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: Identify the change.
A gas has rms speed $V$ at pressure $P$. The pressure is doubled while the temperature is held constant, and we want the new rms speed.
Step 2: Recall the rms speed formula.
$v_{rms} = \sqrt{\dfrac{3RT}{M}}$, where $T$ is the absolute temperature and $M$ the molar mass.
Step 3: See what it depends on.
This expression contains only $T$ and $M$. Pressure does not appear, so at fixed temperature the rms speed cannot change.
Step 4: Cross-check with the density form.
We can also write $v_{rms} = \sqrt{\dfrac{3P}{\rho}}$. At constant temperature, doubling $P$ also doubles the density $\rho$ (since $P \propto \rho$ at fixed $T$).
Step 5: Note the cancellation.
The ratio $\dfrac{P}{\rho}$ therefore stays the same, so $v_{rms}$ is unchanged.
Step 6: Conclude.
The rms speed stays $V$, option (1). The key phrase is at the same temperature, which fixes the molecular speed.
\[ \boxed{v_{rms} = V} \]