A specific gas with a molecular diameter of 2 10⁻¹⁰ m is maintained at a pressure of 1.38 10^5 Pa . If the mean free path of the gas molecules is measured to be 10⁻⁶ 2 m , the absolute temperature of the gas is: (Given k_B = 1.38 10⁻²³ J/K )

Step-by-step solution

The mean free path of a gas molecule is given by the formula: = k_B T 2 d^2 P Rearranging the formula to solve for the absolute temperature T : T = 2 d^2 P k_B Substituting the given values: d = 2 10⁻¹⁰ m P = 1.38 10^5 Pa = 10⁻⁶ 2 m k_B = 1.38 10⁻²³ J/K T = 2 (2 10⁻¹⁰)^2 1.38 10^5 ( 10⁻⁶ 2 ) 1.38 10⁻²³ The terms 2 and 1.38 cancel out: T = 4 10⁻²⁰ 10^5 10⁻⁶ 10⁻²³ T = 4 10⁻²¹ 10⁻²³ = 4 10^2 = 400 K Answer: 400 K