Certain amount of an ideal gas is taken from its initial state 1 to final state 4 through the paths 1 → 2 → 3 → 4 as shown in the figure AB, CD, EF are all isotherms. If $v_p$ is the most probable speed of the molecules, then:
Given: The process involves an ideal gas taken from initial state 1 to final state 4 through the paths 1 → 2 → 3 → 4. The figure shows isotherms AB, CD, and EF. We are asked to compare the most probable speed \( v_p \) at different points of the process.
Key Concept: The most probable speed \( v_p \) of molecules is given by the formula: \[ v_p = \sqrt{\frac{2kT}{m}} \] where: - \( k \) is the Boltzmann constant, - \( T \) is the temperature, and - \( m \) is the mass of a gas molecule. For an isothermal process, the temperature \( T \) remains constant. Hence, the most probable speed \( v_p \) is directly proportional to the temperature. The temperature of the gas is higher at points where the volume is smaller (since for isothermal processes, \( PV = nRT \)).
Analysis: - In the figure, we see that the isotherms (AB, CD, EF) represent constant temperature curves. - For the ideal gas, the temperature is highest at the smallest volume and decreases as the volume increases. From the isotherms: - \( v_p \) at point 3 and point 4 will be greater than at points 2 and 1 because the temperatures at points 3 and 4 are higher (due to the smaller volumes at those points). - \( v_p \) at point 2 will be greater than at point 1 because point 2 corresponds to a higher temperature than point 1. Therefore, the most probable speed follows this order: \[ v_p \text{ at 3} = v_p \text{ at 4} > v_p \text{ at 2} > v_p \text{ at 1} \]
Final Answer: \( v_p \text{ at 3} = v_p \text{ at 4} > v_p \text{ at 2} > v_p \text{ at 1} \).
