Question

The kinetic energy of $1 gram$ mole of the gas at normal temperature and the pressure is ($R = 8.31 J/mol – K$)
A) $0.56 \times 10^4 J$
B) $1.3 \times 10^2 J$
C) $2.7 \times 10^2 J$
D) $3.4 \times 10^3 J$

Answer 1

Hint
Here in the given question we have to find the kinetic energy of $1 gram$ mole of the gas at normal temperature and the pressure. So we simply use the formula-
Kinetic energy = $\dfrac{3}{2}RT$

Complete answer:
As we have to find the kinetic energy of the gas so we simply use the formula
K.E. = $\dfrac{3}{2}RT$
Here, K.E. is the kinetic energy of the gas,
$R$ = it is the gas constant,
The value of the R is given by $R = 8.31 J/mol- K$.
The normal temperature we take $T = 273K$
Hence putting the value in the equation we get
Kinetic energy = $\dfrac{3}{2}\; \times \;8.314 \times 273$
Hence solving the above equation we get the value that -
Kinetic energy = $3.4 \times 10^3 J$.
Correct option is (D).

Note
The ideal gas law also called the general gas equation is the equation of the state of a hypothetical ideal gas. It is a good approximation of the behavior of the many gases under much condition, although it has several limitations. The ideal equation state that-
$P V = nRT$
Here, $P$ = pressure of the gas
$V$ = volume of the gas
$N$ = it represent the number of the mole
$R$ = gas constant
The value of the $R$ is given by $R = 8.31 J/mol- K$
and $T$ = it represents the temperature.