Question

The total kinetic energy of $1$ mole of ${{N}_{2}}$ at $27{}^\circ C$ will be approximately:
A.$1500J$
B.$1500\text{ calorie}$
C.$1500\text{ kilocalorie}$
D.$1500\text{ erg}$

Answer 1

Hint: We know that Nitrogen is a diatomic molecule and hence we can find it kinetic energy by the relation of kinetic energy with the degree of freedom of Nitrogen, the number of moles, the universal gas constant and the temperature of the gas.
Formula used:
$K.E=\dfrac{f}{2}nRT$

Complete answer:
Let us first look at the formula to calculate the kinetic energy of Nitrogen gas:
$K.E=\dfrac{f}{2}nRT$
We first need to find out the degrees of freedom for the gas. The degrees of freedom of a gas is the total number of ways in which gas can take in the energy. Since Nitrogen is a diatomic molecule hence it has three translational degrees of freedom and two rotational degrees of freedom, hence we got the value of degree of freedom as:
$f=5$.
Now it is given that the number of moles of Nitrogen gas is:
$1$ mole,
The temperature of the gas in Kelvin scale will be:
$T=27+273=300K$,
And the value of universal gas constant can be taken as:
$R=2\text{ Cal}\cdot \text{mo}{{\text{l}}^{-1}}\cdot {{\text{K}}^{-1}}$.
Substituting all the values in the equation, we get the total kinetic energy as follows:
$\begin{align}
  & K.E=\dfrac{5}{2}\times 1\times 2\times 300 \\
 & \therefore K.E=1500\text{ calories} \\
\end{align}$
Therefore, the kinetic energy of $1$ mole of ${{N}_{2}}$ gas at $27{}^\circ C$ will be $1500\text{ calorie}$.

Thus the correct option is $B$.

Note:
The equation that we used in the solution only holds true for ideal gases and not for real gases. The assumptions to be considered while using this formula are that the particles of the gas must be non-interactive and the size of the particles of the gas is very small.