Question

A sample of chlorine gas has a volume of \(0.30 \times 10^{-3} \mathrm{~m}^{3}\) at \(273 \mathrm{~K}\) and \(101,325 \mathrm{~Pa}\) pressure. What temperature (in \({ }^{\circ} \mathrm{C}\) ) would be required to increase the volume to \(1.0 \times 10^{-3} \mathrm{~m}^{3}\) ? To decrease the volume to \(0.20 \times 10^{-3} \mathrm{~m}^{3} ?\)

Short answer

To increase the volume to \(1.0 \times 10^{-3} \mathrm{~m}^3\), the temperature must be \(637^{ ext{o}}C\). To decrease it to \(0.20 \times 10^{-3} \mathrm{~m}^3\), it must be \(-91^{ ext{o}}C\).

Step-by-step solution

Understand the Problem

The problem involves changes in volume and temperature of a gas at constant pressure. We need to use the ideal gas law to calculate the required temperatures for the given volume changes.

Use Charles's Law

Charles's Law states \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \) for a gas at constant pressure. Here, \( V_1 = 0.30 \times 10^{-3} \mathrm{~m}^3 \) and \( T_1 = 273 \mathrm{~K} \). We need to find \( T_2 \) when \( V_2 = 1.0 \times 10^{-3} \mathrm{~m}^3 \) and \( T_2 = \ ? \).

Solve for the First Scenario

Rearrange Charles's Law to solve for \( T_2: T_2 = T_1 \times \frac{V_2}{V_1} \). Substitute the known values: \[ T_2 = 273 \times \frac{1.0 \times 10^{-3}}{0.30 \times 10^{-3}} \] Calculate \( T_2 \).

Convert Temperature to Celsius (First Scenario)

The temperature calculated is in Kelvin. Convert it to Celsius using the formula: \( T(^{ ext{o}}C) = T( ext{K}) - 273 \).

Solve for the Second Scenario

Again use Charles's Law for the change to a volume of \( V_2 = 0.20 \times 10^{-3} \mathrm{~m}^3 \). Use the formula: \[ T_2 = 273 \times \frac{0.20 \times 10^{-3}}{0.30 \times 10^{-3}} \] Calculate \( T_2 \).

Convert Temperature to Celsius (Second Scenario)

Convert the Kelvin temperature calculated for the second scenario to Celsius:\( T(^{ ext{o}}C) = T( ext{K}) - 273 \).

Key concepts

Charles's Law

Imagine if you blew up a balloon indoors and then took it outside on a chilly day. You might notice it shrinks in size. That's because Charles's Law describes how gases behave when temperature changes under constant pressure.

In the example provided, this law is crucial. Charles's Law is given by the formula:

• \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \)

where \(V_1\) and \(T_1\) represent the initial volume and temperature, and \(V_2\) and \(T_2\) the final conditions. By using this relationship, we can solve for the unknown temperature \(T_2\) when the gas volume changes.

This relationship helps solve problems where predicting the result of temperature-volume shifts is needed, helping us see the flexibility of gases when heated or cooled.

Temperature Conversion

Once you solve for a gas's new temperature using Charles's Law, you need to convert the temperature from Kelvin to Celsius, because Celsius is more commonly used in everyday life and science.

Fortunately, this conversion is quite straightforward:

• To convert Kelvin to Celsius, subtract 273 from the Kelvin temperature: \( T(^{\circ}C) = T(\text{K}) - 273 \)

This step can often be overlooked, so remember, with gases and their calculations, ensuring you are using the correct units is essential. Keep in mind, values in Kelvin always maintain positive numbers and zero signifies absolute cold, or absolute zero, a crucial concept in scientific measurements.

Gas Volume and Temperature

Gas volume and temperature are interrelated. As one rises, so typically does the other, provided other conditions like pressure remain constant.

Using the previous example, when the initial volume of the gas is given as \(0.30 \times 10^{-3} \mathrm{~m}^3\) at \(273 \mathrm{~K}\), we are tasked with finding what temperature will lead to a new volume. By applying Charles's Law, we see how these two properties of gas shift hand-in-hand.

Understanding this dynamic assists in predicting how gases behave in different environments, whether in labs or natural settings. Recognizing the effects of heating and cooling on volume can help in fields ranging from meteorology to engineering, showing the laws of gas dynamics at work daily.