Question

The kinetic energy of one mole gas at 300 K temperature is $E$ . At 400 K temperature, the kinetic energy is $E'$ . What is the value of $\dfrac{{E'}}{E}$ ?
(A) $1.33$
(B) $\sqrt {\dfrac{4}{3}} $
(C) $\dfrac{{16}}{9}$
(D) $2$

Answer 1

Hint:The kinetic energy of $n$ moles of an ideal gas at a temperature of $T$ Kelvin is given by $K.E. = \dfrac{3}{2}nRT$ . Use this formula to calculate the various kinetic energies of the gas at different temperatures. Then, proceed further to calculate the required ratio.


Formula used:
The kinetic energy of $n$ moles of an ideal gas at a temperature of $T$ Kelvin is given by:
$K.E. = \dfrac{3}{2}nRT$


Complete answer:
Kinetic energy:
$K.E. = \dfrac{3}{2}nRT$ … (1)
Here $R$ is the ideal gas constant.
For one mole, $n = 1$ .
Calculating the kinetic energy of one mole gas at $T = 300{\text{ K}}$ using formula (1),
$K.E. = \dfrac{3}{2}R(300) = \dfrac{{900}}{2}R$
It is given that this value of kinetic energy is $E$ .
Hence, $E = \dfrac{{900}}{2}R$ . … (2)
Similarly, calculating the kinetic energy of one mole gas at $T = 400{\text{ K}}$ using formula (1),
$K.E. = \dfrac{3}{2}R(400) = \dfrac{{1200}}{2}R$
It is given that this value of kinetic energy is $E'$ .
Hence, $E' = \dfrac{{1200}}{2}R$ . … (3)
Dividing equation (3) by equation (2),
$\dfrac{{E'}}{E} = \dfrac{{\dfrac{{1200}}{2}R}}{{\dfrac{{900}}{2}R}}$
On simplifying further, we get:
$\dfrac{{E'}}{E} = \dfrac{{12}}{9} = \dfrac{4}{3} = 1.33$
Hence, $\dfrac{{E'}}{E} = 1.33$ .
Thus, the correct option is A.



Note: The given question can be easily solved using the formula to calculate the kinetic energy of an ideal gas. This formula is given by $K.E. = \dfrac{3}{2}nRT$ , where $n$ is the number of moles of the ideal gas, $T$ is the temperature in Kelvin and $R$ is the ideal gas constant having a value of $8.314{\text{ J}}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}$