Problem 57 - Chemistry and Chemical Reactivity

Question

You have a sample of helium gas at \(-33^{\circ} \mathrm{C},\) and you want to increase the average speed of helium atoms by \(10.0 \% .\) To what temperature should the gas be heated to accomplish this?

Step-by-Step Solution

Answer

The gas should be heated to approximately \( 17.43^{\circ}C \).

Step 1: Understand the Relationship Between Speed and Temperature

The average speed of gas particles is related to temperature through the equation \( v \propto \sqrt{T} \), where \( v \) is the average speed and \( T \) is the absolute temperature in Kelvin. This means that if the average speed increases by 10%, the temperature must increase in such a way to maintain this proportion.

Step 2: Convert Initial Temperature to Kelvin

The initial temperature is given as \(-33^{\circ} \mathrm{C}\). To convert this to Kelvin, use the formula \( T(K) = T(^{\circ}C) + 273.15 \). Thus, the initial temperature in Kelvin is \( T_1 = -33 + 273.15 = 240.15 \, \mathrm{K} \).

Step 3: Setup the Proportional Relationship

Since we want to increase the average speed by 10%, then the new speed \( v_2 = 1.1 \, v_1 \). By the formula \( v \propto \sqrt{T} \), we have:\[v_2 = \sqrt{T_2} \quad \text{and} \quad v_1 = \sqrt{T_1}\]Thus, this means:\[\sqrt{T_2} = 1.1 \, \sqrt{T_1}\]

Step 4: Solve for the New Temperature

From the equation \( \sqrt{T_2} = 1.1 \, \sqrt{T_1} \), square both sides to remove the square root:\[T_2 = (1.1)^2 \, T_1\]Substitute \( T_1 = 240.15 \, \mathrm{K} \) into the equation:\[T_2 = (1.1)^2 \, \times \ 240.15 = 1.21 \, \times \ 240.15\]Thus, \( T_2 \approx 290.58 \, \mathrm{K} \).

Step 5: Convert the Final Temperature to Celsius

To convert \( T_2 \) back to Celsius, use the equation \( T(^{\circ}C) = T(K) - 273.15 \). So,\[T_2(^{\circ}C) = 290.58 - 273.15 = 17.43^{\circ}C\]

Key Concepts

Temperature Conversion

Temperature conversion is crucial when dealing with thermodynamics, especially since physical laws often require temperatures in a specific unit, Kelvin. Converting from Celsius to Kelvin is straightforward by using the formula:
\[T(K) = T(^{\circ}C) + 273.15\]This formula simply adds 273.15 to the Celsius temperature to shift it to the Kelvin scale. This is because the Kelvin scale starts at absolute zero, which is -273.15°C. When converting back to Celsius, you subtract 273.15 from the Kelvin measurement:
\[T(^{\circ}C) = T(K) - 273.15\]Understanding and mastering this conversion is essential:

• Working with gas laws where temperature must be absolute.
• Ensuring consistency in scientific calculations and comparisons.

Kinetic Molecular Theory

The kinetic molecular theory is a fundamental concept in thermodynamics explaining the behavior of gases. It describes gas as composed of numerous small particles, typically atoms or molecules, that are in constant, random motion. Let's delve into what this means:

• **Temperature and Motion:** The theory relates temperature directly to the kinetic energy of these particles. The higher the temperature, the faster the particles move due to increased kinetic energy. This is why in the problem, increasing the average speed of helium atoms requires heating the gas.
• **Particle Interactions:** Assuming ideal gas behavior, which means no intermolecular forces are acting, allows us to simplify the mathematical treatment of gas laws.

Understanding this theory helps analyze how temperature changes, such as in the exercise, affect gas behavior by directly influencing particle speed and thereby kinetic energy.

Gas Laws

Gas laws are scientific laws that describe how gases behave with respect to pressure, volume, and temperature. They are vital in predicting how a gas will respond under various conditions. Several key laws include:

• **Charles's Law:** Volume and temperature of a gas are directly proportional at constant pressure.This means if the temperature of a gas increases, so does its volume.
• **Boyle's Law:** Pressure and volume of a gas are inversely proportional at constant temperature.Thus, increasing the volume of a gas leads to a decrease in pressure, and vice versa.
• **Ideal Gas Law:** This integrates Charles's and Boyle's laws into one formula: \[PV = nRT\]where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles of gas, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.It applies well to ideal gases, providing a broad understanding of gas behavior.

When solving the given exercise, recognizing these laws, especially the relationship between temperature and particle speed, helps in predicting the resultant temperature needed to achieve a certain particle speed change.