Question

The rms speed of a gas molecule is 'V' at pressure 'P'. If the pressure is increased by two times, then the rms speed of the gas molecule at the same temperature will be

• A. $V$
• B. $2V$
• C. $\frac{V}{3}$
• D. $\frac{V}{2}$

Hint:

Always remember: The kinetic energy and velocity parameters of ideal gas particles are functions of temperature only! Changes in volume or pressure have zero influence on $v_{\text{rms}}$ as long as the temperature parameter is anchored.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks how the root-mean-square ($v_{\text{rms}}$) speed of gas molecules changes when the gas pressure is doubled while holding the temperature strictly constant.

Step 2: Key Formula or Approach:
The Maxwell-Boltzmann formula for the root-mean-square velocity of an ideal gas is:
$$v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$$ where $R$ is the universal gas constant, $T$ is the absolute temperature, and $M$ is the molar mass of the gas.

Step 3: Detailed Explanation:
Looking at the formula, $v_{\text{rms}}$ depends exclusively on the absolute temperature $T$ and the molecular properties ($M$) of the gas species.
Even though the ideal gas law can express this as $v_{\text{rms}} = \sqrt{\frac{3P}{\rho}}$, increasing the pressure at a constant temperature cause the density ($\rho$) to increase proportionally by the exact same factor ($\rho \propto P$).
Consequently, the ratio $\frac{P}{\rho}$ remains completely unchanged.
Because the absolute temperature is kept constant ("at the same temperature"), the root-mean-square speed of the molecules remains completely unaltered. Thus, the new speed is equal to its original value, $V$.

Step 4: Final Answer:
The rms speed remains $V$, which corresponds to option (A).