Question 657: Thermodynamics & KTG

Question

The temperature of an ideal gas is increased from 100 K to 400 K . If ' $x$ ' is the root mean square velocity of its molecules at 100 K , r.m.s. velocity becomes

Accepted Answer

Step 1: Formula for r.m.s. speed of a gas
$ v_{rms} = \sqrt{\frac{3kT}{m}} $
where:
$k$ is Boltzmann constant,
$T$ is absolute temperature,
$m$ is mass of one molecule.
Step 2: Relation with temperature
$ v_{rms} \propto \sqrt{T} $
Step 3: Ratio of r.m.s. velocities
Let the r.m.s. velocity at $T_1 = 100\ 	ext{K}$ be $x$. At that point:
$ v_1 = x $
At $T_2 = 400\ 	ext{K}$:
$ \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{400}{100}} = \sqrt{4} = 2 $
So,
$ v_2 = 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 root mean square velocity of its molecules at 100 K , r.m.s. velocity becomes

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