simple linear relationship P = P_0(1 + α_v t).

04The Concept and Extrapolation of Absolute Zero of Temperature

One of the most profound conceptual triumphs arising from the study of the gas laws is the discovery and thermodynamic definition of absolute zero. In the linear relationship P_t = P_0 (1 + α_v t), as temperature decreases below 0 °C, the pressure exerted by the gas progressively diminishes. Setting the absolute pressure to zero (P_t = 0) defines a theoretical lower limit of temperature below which the pressure would become unphysically negative: 0 = P_0 (1 + α_v T_0) ⟹ T_0 = −1 / α_v. Using the experimental value of the pressure coefficient α_v ≈ 0.003661 K⁻¹, the extrapolated temperature where gas pressure vanishes is T_0 = −1 / 0.003661 °C ≈ −273.15 °C. Lord Kelvin (William Thomson) recognized in 1848 that this zero-pressure intercept is not an accidental property of dry air, but a universal thermodynamic property common to all dilute gases. Kelvin utilized Carnot’s reversible heat engine cycle to establish the absolute thermodynamic temperature scale, defining the absolute Kelvin temperature T by the linear transformation T = t + 273.15. In terms of absolute temperature, Charles’ law at constant volume simplifies to the direct proportionality P = (P_0 / T_0) T = C T, or P_1 / T_1 = P_2 / T_2. Microscopic quantum mechanics reveals that even at absolute zero, molecules do not come to a complete halt; instead, they retain irreducible quantum zero-point vibrational energy E_0 = (1/2) ħ ω. However, at absolute zero, all thermal kinetic excitations cease, the entropy of a pure crystalline substance approaches zero in accordance with the Third Law of Thermodynamics (Nernst Heat Theorem), and a classical ideal gas would exert exactly zero pressure because it possesses zero thermal momentum to transfer to the containment walls. Extrapolating the linear regression line P(t) to its horizontal intercept where P = 0 provides a direct experimental determination of absolute zero on the Celsius scale: T_0 = −b / m.

P_t = 0 \implies 1 + \alpha_v T_0 = 0 \implies T_0 = -\frac{1}{\alpha_v} = -\frac{b}{m} \approx -273.15\,^\circ\text{C} \implies T = t + 273.15\text{ K}
Does the extrapolation of Charles’ law to absolute zero imply that real gas molecules stop moving and have zero volume at −273.15 °C?

No. Real gases liquefy and solidify long before reaching absolute zero (e.g., air liquefies near −194 °C). Furthermore, quantum mechanics dictates that atoms retain non-zero quantum zero-point energy at 0 K. Absolute zero represents the state where all thermal kinetic energy is removed, not the annihilation of atomic volume or quantum motion.

05Systematic Corrections: Capillary Dead-Space and Bulb Thermal Dilatation

In high-precision gas thermometry, two systematic physical phenomena cause the observed pressure-temperature slope to deviate slightly from the ideal behavior of a rigid vessel: the capillary dead-space effect and the cubical thermal expansion of the glass bulb envelope. First, the gas contained in the narrow capillary stem between the liquid bath surface and the index pointer does not reside at the bath temperature t, but at an intermediate room temperature t_r. Let V_0 be the bulb volume at bath temperature t and v be the unheated capillary dead-space volume at ambient room temperature t_r. The total number of gas moles is divided into two thermal zones: n = n_bulb + n_cap = (P V_0) / [R (T_0 + t)] + (P v) / [R (T_0 + t_r)]. Because the capillary gas is cooler than the bath, its density is higher, drawing a small fraction of the gas mass out of the hot bulb. The true corrected pressure P_corr that would be exerted if the entire volume V_0 + v were maintained at bath temperature t is given to first order by P_corr = P_obs [1 + (v / V_0) (t − t_r) / (T_0 + t_r)]. For a well-designed Jolly bulb where v / V_0 < 0.005, this dead-space correction increases the calculated pressure by approximately 0.2% to 0.5% at 80 °C. Second, glass is not a perfectly rigid material; upon heating from 0 °C to t, the borosilicate glass bulb expands cubically according to V_t = V_0 (1 + γ_g t), where γ_g = 3 α_g ≈ 9.9 × 10⁻⁶ K⁻¹ is the cubical thermal expansion coefficient of borosilicate glass. Because the containment volume actually increases slightly at elevated temperatures, the observed gas pressure is slightly lower than it would be in an ideally rigid vessel. Applying the bulb expansion correction gives P_corr = P_obs (1 + γ_g t). Combining both corrections yields the fully corrected thermal pressure coefficient: α_v_true = α_v_apparent + γ_g + (v / V_0) (1 / T_0) [1 − t_r / (T_0 + t_r)]. Neglecting glass expansion underestimates α_v by approximately 0.010 × 10⁻³ K⁻¹ (about 0.27%).

P_{\text{corr}} = P_{\text{obs}} \left[1 + \gamma_g t + \frac{v}{V_0} \left(\frac{t - t_r}{T_0 + t_r}\right)\right] \iff \alpha_{v,\text{true}} \approx \alpha_{v,\text{obs}} + \gamma_g
How does the thermal expansion of the glass bulb affect the experimentally determined value of the gas pressure coefficient?

Because the glass bulb expands when heated, the gas occupies a slightly larger volume at higher temperatures than at 0 °C. This volume increase reduces the pressure buildup, causing the observed pressure P to be lower than in a truly rigid vessel. Consequently, the apparent slope m and apparent α_v are underestimated by an amount equal to the cubical expansion coefficient of glass (γ_g ≈ 10 × 10⁻⁶ K⁻¹).

06Virial Equation of State, Real Gas Departures, and the Ideal Gas Scale

While Charles’ law assumes an ideal gas with non-interacting point particles, real gases such as atmospheric air (predominantly nitrogen and oxygen) deviate subtly from ideal behavior due to intermolecular forces and finite molecular excluded volume. In modern thermodynamics, real gas isotherms are described by the virial equation of state expanded in powers of molar density ρ = 1 / V_m: P V_m = R T [1 + B(T) ρ + C(T) ρ² + ...], where B(T) is the second virial coefficient representing pairwise molecular interactions and C(T) represents three-body interactions. At moderate pressures near 1 atmosphere (101.3 kPa), higher-order terms are negligible, and the equation of state reduces to P = (R T / V_m) [1 + B(T) / V_m]. Differentiating with respect to temperature at constant molar volume yields the isochoric pressure derivative: (dP / dt)_V = (R / V_m) [1 + (B(T) + T dB/dT) / V_m]. The experimental pressure coefficient for a real gas therefore depends slightly on density (and hence on the filling pressure P_0): α_v(P_0) = α_v_ideal + (P_0 / R) [dB/dT − B(T_0)/T_0]. For nitrogen and dry air at room temperature, the second virial coefficient B(T) is slightly negative (B ≈ −5 to −10 cm³ mol⁻¹) because attractive van der Waals dispersion forces dominate over repulsive hard-core exclusions. As temperature rises, thermal kinetic energy overcomes these intermolecular attractions, so dB/dT is positive. Consequently, the measured pressure coefficient α_v of dry air at 1 atmosphere (0.003673 K⁻¹) is slightly higher than the true thermodynamic ideal gas limit (0.003661 K⁻¹). To establish the absolute thermodynamic temperature scale independently of gas identity, metrologists measure α_v across several lower filling pressures P_0 (e.g., 0.75, 0.50, and 0.25 atm) and extrapolate linearly to zero initial pressure (P_0 → 0): lim_{P_0 → 0} α_v(P_0) = 1 / T_0 = 1 / 273.15 K⁻¹. This zero-pressure extrapolation defines the international gas thermometric temperature scale.

P V_m = R T \left[1 + \frac{B(T)}{V_m}\right] \implies \alpha_v(P_0) = \alpha_{v,\text{ideal}} + P_0 \cdot \phi(T) \implies \lim_{P_0 \to 0} \alpha_v(P_0) = \frac{1}{273.15\text{ K}}
Why is the pressure coefficient α_v of dry air at 1 atmosphere (~0.003673 K⁻¹) slightly larger than the ideal gas value (0.003661 K⁻¹)?

At atmospheric pressure and near 0 °C, attractive intermolecular van der Waals forces between air molecules slightly diminish the ice-point pressure P_0. As temperature rises, thermal kinetic motion weakens the relative effect of these attractive forces (dB/dT > 0), causing the pressure to increase slightly faster than in an ideal gas, making the ratio α_v = (dP/dt)/P_0 slightly greater.

07Atmospheric Barometric Reduction and Meniscus Capillary Depression

The determination of absolute pressure from a liquid manometer requires careful metrological corrections for atmospheric barometric pressure and capillary surface tension. The absolute pressure of the gas is P = B + Δh. The barometric pressure B is read from a Fortin mercurial barometer whose raw mercury height H_bar must be corrected for: (1) thermal expansion of the mercury column and brass vernier scale, H_0 = H_bar [1 − (β_Hg − α_brass) t_room] / [1 + β_Hg t_room], where β_Hg = 1.818 × 10⁻⁴ K⁻¹ and α_brass = 1.84 × 10⁻⁵ K⁻¹; and (2) standard gravity reduction, B = H_0 (g_local / 9.80665 m s⁻²). Furthermore, liquid mercury does not wet glass (contact angle θ ≈ 130° to 140°), forming a convex curved meniscus that experiences capillary depression due to Laplace surface tension: ΔP_cap = (2 σ cos θ) / r, where σ = 0.485 N m⁻¹ is the surface tension of mercury and r is the internal tube radius. This downward capillary force depresses the mercury column by an amount h_cap = (2 σ cos θ) / (ρ_Hg g r). In a manometer where both the closed and open tubes have identical bore diameters (e.g., both 8.0 mm), the capillary depressions in both limbs are equal in magnitude and oppose each other, canceling out in the height difference Δh = h_open − h_closed. However, if the closed capillary tube is narrower than the open leveling reservoir tube (as is common in Jolly thermometers, where the capillary has a bore of 1.0 mm while the reservoir has a bore of 10.0 mm), the capillary depression in the narrow closed tube is significantly larger (~2.8 mm vs ~0.2 mm). This differential capillary depression artificially raises the measured open column height relative to the closed column, adding an apparent positive gauge pressure. Because the closed tube meniscus is always aligned to the identical index pointer at all temperatures, this capillary depression represents a constant additive offset that contributes solely to the vertical intercept b without altering the slope m = dP/dt. The thermal pressure coefficient α_v = m / P_0 remains robust against constant capillary offsets provided P_0 is evaluated from the true zero-degree reading.

B_{\text{true}} = H_{\text{bar}} \left[1 - (\beta_{\text{Hg}} - \alpha_{\text{brass}}) t_{\text{room}}\right] \frac{g_{\text{local}}}{g_{\text{std}}}, \quad h_{\text{cap}} = \frac{2 \sigma \cos \theta}{\rho_{\text{Hg}} g r}
Why does capillary depression in an unequal-bore manometer not distort the slope of the pressure-temperature graph?

Because the meniscus in the narrow capillary is always brought to the identical vertical fiducial pointer at every temperature, the capillary depression force is constant across all trials. A constant offset shifts the vertical intercept b of the linear graph uniformly without altering the gradient m = dP/dt.

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Connect the equations to a measurement.

The linear relationship P = P_0(1 + α_v t) directly bridges classical pneumatic manometry with Kelvin’s absolute thermodynamic temperature scale. The experimental pressure coefficient α_v = 1/273.15 K⁻¹ demonstrates that gas pressure approaches zero at −273.15 °C, establishing the physical foundation for the absolute scale of temperature where T = t + 273.15.