By how many folds the temperature of a gas would increase when the root mean square velocity of the gas molecule in a container of fixed volume from $5 \times 10^{4} c m / s$ to $10 \times 10^{4} c m / s$ ?

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Solution

The correct option is D 4
$As, V_{r m s} \propto \sqrt{T} \frac{V_{1}}{V_{2}} = \sqrt{\frac{T_{1}}{T_{2}}} \Rightarrow \frac{5 \times 10^{4}}{10 \times 10^{4}} = \sqrt{\frac{T_{1}}{T_{2}}} \Rightarrow \frac{T_{1}}{T_{2}} = \frac{1}{4} \Rightarrow T_{2} = 4 T_{1} \Rightarrow FOUR times$

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Similar questions

Q. The root mean square velocity of an ideal gas in a closed container of fixed volume is increased from

5 \times 10^{4} c m s^{- 1}

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The root mean square velocity of an ideal gas in a closed container of fixed volume is increased from $5 \times 10^{4} c m s^{- 1}$ to $10 \times 10^{4} c m s^{- 1}$ Which of the following statement correctly explains how the change is accomplished

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By how many folds the temperature of a gas would increase when the root mean square velocity of the gas molecule in a container of fixed volume from 5×104cm/s to 10×104cm/s?

The correct option is D 4As, Vrms∝√TV1V2=√T1T2⇒5×10410×104=√T1T2⇒T1T2=14⇒T2=4T1⇒FOUR times

By how many folds the temperature of a gas would increase when the root mean square velocity of the gas molecule in a container of fixed volume from $5 \times 10^{4} c m / s$ to $10 \times 10^{4} c m / s$ ?

The correct option is D 4
$As, V_{r m s} \propto \sqrt{T} \frac{V_{1}}{V_{2}} = \sqrt{\frac{T_{1}}{T_{2}}} \Rightarrow \frac{5 \times 10^{4}}{10 \times 10^{4}} = \sqrt{\frac{T_{1}}{T_{2}}} \Rightarrow \frac{T_{1}}{T_{2}} = \frac{1}{4} \Rightarrow T_{2} = 4 T_{1} \Rightarrow FOUR times$