Zeroth Law of Thermodynamics and Concept of Temperature

Existence of Empirical Temperature Function

Conditions: Three thermodynamic systems A, B, and C with equations of state involving pressures and volumes.

• Let system A have equation of state f_A(P_A, $V_A) = 0$, system B have f_B(P_B, $V_B) = 0$, and system C have f_C(P_C, $V_C) = 0$.
• When A and C are in thermal equilibrium, their coordinates are related by a constraint equation: F_1(P_A, V_A, P_C, $V_C) = 0$, which can be solved to express $P_C = g_1(P_A$, V_A, V_C).
• Similarly, when B and C are in thermal equilibrium, $P_C = g_2(P_B$, V_B, V_C).
• Equating the two expressions for P_C eliminates V_C, yielding a relation between the coordinates of A and B: h(P_A, $V_A) = k(P_B$, V_B).
• The Zeroth law implies that this same relation holds when A and B are in direct thermal equilibrium, meaning both functions can be equated to a universal property called empirical temperature \theta(P, V).

Result: \theta_A(P_A, $V_A) = \theta_B(P_B$, $V_B) = \theta$

Microscopic Temperature Derivation from Kinetic Theory

Conditions: Ideal gas containing N molecules enclosed in a volume V.

• From kinetic theory, the pressure exerted by an ideal gas is given by $P = \frac{1}{3}\frac{N}{V} m \bar{v^2}$.
• Using the ideal gas law equation of state, P$V = N k_B T$, where k_B is the Boltzmann constant.
• Equating the two expressions for PV gives N k_B $T = \frac{1}{3} N m \bar{v^2}$.
• Solving for temperature yields k_B $T = \frac{2}{3} \left(\frac{1}{2} m \bar{v^2}\right)$, where \frac{1}{2} m \bar{v^2} is the average translational kinetic energy per molecule.

Result: $T = \frac{2}{3k_B} E_{k}$, \text{avg}}

Constant-Volume Gas Thermometer Isotherms

Draw: A plot of pressure P versus temperature T (in Celsius) for various low densities of a gas extrapolated to absolute zero.

Interpret: All straight lines intersect the temperature axis at a single point, -273.15 \text{ } ^\circ C, which defines the absolute zero of temperature.

Exam use: Tested in TGT/PGT to identify the universal nature of the absolute zero intercept independent of the working gas density.

Triple Point Cell of Water

Draw: Schematic diagram of a cylindrical glass vessel containing ice, water, and water vapor in thermal equilibrium under its own vapor pressure.

Interpret: Establishes a unique, unvarying temperature reference point (273.16 K at 0.01 \text{ } ^\circ C and 611.65 Pa) used to calibrate modern thermometers.

Exam use: Conceptual and factual questions regarding why the triple point is preferred over ice and steam points.

Thermometric Property vs Temperature Non-Linearity

Draw: A non-linear curve plotting a generic thermometric property X against empirical temperature compared to a straight diagonal line.

Interpret: Different thermometric materials show deviations from one another because their physical properties do not vary linearly in the exact same manner with temperature.

Exam use: Helps explain why gas thermometers are taken as standard over liquid-in-glass thermometers.

On a new scale of temperature (called the N scale), the freezing point of water is 0^\circ N and the boiling point is 150^\circ N. What is the temperature on the Celsius scale corresponding to 60^\circ N?

• Use the linear temperature scale conversion formula: \frac{X - X_0}{X_{100} - $X_0 = \frac{C - 0}{100 - 0}$.
• Substitute the given values for the N scale: X = 60, $X_0 = 0$, $X_{100} = 150$.
• \frac{60 - 0}{150 - 0} = \frac{C}{100}.
• \frac{60}{150} = \frac{C}{100} \implies $C = \frac{2}{5} \times 100 = 40^\circ C$.

Answer: 40^\circ C

The resistance of a platinum resistance thermometer at the ice point (0^\circ C) is 5.00 \Omega and at the steam point (100^\circ C) is 5.23 \Omega. When the thermometer is immersed in a hot liquid, its resistance becomes 5.79 \Omega. Calculate the temperature of the liquid on this platinum scale.

• Assume a linear relation between resistance and temperature: $t = \frac{R_t - R_0}{R_{100} - R_0} \times 100$.
• Substitute the given values: $R_0 = 5.00 \Omega$, $R_{100} = 5.23 \Omega$, and $R_t = 5.79 \Omega$.
• R_t - $R_0 = 5.79 - 5.00 = 0.79 \Omega$.
• R_{100} - $R_0 = 5.23 - 5.00 = 0.23 \Omega$.
• $t = \frac{0.79}{0.23} \times 100 = \frac{790}{23} \approx 34.35^\circ C$.

Answer: 34.35^\circ C

A constant-volume gas thermometer registers a pressure of 1.50 \times 10^4 \text{ Pa} at the triple point of water (273.16 K). What pressure will it register when immersed in a chamber maintained at 350 K?

• For a constant volume gas thermometer, pressure is directly proportional to absolute temperature: \frac{P_1}{$T_1 = \frac{P_2}{T_2}$.
• Identify knowns: $P_1 = 1.50 \times 10^4 \text{ Pa}$, $T_1 = 273.16 \text{ K}$, $T_2 = 350 \text{ K}$.
• Rearrange to solve for P_2: $P_2 = P_1 \times \left(\frac{T_2}{T_1}\right)$.
• $P_2 = 1.50 \times 10^4 \times \left(\frac{350}{273.16}\right) = 1.50 \times 10^4 \times 1.2813$.
• $P_2 = 1.922 \times 10^4 \text{ Pa}$.

Answer: 1.92 \times 10^4 Pa