The root mean square speed of hydrogen molecules of an ideal hydrogen gas kept in a gas chamber at 0°C is 3180 m/s. The pressure on the hydrogen gas is (Density of hydrogen gas is 8.99x10⁻² kg/m³ , 1 atmosphere = 1.01x10⁵ N /m² )

The correct option is 3.0 atm.

To find the pressure on the hydrogen gas, we can use the formula relating the root mean square (rms) speed of gas molecules to the pressure and density of the gas:

$v_{\mathrm{rms}}=\sqrt{\frac{3P}{\rho }}$

Where:
• $v_{\mathrm{rms}}$ is the root mean square speed of the gas molecules ($3180\text{m/s}$).
• $P$ is the pressure on the gas.
• $\rho$ is the density of the gas ($8.99\times {10}^{-2}{\text{kg/m}}^3$).

Squaring both sides of the equation gives:
$v_{\mathrm{rms}}^2=\frac{3P}{\rho }$

Rearranging the equation to solve for the pressure ($P$):
$P=\frac{\rho v_{\mathrm{rms}}^2}{3}$

Substituting the given values into the formula:
$P=\frac{(8.99\times {10}^{-2})\times {\left(3180\right)}^2}{3}$

Calculating the square of the speed:
${\left(3180\right)}^2=10112400{\text{m}}^2/{\text{s}}^2=1.01124\times {10}^7{\text{m}}^2/{\text{s}}^2$

Now, compute the pressure in units of ${\text{N/m}}^2$:
$P=\frac{8.99\times {10}^{-2}\times 1.01124\times {10}^7}{3}$
$P=\frac{9.091\times {10}^5}{3}$
$P\approx 3.03\times {10}^5{\text{N/m}}^2$

To convert this pressure into atmospheres (atm), divide by the value of 1 atmosphere ($1\text{atm}=1.01\times {10}^5{\text{N/m}}^2$):
$P_{\text{atm}}=\frac{3.03\times {10}^5{\text{N/m}}^2}{1.01\times {10}^5{\text{N/m}}^2}$
$P_{\text{atm}}\approx 3.0\text{atm}$