Question

At what temperature the translational kinetic energy of 14g of nitrogen will be the same as that of 32 gm of oxygen at 300 k

Answer 1

Translational kinetic energy of 14g of nitrogen can be calculated as :

$K.E.=\frac{3}{2}\times n\times R\times T$

Where K.E. Denotes kinetic energy, n denotes number of moles , R denotes gas constant, T denotes temperature.

As here 14 g of nitrogen is used so number of moles of nitrogen will be:

Number of moles of a substance=$\frac{Given mass of the substance}{Molar mass of the substance}$

Number of moles of nitrogen= $\frac{Given mass of the nitrogen}{Molar mass of the nitrogen}$

= $\frac{14}{14}$

=1 mol

As here 32 g of oxygen is used so number of moles of oxygen will be:

Number of moles of a substance= $\frac{Given mass of the substance}{Molar mass of the substance}$

Number of moles of oxygen= $\frac{Given mass of the oxygen}{Molar mass of the oxygen}$

= $\frac{32}{16}$

=2 mol

As translational kinetic energy of 14g of nitrogen will be the same as that of 32 gm of oxygen at 300 k

So , kinetic energy will be:

$\frac{3}{2}\times 1\times 8.314\times T=\frac{3}{2}\times 2\times 8.314\times 300$

So T comes out to be:

600 K

At 600 K, the translational kinetic energy of 14 g of nitrogen will be the same as that of 32 g of oxygen at 300 K.