Question

State whether the following statements are true or false. When a \(200 \mathrm{ml}\) gas, in a closed vessel, is heated to \(50^{\circ} \mathrm{C}\) from \(25^{\circ} \mathrm{C}\), the volume becomes doubled.

Short answer

False

Step-by-step solution

1. Write down the initial parameters

First, write down the given initial parameters: Initial volume (\(V_1\)) = \(200 \,\mathrm{ml}\) Initial temperature (\(T_1\)) = \(25^{\circ}\mathrm{C}\) Final temperature (\(T_2\)) = \(50^{\circ}\mathrm{C}\)

2. Convert the temperature to Kelvin scale

To work with the ideal gas law, we need to convert the given temperatures to Kelvin scale. \(T_1\) in Kelvin = \(25^{\circ}\mathrm{C} + 273.15 = 298.15\,\mathrm{K}\) \(T_2\) in Kelvin = \(50^{\circ}\mathrm{C} + 273.15 = 323.15\,\mathrm{K}\)

3. Use the ideal gas law relationship

We assume the gas behaves ideally and the pressure remains constant. In this situation, we can use Charles' Law, which states that \((V_1/T_1) = (V_2/T_2)\) where, \(V_2\) is the final volume. We need to find \(V_2\) and see if it is twice \(V_1\).

4. Solve the equation to find the final volume

Now, solve for \(V_2\) using the given parameters: \((V_1/T_1) = (V_2/T_2)\) \(\frac{200\,\mathrm{ml}}{298.15\,\mathrm{K}} = \frac{V_2}{323.15\,\mathrm{K}}\) Multiply both sides by \(323.15\,\mathrm{K}\) to isolate \(V_2\): \(V_2 = \frac{(200\,\mathrm{ml})(323.15\,\mathrm{K})}{298.15\,\mathrm{K}}\) Now calculate \(V_2\): \(V_2 \approx 216.43\,\mathrm{ml}\)

5. Compare the final volume with the doubled initial volume

We found that the final volume of the gas, \(V_2\), is approximately \(216.43\,\mathrm{ml}\). Now, let's check if this value is equal to double the initial volume: Doubled initial volume = \(2 \times 200\,\mathrm{ml} = 400\,\mathrm{ml}\) Since \(216.43\,\mathrm{ml} \ne 400\,\mathrm{ml}\), the statement is false.

Key concepts

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry and physics that expresses the relationship between the pressure, volume, and temperature of an ideal gas. It is commonly formulated as \( PV = nRT \), where:

• \( P \) is the pressure of the gas.
• \( V \) is the volume of the gas.
• \( n \) is the number of moles of the gas.
• \( R \) is the ideal gas constant, and
• \( T \) is the temperature in Kelvin.

Despite its name, the ideal gas law applies very well to real gases under many conditions. When dealing with changes in conditions, such as volume changes at constant pressure, Charles' Law is often derived from the ideal gas equation. Charles' Law focuses specifically on the volume-temperature relationship, considering constant pressure to analyze the effect of temperature changes on a gas's volume. Remember that the ideal gas law helps in predicting the behavior of gases, especially when dealing with different variables involved in the process.

Temperature Conversion to Kelvin

To apply the ideal gas law, all temperature measurements must be converted to the Kelvin scale. The Kelvin scale is essential because it starts at absolute zero, the point where all molecular motion ceases, ensuring that the temperature is always positive.

• The formula for conversion is: Temperature in Kelvin \( T(K) = T(°C) + 273.15 \).

In our exercise, the temperatures given in degrees Celsius were converted as follows:

• For the initial temperature, \( 25^{\circ} \mathrm{C} + 273.15 = 298.15\,\mathrm{K} \).
• For the final temperature, \( 50^{\circ} \mathrm{C} + 273.15 = 323.15\,\mathrm{K} \).

These conversions allow us to correctly use Charles' Law in calculating the new volume of the gas when the temperature is changed, affirming the importance of using Kelvin in gas law equations. Without this conversion, equations such as \( V_1/T_1 = V_2/T_2 \) would not hold true.

Volume-Temperature Relationship

In a closed system where the pressure remains constant, the volume of a gas is directly proportional to its temperature when measured in Kelvin. This relationship is known as Charles' Law, which is a derivative of the ideal gas law.

• The formula is \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \), indicating that when temperature increases, volume also increases, provided pressure and the amount of gas remain constant.

In the specific problem we addressed:

• Initially, the gas volume \( V_1 \) was 200 \mathrm{ml} at 298.15 \mathrm{K}.
• The final condition calculates \( V_2 \), which results in approximately 216.43 \mathrm{ml} at 323.15 \mathrm{K}.

This shows that even though the temperature increased, the volume did not double, demonstrating that the rising temperature affects the volume, but not to the extent of doubling it in this case. Understanding Charles' Law highlights the crucial links between volume and temperature when assessing gas behaviors.

Gas Behavior in Closed Systems

Gases in closed systems are contained and isolated from external pressure changes, making the study of their behaviors more predictable using gas laws. In these settings, we consider the effects of temperature and volume changes primarily. With pressure assumed to be constant, such as in the exercise scenario, the relationship between volume and temperature becomes particularly relevant.

• Heating the gas increases molecular motion, leading to an increase in volume.
• In our closed system example, converting temperature into Kelvin and applying Charles' Law allowed us to predict the volume increase accurately.
• The exercise demonstrates that, in closed systems, although the volume will increase with temperature, it won’t necessarily double unless specific conditions are met.

This serves as a valuable reminder that while the behaviors of gases in closed systems can be modeled with straightforward laws, real-world interactions may yield results that require careful calculation and deeper analysis of related physical conditions.