Given below are two statements : Statement I : The temperature of a gas is -73^\circ C. When the gas is heated to 527^\circ 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 option given below :

Statement I is true but Statement II is false

Both Statement I and Statement II are true

Statement I is false but Statement II is true

Both Statement I and Statement II are false

Heat and Thermodynamics PYQ — JEE Main 2023 · 24 Jan · Shift 1 · Q59 with Solution

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Correct answer: A

Check Statement I

For an ideal gas, the root mean square speed is $v_{\text{rms}} = \sqrt{\frac{3kT}{m}}$ So, $v_{\text{rms}} \propto \sqrt{T}$ where $T$ must be in kelvin.

Initial temperature: $T_1 = -73^\circ C = 200\,K$
Final temperature: $T_2 = 527^\circ C = 800\,K$

Therefore, $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{800}{200}} = \sqrt{4} = 2$

Hence, the rms speed is doubled.

So, Statement I is true.

Check Statement II

For an ideal gas, $PV = nRT$ Also, the total translational kinetic energy of $n$ moles of an ideal gas is $K = \frac{3}{2}nRT$ Thus, $K = \frac{3}{2}PV$

So, $PV$ is not equal to the translational kinetic energy; rather, $PV = \frac{2}{3}K$

Hence, Statement II is false.

Conclusion

Statement I: True
Statement II: False

Therefore, the correct option is: $\boxed{\text{A}}$

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