91Ó°ÊÓ

Temperature is a fundamental aspect of thermodynamics and plays a key role in understanding gas behavior. In this exercise, the temperatures of gases are given in degrees Celsius, a common unit of temperature. However, for calculations involving the Ideal Gas Law, we must convert these temperatures to Kelvin. This is because Kelvin is the standard SI unit of temperature, starting from absolute zero, which makes mathematical manipulation involving temperature more accurate and meaningful.
To convert from Celsius to Kelvin, you add 273 to the Celsius temperature. The conversion formula is:

• For box A: \(T_A = 50^{\circ}C + 273 = 323 K\)
• For box B: \(T_B = 10^{\circ}C + 273 = 283 K\)

This conversion is crucial because it allows us to use the temperature to determine other properties of the gas, like kinetic energy and molecular speed.

Kinetic energy in the context of gases is an indicator of how energetically the molecules in the gas are moving. It directly relates to the temperature of the gas, and can be described by the formula \( KE = \frac{3}{2}kT \). Here, \( k \) is the Boltzmann constant and \( T \) is the temperature in Kelvin.
The formula illustrates that kinetic energy is directly proportional to temperature. Therefore, an increase in the temperature of a gas leads to an increase in the average kinetic energy of its molecules. In the given exercise, since box A has a higher temperature than box B after conversion to Kelvin, we can conclude:

• The average kinetic energy of molecules in box A is higher than in box B.
• Statement (d) in the exercise is thus true because temperature \(T_A\) is greater than \(T_B\).

Understanding this relationship is crucial for analyzing the energetic dynamics of gas molecules and their behaviors under different temperature conditions.

The speed of molecules in a gas is another important factor to consider when determining the physical state and behavior of the gas. Molecular speed is linked with kinetic energy and mass by the equation \( v = \sqrt{\frac{3kT}{m}} \). This provides the root-mean-square speed of molecules, where \( m \) represents the molecular mass.
In our scenario, since temperature \( T \) is greater in box A, the kinetic energy and speed of molecules could be higher if the gas type (and thus molecular mass) remains constant between the boxes. Key points include:

• Molecules in box A might move faster than in box B, provided their masses are similar.
• Therefore, statement (e) could be true, based on these assumptions.

It’s essential to understand that while temperature gives us insights into probable speeds, actual molecular speed depends also on the molecule’s mass, so assumptions need validation with more data.