Answer:
- The average velocity of gas molecules at T₂ is less than the average velocity of gas molecules at T₁.
Explanation:
Particles A and C appear aligned vertically, suggesting they share the same kinetic energy. Given that both are positioned to the left of particle B, it indicates that A and C possess lesser kinetic energy than B.
The likelihood of a particle engaging in a reaction is directly proportional to its kinetic energy, meaning that particle B is more prone to react compared to A and C. Thus, the first statement is incorrect.
The graph resembles a bell curve, demonstrating a mixture of many molecules with low and high kinetic energy. Therefore, one cannot conclude that most particles in both gases have high velocities. As a result, the second statement is also incorrect.
Examining the higher kinetic energy values (to the right of the curve), the line for T₁ exceeds that for T₂, indicating that more molecules possess high kinetic energy at T₁ compared to T₂.
Conversely, for lower kinetic energy values (to the left of the curve), T₂'s line is above T₁'s, which implies that at T₂ there are more molecules with low kinetic energy than at T₁.
Thus, the observations from the previous two paragraphs suggest that the average kinetic energy of gas particles at T₂ is lower than the average kinetic energy of particles at T₁.
Since average speed is proportional to the square root of temperature, the relationship that applies to average kinetic energy equally pertains to average speed, leading us to conclude that the last statement holds true: "The average speed of gas particles at T₂ is lower than at T₁.".
Additionally, given that there are more gas molecules at T₁ with higher kinetic energy compared to those at T₂, it indicates that gas particles at T₁ are more likely to react than those at T₂, marking the third statement as incorrect.
