Question

The kinetic energy of n moles of nitrogen gas at 127 is (R= 2 cal $mo{{l}^{-1}}$ ${{K}^{-1}}$) is
a) 4400 cal
b) 3200 cal
c) 4800 cal
d) 1524 cal

Answer 1

According to kinetic energy of gases, the average kinetic energy of the molecules of a gas is directly proportional to the absolute temperature of the gas.
Therefore, from the kinetic gas equation, we derive kinetic energy of a gas containing n molecules at temperature T as
$K=\dfrac{3}{2}nRT$
where, R is gas constant.

Complete answer:
We know the expression for the kinetic energy of a gas containing n molecules, each of mass m, moving with root mean square velocity, u is given as
$K=\dfrac{1}{2}mn{{u}^{2}}$
We can find the kinetic energy of a gas by modifying the kinetic gas equation given for a gas containing n molecules at pressure, P and volume, V
$PV=\dfrac{1}{3}mn{{u}^{2}}$
Now, diving the above two equations, we get
$\begin{aligned} & \dfrac{K}{PV}=\dfrac{3mn{{u}^{2}}}{2mn{{u}^{2}}} \\\ & K=\dfrac{3}{2}PV \\\ \end{aligned}$
Further, n mole of an ideal gas, we have, $PV=nRT$
Substituting the $PV=nRT$ in the above expression, we obtain
$K=\dfrac{3}{2}nRT$
We can now calculate the kinetic energy of the 4 moles of ${{N}_{2}}$ gas at temperature 127$^{o}C$.
To convert temperature in Celsius ($^{o}C$) to Kelvin (K),
T (K) =T ($^{o}C$ ) + 273.15
T(K) = 127 + 273.15
= 400.15 K
Let us take T= 400 K for our convenience.
Given value of the gas constant, R = 2 cal $mo{{l}^{-1}}$ ${{K}^{-1}}$
Substituting n = 4, T = 400K and R = 2 cal $mo{{l}^{-1}}$ ${{K}^{-1}}$ in the kinetic energy expression, we get
$\begin{aligned} & K=\dfrac{3}{2}nRT \\\ & K=\dfrac{3}{2}\times 4mol\times 2calmo{{l}^{-1}}{{K}^{-1}}\times 400K \\\ & K=4800cal \\\ \end{aligned}$

Therefore, the correct option is (C).

Note:
Note that molecules in a gas are moving with different velocities and hence, have different kinetic energies. However, the average kinetic energy of a gas depends only on the absolute temperature of the gas. It is independent of the nature of the gas.