Question

The Kinetic energy of 1g molecules of a gas, at normal temperature and pressure, is?
A) $0.56 \times {10^4}J$
B) $2.7 \times {10^2}J$
C) $1.3 \times {10^2}J$
D) $3.4 \times {10^3}J$

Answer 1

Hint: Gas is considered to be made up of molecules that are in random motion. These molecules undergo collisions with one another and they also collide with the walls of the container. The velocity of the molecules increases with the increase in temperature, which in turn leads to more collisions. The kinetic energy of the molecule depends on temperature. The gas is at normal pressure, and the temperature of the gas is also given as normal.

Formula used:
The kinetic energy of the gas,
$K.E. = \dfrac{3}{2}RT$
Where $K.E.$ is the kinetic energy of the gas, $R$ stands for the universal gas constant, $T$ stands for the temperature of the gas.

Complete step by step solution:
We use the kinetic energy of gas that we derive from the ideal gas equation, to solve this problem
The normal temperature of the gas is given, $T = 273K$
The value of universal gas constant $R = 8.314J.mo{l^{ - 1}}{K^{ - 1}}$
The kinetic energy of the gas, $K.E. = \dfrac{3}{2}RT$
Substituting the values of $T$and$R$in the equation of kinetic energy,
$K.E. = \dfrac{3}{2} \times 8.3145 \times 273$
$K.E. = 3.40 \times {10^{ - 3}}J $

Therefore, The answer is option (D), $3.4 \times {10^3}J.$

Note: The number of molecules per unit volume is the same for all gases at fixed temperature and pressure, this is the Avogadro hypothesis. That number is given by the Avogadro number, ${N_A} = 6.02 \times {10^{23}}$. The amount of substance containing $6.02 \times {10^{23}}$ molecules is called one mole. The mass in grams of $22.4$ liters of any gas at S.T.P is called the gram molecular weight. The value of the universal gas constant is very important in the kinetic theory of gases, the value should be known to do problems. The internal energy of the gas is the total kinetic energy of the molecules.