Question

What is the kinetic energy of 1 g ${O_2}$ at $47^\circ C$?
A. $1.24 \times {10^2}J$
B. $2.24 \times {10^2}J$
C. $1.24 \times {10^3}J$
D. $3.24 \times {10^2}J$

Answer 1

Hint:From the kinetic molecular theory of gases, the kinetic energy for n moles is given by $K.E = \dfrac{3}{2}nRT$. According to kinetic molecular theory of gas particles the average kinetic energy of the gas particles is proportional to the absolute temperature of the gas.

Complete step by step answer:
The mass of ${O_2}$ is 1g.
The temperature is $47^\circ C$
The temperature in degree Celsius is converted to kelvin scale.
$ \Rightarrow T = 47 + 273$
$ \Rightarrow T = 230K$
The formula to calculate the number of moles is shown below.
$n = \dfrac{m}{M}$
Where,
n is the number of moles
m is the mass
M is the molar mass
The molar mass of oxygen ${O_2}$ is 32 g/mol.
To calculate the moles of oxygen, substitute the values in the above equation.
$ \Rightarrow n = \dfrac{1}{{32}}$
According to the kinetic molecular theory of gases, the kinetic energy of n number of moles is given by the equation as shown below.
$K.E = \dfrac{3}{2}nRT$
Where,
K.E is the kinetic energy of n moles.
n is the number of moles.
R is the universal gas constant.
T is the temperature.
The value of universal gas constant (R) is 8.314 J ${K^{ - 1}}$
To calculate the kinetic energy substitute the given values in the equation.
$ \Rightarrow K.E = \dfrac{3}{2}\left( {\dfrac{1}{{32}}} \right)(8.314J{K^{ - 1}}mo{l^{ - 1}})(320K)$
$ \Rightarrow K.E = 124.71J$
$ \Rightarrow K.E = 1.24 \times {10^2}J$
Thus, the kinetic energy of 1 g ${O_2}$ at $47^\circ C$is $1.24 \times {10^2}J$.
Therefore, the correct option is A.

Note:
The kinetic molecular theory of gases is applied only when the gases particles have no definite volume but have defined mass. There is no intermolecular attraction or repulsion between gas particles. The average kinetic energy is similar for all gases at the given temperature.