The root mean square speed of hydrogen molecules at 300 K is 1930 m/s. Then the root mean square speed of oxygen molecules at 900 K will be

The correct option is 836 m/s.

Step-by-step Explanation:

The root mean square (rms) speed of a gas molecule is given by the formula:
$v_{\mathrm{rms}}=\sqrt{\frac{3RT}{M}}$
where:
- $R$ is the universal gas constant,
- $T$ is the absolute temperature in Kelvin,
- $M$ is the molar mass of the gas.

From this relation, we can see that the rms speed is directly proportional to the square root of the temperature and inversely proportional to the square root of the molar mass:
$v_{\mathrm{rms}}\propto \sqrt{\frac{T}{M}}$

Let the subscripts 1 and 2 represent hydrogen (${\text{H}}_2$) and oxygen (${\text{O}}_2$) respectively.
For hydrogen (${\text{H}}_2$):
- Molar mass, $M_1=2\text{g/mol}$
- Temperature, $T_1=300\text{K}$
- RMS speed, $v_{\mathrm{rms},1}=1930\text{m/s}$

For oxygen (${\text{O}}_2$):
- Molar mass, $M_2=32\text{g/mol}$
- Temperature, $T_2=900\text{K}$

Taking the ratio of their rms speeds:
$\frac{v_{\mathrm{rms},2}}{v_{\mathrm{rms},1}}=\sqrt{\frac{T_2}{T_1}\cdot \frac{M_1}{M_2}}$

Substitute the given values into the ratio:
$\frac{v_{\mathrm{rms},2}}{1930}=\sqrt{\frac{900}{300}\cdot \frac{2}{32}}$

Simplify the expression under the square root:
$\frac{v_{\mathrm{rms},2}}{1930}=\sqrt{3\cdot \frac{1}{16}}=\frac{\sqrt{3}}{4}$

Solve for $v_{\mathrm{rms},2}$:
$v_{\mathrm{rms},2}=1930\cdot \frac{\sqrt{3}}{4}$

Using $\sqrt{3}\approx 1.732$:
$v_{\mathrm{rms},2}\approx 1930\cdot \frac{1.732}{4}=1930\cdot 0.433\approx 835.7\text{m/s}$

Rounding to the nearest whole number, the root mean square speed of oxygen molecules at 900 K is approximately 836 m/s.