Question

Calculate the new temperature, in degrees Celsius, for each of the following with \(n\) and \(V\) constant: a. A tank of helium gas with a pressure of 250 torr at \(0{ }^{\circ} \mathrm{C}\) is heated to give a pressure of 1500 torr. b. A sample of air at \(40 .{ }^{\circ} \mathrm{C}\) and \(740 \mathrm{mmHg}\) is cooled to give a pressure of \(680 \mathrm{mmHg}\).

Short answer

a) 1365.75 °C, b) 14.35 °C

Step-by-step solution

Understand the Problem

Two scenarios are given where the pressure changes while the number of moles (n) and the volume (V) are constant. We need to find the new temperature for each case.

Use the Ideal Gas Law (Boyle's Law)

Since the number of moles and volume are constant, the relationship between pressure and temperature can be expressed using the ideal gas law: \ \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \)

Convert Temperatures to Kelvin

For both parts, we need to convert the initial temperatures from Celsius to Kelvin using the formula: \ \( T(K) = T(\degree C) + 273.15 \)

Calculate the Final Temperature for Part (a)

Given: \ \( P_1 = 250 \text{ torr} \) and \ \( T_1 = 0 \degree C = 273.15 \text{ K} \), \ \( P_2 = 1500 \text{ torr} \). \ Using the formula: \ \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \) \ Rearranging for \( T_2 \): \ \( T_2 = \frac{P_2 \times T_1}{P_1} \) \ \( T_2 = \frac{1500 \times 273.15}{250} \approx 1638.9 \text{ K} \)\ Converting back to Celsius: \ \( T_2 = 1638.9 - 273.15 \approx 1365.75 \degree C \)

Calculate the Final Temperature for Part (b)

Given: \ \( P_1 = 740 \text{ mmHg} \) and \ \( T_1 = 40 \degree C = 313.15 \text{ K} \), \ \( P_2 = 680 \text{ mmHg} \). \ Using the formula: \ \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \) \ Rearranging for \( T_2 \): \ \( T_2 = \frac{P_2 \times T_1}{P_1} \) \ \( T_2 = \frac{680 \times 313.15}{740} \approx 287.5 \text{ K} \) \ Converting back to Celsius: \ \( T_2 = 287.5 - 273.15 \approx 14.35 \degree C \)

Key concepts

Boyle's Law

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