The temperature of an ideal gas is increased from 200 K to 800 K. If r.m.s. speed of gas at 200 K is v0, then the r.m.s. speed of the gas at 800 K will be:

Given: The temperature of an ideal gas changes from 200 K to 800 K. The r.m.s. speed at 200 K is v0. Find: The r.m.s. speed at 800 K. Using v rms = 3RTm At 200 K, the r.m.s. speed is v0: v0 = 3R 200m (1) At 800 K, the r.m.s. speed is v': v' = 3R 800m (2) Dividing equation (2) by equation (1): v'v0 = 800200 = 4 = 2 Therefore, v' = 2v0 So, the correct option is B.

The temperature of an ideal gas is increased from 200 K to 800 K.… — Solution | JEE Physics

MCQEasyJEE 2023Molecular Speeds (rms, Average, Most Probable)

The temperature of an ideal gas is increased from $200 \, \text{K}$ to $800 \, \text{K}$ . If r.m.s. speed of gas at $200 \, \text{K}$ is $v_0$ , then the r.m.s. speed of the gas at $800 \, \text{K}$ will be:

A
$4v_0$
B
$2v_0$
C
$v_0$
D
$\frac{v_0}{4}$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The temperature of an ideal gas changes from $200 \, \text{K}$ to $800 \, \text{K}$ . The r.m.s. speed at $200 \, \text{K}$ is $v_0$ .

Find: The r.m.s. speed at $800 \, \text{K}$ .

Using

v_{\text{rms}} = \sqrt{\frac{3RT}{m}}

At $200 \, \text{K}$ , the r.m.s. speed is $v_0$ :

v_0 = \sqrt{\frac{3R \times 200}{m}} \quad \cdots (1)

At $800 \, \text{K}$ , the r.m.s. speed is $v'$ :

v' = \sqrt{\frac{3R \times 800}{m}} \quad \cdots (2)

Dividing equation $(2)$ by equation $(1)$ :

\frac{v'}{v_0} = \sqrt{\frac{800}{200}} = \sqrt{4} = 2

Therefore,

v' = 2v_0

So, the correct option is B.

Common mistakes

Using a linear relation between r.m.s. speed and temperature is incorrect because $v_{\text{rms}} \propto \sqrt{T}$ , not $T$ . Use the square-root dependence before comparing speeds.
Substituting the temperatures and then forgetting to divide the new expression by the old one can lead to unnecessary algebra. Take the ratio $\frac{v'}{v_0}$ so the constants cancel directly.
Taking $\frac{800}{200} = 4$ and concluding the speed becomes $4v_0$ is wrong because the square root must still be applied. After simplification, evaluate $\sqrt{4} = 2$ .

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Answer

Step-by-step solution

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