Question

The kinetic energy of ‘N’ molecules of ${H_2}$ is $3J$ at $ - 73^\circ C$ the kinetic energy of the same sample of ${H_2}$ at $ - 127^\circ C$ is?

Answer 1

Hint: We have to know that the temperature and average kinetic energy are directly proportional to each other. As the temperature goes up, the kinetic energy also increases, so the molecules start to move at higher speed. The equation relating average kinetic energy and temperature is,
$KE \propto T$
Here, we represent the kinetic energy as KE.
We represent the temperature as T.

Complete answer:
As per the kinetic theory of gases, all gases are made of tiny atoms that move in straight lines until they chance upon another gas particle or item. The exchange of energy makes them move around quicker and catch each other more.
Kinetic energy is corresponding to the speed of the particles. As the speed of the impacting atoms increments, so does the total kinetic energy of the relative multitude of gas particles. It's really hard to gauge the speed of an individual gas particle. All things being equal, temperature can be utilized as a proportion of the average kinetic energy of the relative multitude of particles in the gas. As the gas atoms acquire energy and move quicker, the temperature goes up.
To decide the average kinetic energy of gas atoms, we need to know the temperature of the gas, the general gas steady (R), and Avogadro's number (NA). We can relate the temperature and the average kinetic energy of the molecules using the expression,
$K = \dfrac{3}{2}\dfrac{R}{{{N_a}}}T$
We can write the simple equation relating the kinetic energy and temperature as,
$KE \propto T$
$\dfrac{{K{E_1}}}{{K{E_2}}} = \dfrac{{{n_1}{T_1}}}{{{n_2}{T_2}}}$
Now we can substitute the known values we get,
$\dfrac{3}{{K{E_2}}} = \dfrac{{N \times 200}}{{N \times 400}}$
On simplification we get,
$KE = 6J$
The kinetic energy of ‘N’ molecules of ${H_2}$ is $3J$ at $ - 73^\circ C$ the kinetic energy of the same sample of ${H_2}$ at $ - 127^\circ C$ is $6\,J$.

Note:
 We have to know that while performing the calculations using temperature. We have to convert the value of degree Celsius into kelvin. We can use the conversion factor to carry out this conversion. The equation for converting degree Celsius into kelvin is $K = ^\circ C + 273.15\,$. If we don’t convert the value of temperature in kelvin, there would be a possible error in the final answer.