A constant-volume gas thermometer registers a pressure of 60\,\text{kPa} at the ice point … — JAMB UTME

A constant-volume gas thermometer registers a pressure of $60\,\text{kPa}$ at the ice point ( $0^\circ\text{C}$ ) and $84\,\text{kPa}$ at the steam point ( $100^\circ\text{C}$ ). What is the temperature in degrees Celsius when the pressure registered by the thermometer is $72\,\text{kPa}$ ?

Answer: 50 °C

Answer

The temperature corresponding to a pressure reading of

72\,\text{kPa}

50^\circ\text{C}

Using the linear relation for a constant-volume gas thermometer:

T = \frac{P_T - P_0}{P_{100} - P_0} \times 100^\circ\text{C}

. Substituting

P_T = 72\,\text{kPa}

P_0 = 60\,\text{kPa}

, and

P_{100} = 84\,\text{kPa}

yields

T = \frac{72 - 60}{84 - 60} \times 100 = \frac{12}{24} \times 100 = 50^\circ\text{C}

Step-by-Step Solution

Identify the thermometric property values at the fixed points and target state.

P_0 = 60\,\text{kPa}

P_{100} = 84\,\text{kPa}

, and

P_T = 72\,\text{kPa}

These represent the pressure values corresponding to

0^\circ\text{C}

100^\circ\text{C}

, and the unknown temperature

T

respectively.

Set up the linear scale conversion equation.

T = \frac{P_T - P_0}{P_{100} - P_0} \times 100^\circ\text{C}

Temperature changes linearly with the thermometric property (gas pressure at constant volume).

Calculate the numerical value.

T = \frac{12}{24} \times 100 = 50^\circ\text{C}

Simplifying the fraction

\frac{12}{24} = 0.5

and multiplying by

100

gives

50

Key Concept

Temperature measurement using constant-volume gas pressure as a thermometric property

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