91Ó°ÊÓ

The volume of the gas is\(V = 8.5\;{{\rm{m}}^3}\).

The pressure is\(P = 2.9\;{\rm{atm}}\).

The number of moles is\(n = 2100\;{\rm{mol}}\).

The molecular mass of nitrogen is \(M = 28\;{\rm{g/mol}} = 0.028\;{\rm{kg/mol}}\).

The root-mean-square speed of gas molecules depends on the molecular mass and absolute temperature of the gas.

The root-mean-square speed is given as follows:

\({v_{{\rm{rms}}}} = \sqrt {\frac{{3RT}}{M}} \) … (i)

Here, R is the universal gas constant, T is the temperature, and M is the molecular mass.

The ideal gas equation is given by the following equation:

\(\begin{aligned}{c}PV &= nRT\\T &= \frac{{PV}}{{nR}}\end{aligned}\)

Here, P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the temperature.

Substitute the values in the above equation.

\(\begin{aligned}{l}T &= \frac{{2.9\;{\rm{atm}} \times \frac{{1.013 \times {{10}^5}\;{\rm{Pa}}}}{{1\;{\rm{atm}}}} \times 8.5\;{{\rm{m}}^{\rm{3}}}}}{{2100\;{\rm{mol}} \times 8.314\;{\rm{J/mol - K}}}}\\T &= 143.02\;{\rm{K}}\end{aligned}\)

The rms speed is given by the following equation:

\({v_{{\rm{rms}}}} = \sqrt {\frac{{3RT}}{M}} \)

Here, Mis the molecular mass.

Substitute the values in the above equation.

\(\begin{aligned}{l}{v_{{\rm{rms}}}} &= \sqrt {\frac{{3\left( {8.314\;{\rm{J/mol}} \cdot {\rm{K}}} \right) \times 143.02\;{\rm{K}}}}{{0.028\;{\rm{kg/mol}}}}} \\{v_{{\rm{rms}}}} &= 356.93\;{\rm{m/s}}\end{aligned}\)

Thus, the rms speed of nitrogen molecules is \(356.93\;{\rm{m/s}}\).