Medium
For a given gas at $1\,atm$ pressure,the $rms$ speed of the molecules is $200\,m/s$ at $127\,^oC$. At $2\,atm$ pressure and at $227\,^oC$,the $rms$ speed of the molecules will be:
- A$80\,m/s$
- B$100\sqrt{5}\,m/s$
- C$100\,m/s$
- D$80\sqrt{5}\,m/s$
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Similar Questions
Column-$I$ represents the formula for ${v_{rms}}$ and Column-$II$ represents the corresponding condition (phenomena). Match them correctly:
Column-$I$ Column-$II$ $(a)$ ${v_{rms}} = \sqrt {\frac{3P}{\rho}}$ $(i)$ For $1 \text{ mole ideal gas}$ $(b)$ ${v_{rms}} = \sqrt {\frac{3RT}{M_0}}$ $(ii)$ For one molecule of gas $(c)$ ${v_{rms}} = \sqrt {\frac{3{k_B}T}{m}}$ $(iii)$ On the basis of kinetic theory
| Column-$I$ | Column-$II$ |
| $(a)$ ${v_{rms}} = \sqrt {\frac{3P}{\rho}}$ | $(i)$ For $1 \text{ mole ideal gas}$ |
| $(b)$ ${v_{rms}} = \sqrt {\frac{3RT}{M_0}}$ | $(ii)$ For one molecule of gas |
| $(c)$ ${v_{rms}} = \sqrt {\frac{3{k_B}T}{m}}$ | $(iii)$ On the basis of kinetic theory |
Medium
View SolutionIf the temperature of a gas is increased from $27^{\circ} C$ to $159^{\circ} C$, the increase in the rms speed of the gas molecules is (in $\%$)
The root mean square velocity of molecules of a gas is
$A$ gas is at a temperature of $0^{\circ}C$. To what temperature in $^{\circ}C$ must the gas be heated so that the $rms$ speed of its molecules becomes double?
Difficult
View SolutionTwo boxes are at the same temperature. The first box contains gas with molecular mass $m_1$ and rms speed $v_1$. The second box contains gas with molecular mass $m_2$ and average speed $v_2$. If $v_1 = 1.5 v_2$,then $\frac{m_1}{m_2}$ is
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