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Root Mean Square Velocity

For CO2, give...

Question

For $C O_{2}$ , given that average velocity at $T_{1}$ is equal to most probable velocity at $T_{2}$ . Thus, $\frac{T_{1}}{T_{2}}$

2.26

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1.27

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0.78

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1.13

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Solution

The correct option is C 0.78
As $u_{av}$ at $T_{1} = u_{mp}$ at $T_{2}$ , so

$\sqrt{\frac{8 R T_{1}}{π M}} = \sqrt{\frac{2 R T_{2}}{M}} \Rightarrow \sqrt{\frac{T_{1}}{T_{2}}} = \sqrt{\frac{2 π}{8}} \Rightarrow \frac{T_{1}}{T_{2}} = \frac{2 π}{8} = 0.78$

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Similar questions

Q. For

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, given that average velocity at

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The average speed at $T_{1} K$ and most probable speed at $T_{2} K$ of $C O_{2}$ gas is $9 \times 10^{4}$ cm/sec. The values of $T_{1}$ and $T_{2}$ are

From above graph choose correct option for $t_{1}$ to $t_{2}$ where $t_{1}$ is the time at which velocity of particle = 0 $t_{2}$ is the time at which velocity of particle is negative.

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Root Mean Square Velocity

Standard XI Chemistry

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For CO2, given that average velocity at T1 is equal to most probable velocity at T2. Thus, T1T2

The correct option is C 0.78As uav at T1=ump at T2, so √8RT1πM=√2RT2M⇒√T1T2=√2π8⇒T1T2=2π8=0.78

For $C O_{2}$ , given that average velocity at $T_{1}$ is equal to most probable velocity at $T_{2}$ . Thus, $\frac{T_{1}}{T_{2}}$

The correct option is C 0.78
As $u_{av}$ at $T_{1} = u_{mp}$ at $T_{2}$ , so

$\sqrt{\frac{8 R T_{1}}{π M}} = \sqrt{\frac{2 R T_{2}}{M}} \Rightarrow \sqrt{\frac{T_{1}}{T_{2}}} = \sqrt{\frac{2 π}{8}} \Rightarrow \frac{T_{1}}{T_{2}} = \frac{2 π}{8} = 0.78$