At \(25^{\circ} \mathrm{C}\) (approximately room temperature) the rms velocity of an Ar atom in air is \(1553 \mathrm{~km} / \mathrm{h} .\) (a) What is the rms speed in \(\mathrm{m} / \mathrm{s}\) ? (b) What is the kinetic energy (in J) of an Ar atom moving at this speed? (c) What is the total kinetic energy of $1 \mathrm{~mol}$ of Ar atoms moving at this speed?

First, we need to convert the given velocity from km/h to m/s. To do this, we use the following conversion factors: 1 km = 1000 meters (m) 1 hour = 3600 seconds (s) So, we can write the expression to convert the given velocity: \(v_{rms} = 1553 \cdot \frac{1000 \, \text{m}}{3600 \, \mathrm{s}}\)

Now, calculate the new value for the rms velocity: \(v_{rms} = 1553 \cdot \frac{1000}{3600} = 431.39 \, \text{m/s}\) Thus, the rms speed of an Ar atom at room temperature in meters per second is approx. 431.39 m/s.

We now need to calculate the kinetic energy of an Ar atom. We use the formula for kinetic energy: \(K.E = \frac{1}{2}mv^2\) where m is the mass of the Ar atom and v is the rms velocity. The atomic mass of Ar is about 39.95 g/mol. We need to first convert it to kg: \(m = \frac{39.95}{1000} = 0.03995 \, \mathrm{kg}\) Note that this mass corresponds to one mole of Ar atoms. To find the mass of a single atom, we can use Avogadro's number (6.022 × 10^23 atoms/mol): \(m_{atom} = \frac{0.03995 \, \mathrm{kg}}{6.022 \times 10^{23} \, \mathrm{atoms}}\) Now, we can plug in the value of the mass of an Ar atom and the rms velocity into the kinetic energy formula: \(K.E = \frac{1}{2}(m_{atom})(431.39 \, \mathrm{m/s})^2\)

Lastly, to find the total kinetic energy of one mole of Ar atoms, we multiply the kinetic energy of a single atom by Avogadro's number: \(K.E_{total} = K.E \times 6.022 \times 10^{23}\) Now, calculate the values and find the kinetic energy of one Ar atom and one mole of Ar atoms.