$A$ flask contains argon and chlorine in the ratio of $2: 1$ by mass. The temperature of the mixture is $27\,^{\circ}C$. Obtain the ratio of
$(i)$ average kinetic energy per molecule,and
$(ii)$ root mean square speed $v_{rms}$ of the molecules of the two gases.
Atomic mass of argon $= 39.9\,u$; Molecular mass of chlorine $= 70.9\,u$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(A) The important point to remember is that the average kinetic energy (per molecule) of any ideal gas (whether monatomic like argon,diatomic like chlorine,or polyatomic) is always equal to $\frac{3}{2}k_{B}T$. It depends only on temperature and is independent of the nature of the gas.
$(i)$ Since argon and chlorine both have the same temperature in the flask,the ratio of average kinetic energy per molecule of the two gases is $1:1$.
$(ii)$ The root mean square speed is given by $v_{rms} = \sqrt{\frac{3k_{B}T}{m}}$,where $m$ is the mass of a molecule. Therefore,the ratio of $v_{rms}$ for argon to chlorine is:
$\frac{(v_{rms})_{Ar}}{(v_{rms})_{Cl}} = \sqrt{\frac{m_{Cl}}{m_{Ar}}} = \sqrt{\frac{M_{Cl}}{M_{Ar}}} = \sqrt{\frac{70.9}{39.9}} \approx \sqrt{1.776} \approx 1.33$.
Thus,the ratio is $1.33:1$.

Explore More

Similar Questions

Fill in the blanks:
$(i)$ At low $......$ and high $......$ temperature,real gases behave as an ideal gas.
$(ii)$ $......$ is a measure of the average kinetic energy of a gas.
$(iii)$ The kinetic energy of a gas with mass $m$ is $E$. Its momentum will be $......$.
$(iv)$ At $......$ temperature,$v_{rms}$ will be double that of $v_{rms}$ at $0^{\circ} C$.

$A$ vessel containing nitrogen gas is supplied a heat of $498 \,J$, so as to raise the temperature of the gas by $40^{\circ} C$ at constant pressure. The mass of nitrogen gas in the vessel is (Molecular mass of nitrogen $= 28 \,g/mol$; Universal gas constant $= 8.3 \,J \,mol^{-1} \,K^{-1}$) (in $\,g$)

From a certain apparatus,the diffusion rate of hydrogen has an average value of $28.7 \; cm^3 s^{-1}$. The diffusion of another gas under the same conditions is measured to have an average rate of $7.2 \; cm^3 s^{-1}$. Identify the gas.

As shown schematically in the figure,two vessels contain water solutions (at temperature $T$) of potassium permanganate $(KMnO_4)$ of different concentrations $n_1$ and $n_2$ $(n_1 > n_2)$ molecules per unit volume with $\Delta n = (n_1 - n_2) \ll n_1$. When they are connected by a tube of small length $\ell$ and cross-sectional area $S$,$KMnO_4$ starts to diffuse from the left to the right vessel through the tube. Consider the collection of molecules to behave as dilute ideal gases and the difference in their partial pressure in the two vessels causing the diffusion. The speed $v$ of the molecules is limited by the viscous force $-\beta v$ on each molecule,where $\beta$ is a constant. Neglecting all terms of the order $(\Delta n)^2$,which of the following is/are correct? ($k_B$ is the Boltzmann constant)
$(A)$ the force causing the molecules to move across the tube is $\Delta n k_B T S$
$(B)$ force balance implies $n_1 \beta v \ell = \Delta n k_B T$
$(C)$ total number of molecules going across the tube per sec is $\left(\frac{\Delta n}{\ell}\right)\left(\frac{k_B T}{\beta}\right) S$
$(D)$ rate of molecules getting transferred through the tube does not change with time

In the isothermal expansion of $10\,g$ of gas from volume $V$ to $2V$,the work done by the gas is $575\,J$. What is the root mean square speed of the molecules of the gas at that temperature (in $m/s$)?

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo