Question

The Kinetic energy of 'N' molecules of ${\mathrm{H}}_2\mathrm{is}3\mathrm{J}\mathrm{at}-{73}^0\mathrm{C}$ the Kinetic energy of the same sample of ${\mathrm{H}}_2\mathrm{at}-{127}^0\mathrm{C}$ is ?

Answer 1

Step 1: Given data:

The Kinetic energy of 'N' molecules of ${\mathrm{H}}_2$is given as $\mathrm{K}.{\mathrm{E}}_1=3\mathrm{J}$

The initial temperature is given as = ${\mathrm{T}}_1=-{73}^{°}\mathrm{C}=-73+273=200\mathrm{K}$

The final temperature is given as = ${\mathrm{T}}_2=-{127}^{°}\mathrm{C}=-127+273=146\mathrm{K}$

The Kinetic energy of the same sample of ${\mathrm{H}}_2\mathrm{at}-{127}^{°}\mathrm{C}$ is given as $={\mathrm{K}.\mathrm{E}}_2$$=?$

Step 2: Formula for calculating kinetic energy:
Kinetic energy is basically corresponding to the speed of the particles.

We can relate the temperature and the average kinetic energy of the molecules by the expression,
$\mathrm{K}=\frac{3}{2}\frac{\mathrm{R}}{{\mathrm{N}}_{\mathrm{a}}}\mathrm{T}$

Where, $\mathrm{T}=$the temperature of the gas,

$\mathrm{R}=$ Gas constant,

${\mathrm{N}}_{\mathrm{A}}=$ Avogadro's number.

Step 3: Formula relating the kinetic energy and temperature :
The simple equation relating the kinetic energy and temperature can be given as:
$\mathrm{KE}\propto \mathrm{T}$
Hence,

$\frac{\mathrm{K}{\mathrm{E}}_1}{\mathrm{K}{\mathrm{E}}_2}=\frac{{\mathrm{n}}_1{\mathrm{T}}_1}{{\mathrm{n}}_2{\mathrm{T}}_2}$

Step 4: Calculating the kinetic energy of ${\mathrm{H}}_2$ at $-{127}^{°}C$:
By substituting all the given values, we can calculate the kinetic energy of ${\mathrm{H}}_2$ at $-{127}^{°}C$ as:
$\frac{3}{\mathrm{K}{\mathrm{E}}_2}=\frac{\mathrm{N}\times 200}{\mathrm{N}\times 146}$
$\Rightarrow {\mathrm{KE}}_2=2.19\mathrm{J}$
Thus the kinetic energy of ‘N’ molecules of ${\mathrm{H}}_2$ at $-{127}^{°}C$ is $2.19\mathrm{J}$.