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Heat and Thermodynamics question

2021 · 20 Jul · Shift 1 · Q49

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Heat and Thermodynamics
2021 · 20 Jul · Shift 1 · Q49

Heat and Thermodynamics question

2021 · 20 Jul · Shift 1 · Q49

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1

Consider a mixture of gas molecule of types A, B and C having masses mA < mB < mC. The ratio of their root mean square speeds at normal temperature and pressure is :

A
${v_A} = {v_B} e {v_C}$
B
${1 \over {{v_A}}} \gt {1 \over {{v_B}}} \gt {1 \over {{v_C}}}$
C
${1 \over {{v_A}}} \lt {1 \over {{v_B}}} \lt {1 \over {{v_C}}}$
D
${v_A} = {v_B} = {v_C} = 0$

View written solutionFree

Correct answer: C

Use the formula for rms speed

For any gas at temperature $T$ , the root mean square speed is

$v_{\text{rms}} = \sqrt{\frac{3kT}{m}}$

where:

$k$ is Boltzmann constant
$T$ is absolute temperature
$m$ is mass of one molecule

Compare gases at the same temperature

Since the gases A, B, and C are all at NTP, they have the same temperature. Hence,

$v_{\text{rms}} \propto \frac{1}{\sqrt{m}}$

Given:

$m_A < m_B < m_C$

therefore,

$\frac{1}{\sqrt{m_A}} > \frac{1}{\sqrt{m_B}} > \frac{1}{\sqrt{m_C}}$

So,

$v_A > v_B > v_C$

Now compare reciprocals

Taking reciprocals reverses the inequality:

$\frac{1}{v_A} < \frac{1}{v_B} < \frac{1}{v_C}$

Match with options

Option A: $v_A = v_B = v_C$ → false
Option B: $\dfrac{1}{v_A} > \dfrac{1}{v_B} > \dfrac{1}{v_C}$ → false
Option C: $\dfrac{1}{v_A} < \dfrac{1}{v_B} < \dfrac{1}{v_C}$ → true
Option D: $v_A = v_B = v_C = 0$ → false

Hence, the correct option is C.

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