Question 690: Thermodynamics & KTG

Question

The temperature of an ideal gas is increased from 100 K to 400 K . If ' $x$ ' is the R.M.S. velocity of its molecules at 100 K , it becomes

Accepted Answer

Step 1: Recall formula for RMS velocity
The root-mean-square (RMS) velocity is given by:
$ v_{rms} = \sqrt{\frac{3k_B T}{m}} $
So,
$ v_{rms} \propto \sqrt{T}. $
Step 2: Compare velocities at two temperatures
If RMS velocity is $x$ at $T_1 = 100\,K$, then:
$ x \propto \sqrt{100}. $
At $T_2 = 400\,K$:
$ v_{rms}' \propto \sqrt{400}. $
Step 3: Ratio
$ \frac{v_{rms}'}{x} = \frac{\sqrt{400}}{\sqrt{100}}  = \frac{20}{10} = 2. $
So the new velocity at 400 K = $2x$.
Ã¢ÂÂ Final Answer:
Option B:
$2x$

Answer

$\frac{x}{4}$

Answer

$2 x$

Answer

$3 x$

Answer

$4 x$

Thermodynamics & KTG

The temperature of an ideal gas is increased from 100 K to 400 K . If ' x ' is the R.M.S. velocity of its molecules at 100 K , it becomes

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