Question

A 22.5 L sample of gas is cooled from \(145^{\circ} \mathrm{C}\) to a temperature at which its volume is \(18.3 \mathrm{L}\). What is the new temperature? Assume no change in pressure of the gas.

Short answer

Answer: The final temperature of the gas is approximately 69.18°C.

Step-by-step solution

Write down Charles's Law formula.

Charles's law is given by: \[V_1/T_1 = V_2/T_2\] where: - \(V_1\) is the initial volume, - \(T_1\) is the initial temperature in Kelvin, - \(V_2\) is the final volume, and - \(T_2\) is the final temperature in Kelvin.

Convert the given temperatures to Kelvin.

Convert the initial temperature from Celsius to Kelvin using the formula: \[T_1(K) = 145^{\circ} \mathrm{C} + 273.15\] \[T_1 = 418.15 \mathrm{K}\]

Plug the given values into Charles's Law equation and solve for the final temperature.

We have \(V_1 = 22.5 \mathrm{L}\), \(T_1 = 418.15 \mathrm{K}\), and \(V_2 = 18.3 \mathrm{L}\). We need to find \(T_2\). \[22.5/418.15 = 18.3/T_2\] Now, we need to isolate the variable, \(T_2\), to find the final temperature.

Solve for the final temperature in Kelvin.

Isolate \(T_2\) and solve the equation: \[T_2 = \frac{18.3 \times 418.15}{22.5}\] \[T_2 = 342.33 \mathrm{K}\]

Convert the final temperature back to Celsius.

Convert the final temperature in Kelvin to Celsius using the formula: \[T_2(^{\circ}\mathrm{C}) = 342.33 \mathrm{K} - 273.15\] \[T_2 = 69.18 ^{\circ} \mathrm{C}\] The final temperature of the gas is approximately \(69.18^{\circ} \mathrm{C}\) when its volume is \(18.3 \mathrm{L}\).

Key concepts

Gas Laws

Gas laws are essential principles in understanding the behavior of gases under varying conditions. One of the fundamental gas laws is Charles's Law. Charles's Law states that the volume of a fixed amount of gas is directly proportional to its absolute temperature, provided the pressure remains constant. This means that if you increase the temperature of a gas, its volume will expand. Conversely, if you cool the gas, its volume will contract. This relationship is expressed mathematically as: \[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \] Where:

• \(V_1\) is the initial volume
• \(T_1\) is the initial temperature in Kelvin
• \(V_2\) is the final volume
• \(T_2\) is the final temperature in Kelvin

This law helps us predict how a gas will behave when its temperature changes, which is crucial in various scientific and industrial applications.

Temperature Conversion

Temperature conversion is a vital step when working with Charles's Law. This is because the law requires temperatures to be in Kelvin, not Celsius or Fahrenheit. The Kelvin scale is directly related to the movement of particles in a gas, as it starts at absolute zero—where particle motion virtually stops. To convert Celsius to Kelvin, you add 273.15 to the Celsius temperature. For example, an initial temperature of \(145^{\circ} \mathrm{C}\) converts to Kelvin as: \[ \text{Initial } T_1 = 145 + 273.15 = 418.15 \mathrm{K} \] By using Kelvin, we ensure accuracy in calculations related to gas behaviors. After calculations, you may need to convert temperatures back to Celsius for more familiar interpretation: \[ T_2(^{\circ}\mathrm{C}) = T_2(\mathrm{K}) - 273.15 \] This understanding of temperature conversion ensures that our computational procedures align with Charles's Law.

Volume and Temperature Relationship

The volume and temperature relationship is a critical concept in understanding gas behaviors. According to Charles's Law, this relationship is direct: as the temperature increases, the volume increases, given that pressure is constant. This can be explained through particle movement; as temperature rises, gas particles move faster and take up more space. A practical example: Let's say a gas initially at a certain volume is heated. If its volume increases from 22.5 L to 18.3 L, we can apply Charles's Law as follows: \[ \frac{22.5}{418.15} = \frac{18.3}{T_2} \] To find \(T_2\), rearrange: \[ T_2 = \frac{18.3 \times 418.15}{22.5} \approx 342.33 \mathrm{K} \] This simple calculation shows that understanding the direct relationship of volume and temperature helps predict the state of the gas in real-world scenarios.