Question

A tank is filled with gas to a pressure of \(875 \mathrm{~mm} \mathrm{Hg}\) at \(25^{\circ} \mathrm{C}\). The gas is transferred without loss to a tank twice the size of the original tank. If the pressure is to remain constant, at what temperature (in \(^{\circ} \mathrm{C}\) ) should the tank be kept?

Short answer

Answer: The gas should be kept at approximately 323.15°C for the pressure to remain constant in the new tank.

Step-by-step solution

Convert temperature and pressure to appropriate units

Convert the initial temperature from Celsius to Kelvin (K) and pressure from mm Hg to atmospheres (atm): T1 = 25°C + 273.15 ≈ 298.15 K P1 = 875 mm Hg × (1 atm / 760 mm Hg) ≈ 1.15 atm

Find the relationship between the initial and final volume

The final volume of the tank (V2) is twice the size of the initial volume (V1). \(V_2 = 2V_1\)

Apply the ideal gas law relationship

Using the modified ideal gas law equation, substitute the given values and solve for T2: \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\) \(\frac{V_1}{298.15 \mathrm{~K}} = \frac{2V_1}{T_2}\) Divide both sides by \(V_1\) to eliminate it: \(\frac{1}{298.15 \mathrm{~K}} = \frac{2}{T_2}\) Now, solve for \(T_2\): \(T_2 = \frac{2}{\frac{1}{298.15 \mathrm{~K}}} = 2 \times 298.15 \mathrm{~K} = 596.3 \mathrm{~K}\)

Convert the temperature back to Celsius

Convert the final temperature from Kelvin back to Celsius: \(T_2 = 596.3 \mathrm{~K} - 273.15 = 323.15^{\circ} \mathrm{C}\) The tank should be kept at a temperature of approximately \(323.15^{\circ} \mathrm{C}\) for the pressure to remain constant.

Key concepts

Gas Pressure

Understanding gas pressure is essential when working with gases in various contexts, such as chemistry and physics. Gas pressure is defined as the force applied per unit area by gas molecules as they collide with the surfaces of their container.

Imagine a tank filled with gas molecules, all moving in random directions. When these molecules strike the walls of the tank, they exert a force on the walls. The collective force from millions of such collisions results in gas pressure. This pressure can be measured in various units, including millimeters of mercury (mm Hg), atmospheres (atm), or pascals (Pa).

The initial problem given states that a tank has a gas pressure of 875 mm Hg. To use this pressure with the Ideal Gas Law, it's often necessary to convert it to atmospheres, as the law typically uses the SI unit system where pressure is measured in atmospheres. Conversions like this are crucial for maintaining consistency across different measurements and calculations.

Temperature Conversion

Temperature conversion is a fundamental concept in science, especially when working with the Ideal Gas Law, which requires temperatures to be in the Kelvin scale. Kelvin is the SI base unit for temperature, and it is crucial in scientific calculations because it allows for the direct proportional relationship between temperature and other variables like volume and pressure to hold true.

Converting from Celsius to Kelvin is simple: add 273.15 to the Celsius temperature. For instance, the step by step solution demonstrates this with the initial temperature, 25°C, which becomes 298.15 K upon conversion. It's important to note that the Kelvin scale does not have degrees, so temperatures are simply stated as kelvins (e.g., 298.15 K).

Similarly, converting from Kelvin back to Celsius involves subtracting 273.15 from the Kelvin temperature. This reverse conversion is useful for expressing the temperature in a form that’s more intuitive for everyday situations, as observed in the final step of the solution when the answer is reverted to Celsius for practicality.

Volume-Temperature Relationship

The volume-temperature relationship, also known as Charles's Law, is an important principle that describes how gases tend to expand when heated. In the context of the Ideal Gas Law, a constant pressure scenario implies that the volume of a gas is directly proportional to its temperature in Kelvin.

The relationship is represented by the equation \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\), where \(V_1\) and \(V_2\) are the initial and final volumes, and \(T_1\) and \(T_2\) are the initial and final temperatures, respectively. In the example provided, we know the final volume is twice the initial volume, so \((V_2 = 2V_1)\). When you plug in the values and calculate for \(T_2\), you get a value that shows the gas will need to be at a higher temperature to keep the pressure constant if the volume is increased.

This volume-temperature relationship is vital because it informs decisions in various practical applications, such as determining the necessary temperature for a gas to fill a certain space or to maintain a specific pressure when the volume changes, as in the exercise we’re considering.