Question

The noble gases neon (atomic mass 20.1797 u) and krypton (atomic mass 83.798 u) are accidentally mixed in a vessel that has a temperature of 79.2°C. What are the average kinetic energies and rms speeds of neon and krypton molecules in the vessel?

(a) average kinetic energies Kav, Ne = J Kav, Kr = J

(b) rms speeds vrms, Ne = m/s vrms, Kr = m/s

Answer 1

Answer:

(a) Kav Ne = Kav Kr = 7.29x10⁻²¹J

(b) v(rms) Ne= 659.6m/s and v(rms) Kr= 323.7m/s

Explanation:

(a) According to the kinetic theory of gases the average kinetic energy of the gases can be calculated by:

$K_{av} = \frac{3}{2}kT$ (1)        

where $K_{av}$: is the kinetic energy, k: Boltzmann constant = 1.38x10⁻²³J/K, and T: is the temperature

From equation (1), we can calculate the average kinetic energies for the krypton and the neon:

$K_{av} = \frac{3}{2} (1.38\cdot 10^{-23} \frac{J}{K})(352.2K) = 7.29\cdot 10^{-21}J$  

(b) The rms speeds of the gases can be calculated by:

$K_{av} = \frac{1}{2}mv_{rms}^{2} \rightarrow v_{rms} = \sqrt \frac{2K_{av}}{m}$  

where m: is the mass of the gases and $v_{rms}$: is the root mean square speed of the gases

For the neon:

$v_{rms} = \sqrt \frac{2(7.29 \cdot 10^{-21}J)}{20.1797 \frac{g}{mol} \cdot \frac {1mol}{6.022\cdot 10^{23}molecules} \cdot \frac{1kg}{1000g}} = 659.6 \frac{m}{s}$          

For the krypton:

$v_{rms} = \sqrt \frac{2(7.29 \cdot 10^{-21}J)}{83.798 \frac{g}{mol} \cdot \frac {1mol}{6.022\cdot 10^{23}molecules} \cdot \frac{1kg}{1000g}} = 323.7 \frac{m}{s}$  

Have a nice day!