Question

Kinetic energy possessed by 1g of helium gas at NTP is:

1. 850.7J
2. 750.7J
3. 831J
4. 663J

Answer 1

Here the kinetic energy of the gaseous atom is given in relation with the Boltzmann constant${K_B}$. Here, we have to find out the kinetic energy in at normal temperature and pressure and to do that put ${K_B} = nRT = PV$and solve for K.E.

Complete step by step solution:
Find the Kinetic energy:
The formula for the kinetic energy possessed by a gaseous atom.
$K = \dfrac{3}{2}{K_B}T$;
K = Kinetic Energy;
${K_B}$= Boltzmann constant;
T = Temperature;
The internal energy of an ideal monatomic gas U is just the kinetic energy. So,
$U = \dfrac{3}{2}{K_B}T = \dfrac{3}{2}nRT$;
Here:
U = Internal Change in the energy;
n = Number of moles;
R = Gas Constant;
T = Temperature;

We know that PV = nRT, Put this relation in the above equation:
$U = \dfrac{3}{2}PV$; ….(Here P = Pressure; V = Volume)
For 1g of Helium atom we have to divide it by its mass M (M = 4):
$U = \dfrac{3}{2}\dfrac{{PV}}{M}$;
At Normal temperature and pressure, the volume of the gas and the pressure of the gas is:
$P = 1.013 \times {10^5}N/{m^2}$;
$V = 22.4l$;
Put the above value in the equation $U = \dfrac{3}{2}\dfrac{{PV}}{M}$.
$U = \dfrac{3}{2}\left( {\dfrac{{1.013 \times {{10}^5} \times 22.4 \times {{10}^{ - 3}}}}{4}} \right)$;
Do the needed calculation:
$\Rightarrow U = 850.92J$;
$\Rightarrow U \approx 850.7J$;

Option “1” is correct. Kinetic energy possessed by 1g of helium gas at NTP is $850.7J$.

Note: Here we need to know that for ideal gas the internal change in the energy is only in terms of kinetic energy and not potential energy as there would be no energy added in due to the rotational, vibrational and intermolecular interactions.