Correct answer is (C)

The relationship between the root-mean-square (rms) speed ($v_{rms}$) and the average speed ($v_{avg}$) of molecules in a gas can be found using the Maxwell-Boltzmann distribution. The rms speed and average speed are related as follows:

$v_{rms}=\sqrt{\frac{3RT}{M}}$ $v_{avg}=\sqrt{\frac{8RT}{\pi M}}$

Where:

• $R$ is the ideal gas constant
• $T$ is the temperature in Kelvin
• $M$ is the molar mass of the gas
• $\pi$ is the mathematical constant pi

In this problem, the rms speed of the oxygen molecule is given by:

$v_{rms}={(1+\frac{5}{x})}^{\frac{1}{2}}{v}_{avg}$

Now, let's divide the expression for $v_{rms}$ by the expression for $v_{avg}$:

$\frac{v_{rms}}{v_{avg}}=\frac{\sqrt{\frac{3RT}{M}}}{\sqrt{\frac{8RT}{\pi M}}}={(1+\frac{5}{x})}^{\frac{1}{2}}$

By simplifying the expression, we get:

$\frac{v_{rms}}{v_{avg}}=\frac{\sqrt{3}}{\sqrt{\frac{8}{\pi }}}={(1+\frac{5}{x})}^{\frac{1}{2}}$

Square both sides of the equation:

$\frac{3}{\frac{8}{\pi }}=1+\frac{5}{x}$

Now we will substitute the provided value of $\pi =\frac{22}{7}$:

$\frac{3}{\frac{8}{\frac{22}{7}}}=1+\frac{5}{x}$

By simplifying the expression, we get:

$\frac{3\cdot \frac{22}{7}}{8}=1+\frac{5}{x}$

Now let's solve for $x$:

$\frac{66}{56}-1=\frac{5}{x}$ $\frac{10}{56}=\frac{5}{x}$

Multiplying both sides by $x$:

$\frac{10}{56}x=5$

Finally, solving for $x$:

$x=\frac{5\cdot 56}{10}=28$

So, the value of $x$ is $\boxed{{\displaystyle 28}}$.

Correct answer is (D)

The correct relationship between the rms speeds of the two gases is:

$\frac{v_{\mathrm{Ar}}}{v_{\mathrm{Cl}}}=\sqrt{\frac{M_{\mathrm{Cl}}}{M_{\mathrm{Ar}}}}$

Given the molar masses for argon and chlorine:

$M_{\mathrm{Ar}}=39.9\mathrm{u}$

$M_{{\mathrm{Cl}}_2}=70.9\mathrm{u}$

And the rms speed of chlorine molecules:

$v_{\mathrm{Cl}}=490\mathrm{m}/\mathrm{s}$

We can now solve for the rms speed of argon molecules:

$v_{\mathrm{Ar}}=\sqrt{\frac{70.9}{39.9}}\times 490$

$v_{\mathrm{Ar}}\approx 651.7\mathrm{m}/\mathrm{s}$

The rms speed of argon molecules at the same temperature as the chlorine molecules is approximately $651.7\mathrm{m}/\mathrm{s}$.

Correct answer is (D)

To find the root mean square speed of molecules of nitrogen gas, we can use the formula:

$v_{rms}=\sqrt{\frac{3k_{B}T}{m}}$

where $v_{rms}$ is the root mean square speed, $k_B$ is the Boltzmann constant, $T$ is the temperature in Kelvin, and $m$ is the mass of a nitrogen molecule.

First, we need to convert the temperature from Celsius to Kelvin:

$T={27}^{\circ }\mathrm{C}+273=300\mathrm{K}$

Now, substitute the given values of $k_B$, $T$, and $m$ into the formula:

$v_{rms}=\sqrt{\frac{3(1.4\times {10}^{-23}{\mathrm{JK}}^{-1})(300\mathrm{K})}{4.6\times {10}^{-26}\mathrm{kg}}}$

Simplify and calculate the root mean square speed:

$v_{rms}=523\mathrm{m}/\mathrm{s}$

The root mean square speed of molecules of nitrogen gas at ${27}^{\circ }\mathrm{C}$ is 523 m/s.

Correct answer is (B)

The root mean square speed ($v_{rms}$) of a gas is given by the formula:

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

where $k$ is the Boltzmann constant, $T$ is the temperature, and $m$ is the molar mass of the gas molecules.

The vessels contain neon (monoatomic), chlorine (diatomic), and uranium hexafluoride (polyatomic) gases. Their molar masses are:

1. Neon: $20.18\text{g/mol}$ (monoatomic)
2. Chlorine: $2\times 35.45=70.90\text{g/mol}$ (diatomic)
3. Uranium hexafluoride: $238.03+6\times 18.998=352.03\text{g/mol}$ (polyatomic)

Since the temperature and the Boltzmann constant are the same for all gases, the root mean square speed is inversely proportional to the square root of the molar mass:

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

The lighter the gas, the higher its root mean square speed. Comparing the molar masses of the gases, we find that neon is the lightest, followed by chlorine, and uranium hexafluoride is the heaviest. Therefore, the root mean square speeds will be:

$vrms(\text{mono})>vrms(\text{dia})>v_{rms}(\text{poly})$

Correct answer is (B)

The root-mean-square (r.m.s) speed of an ideal gas is given by the formula:

$v_{\mathrm{rms}}=\sqrt{\frac{3RT}{M}}$

where R is the gas constant, T is the temperature in Kelvin, and M is the molar mass of the gas.

In this case, we are given that the initial temperature is $200K$ and the final temperature is $800K$. Let the r.m.s speed at $200K$ be $v_0$, then:

$v_0=\sqrt{\frac{3R\cdot 200}{M}}$

Now, we want to find the r.m.s speed at $800K$, let's call this $v_1$:

$v_1=\sqrt{\frac{3R\cdot 800}{M}}$

Now, divide $v_1$ by $v_0$:

$\frac{v_1}{v_0}=\frac{\sqrt{\frac{3R\cdot 800}{M}}}{\sqrt{\frac{3R\cdot 200}{M}}}$

Simplify the expression:

$\frac{v_1}{v_0}=\sqrt{\frac{800}{200}}=\sqrt{4}=2$

So, $v_1=2v_0$.

Hence, the r.m.s speed of the gas at $800K$ will be 2 times the r.m.s speed of the gas at $200K$.