Sign in
Please select an account to continue using cracku.in
↓ →
Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
Given below are two statements:
Statement I: The temperature of a gas is -73 °C. When the gas is heated to 527 °C, the root mean square speed of the molecules is doubled.
Statement II: The product of pressure and volume of an ideal gas will be equal to translational kinetic energy of the molecules.
In the light of the above statements, choose the correct answer from the options given below:
Statement I: At $$T_1 = -73°C = 200$$ K and $$T_2 = 527°C = 800$$ K.
Since $$v_{rms} \propto \sqrt{T}$$: $$\frac{v_2}{v_1} = \sqrt{\frac{800}{200}} = \sqrt{4} = 2$$
The rms speed is doubled. TRUE.
Statement II: PV = nRT and translational KE = $$\frac{3}{2}nRT$$.
So $$PV = \frac{2}{3} \times KE_{translational}$$, NOT equal to KE.
FALSE.
Therefore, Statement I is true but Statement II is false.
Click on the Email ☝️ to Watch the Video Solution
Educational materials for JEE preparation