Question

Kinetic energy per molecule of a gas at N.T.P. is: $ \left( {k = 1.38 \times {{10}^{ - 23}}J/K} \right) $
(A) $ 5.65 \times {10^{ - 21}}J $
(B) $ 1.38 \times {10^{ - 21}}J $
(C) $ 2.07 \times {10^{ - 23}}J $
(D) $ 3.77 \times {10^{ - 23}}J $

Answer 1

Hint : To solve this question, we need to use the formula of the kinetic energy of a molecule of a gas. Then we have to put the value of the normal temperature and the Boltzmann’s constant in that to get the final answer.

Formula used: The formula used to solve this question is given by
 $ E = \dfrac{3}{2}kT $ , here $ E $ is the kinetic energy of a gas molecule at an absolute temperature of $ T $ , $ k $ is the Boltzmann’s constant.

Complete step by step answer:
We know that the kinetic energy of a molecule of gas due to its thermal motion is given by
 $ E = \dfrac{3}{2}kT $ (1)
Now, according to the question the gas is at the normal temperature and pressure. We know that the normal pressure is equal to $ 1atm $ , and the normal temperature is equal to $ {20^ \circ }C $ . Therefore we have
 $ T = {20^ \circ }C $
We know that $ K{ = ^ \circ }C + 273 $ . So we gave
 $ T = \left( {20 + 273} \right)K $
 $ T = 293K $ (2)
Also, according to the question, we have the value of the Boltzmann’s constant as
 $ k = 1.38 \times {10^{ - 23}}J/K $ (3)
Substituting (2) and (3) in (1) we get
 $ E = \dfrac{3}{2} \times 1.38 \times {10^{ - 23}} \times 293 $
On solving we get
 $ E = 6.06 \times {10^{ - 21}}J $
From the options given in the question, the nearest matching value of the kinetic energy is equal to $ 5.65 \times {10^{ - 21}}J $ .
Hence, the correct answer is option (A).

Note:
The value of the kinetic energy per molecule is the same for all the gases. It does not depend on the atomicity of the gas. Also, do not confuse between the values of the standard temperature and pressure (STP), and the normal temperature and pressure (NTP). The standard pressure is the same as the normal pressure, which is equal to one atmosphere. But the standard temperature is equal to $ {0^ \circ }C $ or $ 273K $ , while the normal temperature is equal to $ {20^ \circ }C $ or $ 293K $ .