Question

At \(T(K)\), the rms velocity of an ideal gas is \(x\,\text{m s}^{-1}\). At what temperature (in K), the rms velocity becomes \(3x\,\text{m s}^{-1}\)?

• A. \(3T\)
• B. \(4T\)
• C. \(9T\)
• D. \(18T\)

Hint:

For an ideal gas: \[ v_{rms} \propto \sqrt{T} \] If speed becomes \(n\) times, temperature becomes \(n^2\) times.

Answer 1

The Correct Option is C

Step 1: Recall the rms speed formula.
The root mean square speed of an ideal gas is \[ v_{rms} = \sqrt{\frac{3RT}{M}}. \] Here $R$ and $M$ are fixed for a given gas, so only temperature $T$ changes the speed.
Step 2: See the key proportionality.
Because the constants stay the same, $v_{rms} \propto \sqrt{T}$. This means the speed grows with the square root of temperature. So if speed goes up, temperature must go up faster.
Step 3: Write the ratio for two states.
Comparing the start and end states, \[ \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}. \] We will plug the given speeds into this.
Step 4: Put in the speeds.
Start: $v_1 = x$ at $T_1 = T$. End: $v_2 = 3x$. So \[ \frac{3x}{x} = \sqrt{\frac{T_2}{T}} \ \Rightarrow \ 3 = \sqrt{\frac{T_2}{T}}. \]
Step 5: Square to remove the root.
Squaring both sides gives \[ 9 = \frac{T_2}{T}. \] So $T_2 = 9T$.
Step 6: State the answer.
To triple the rms speed we must raise the temperature nine times. \[ \boxed{T_2 = 9T} \]