QJEEM23Ph841d: JEE Main 2023 Physics - Class 11 JEEM23Ph841d Solution: The temperature at which the kinetic energy of oxygen molecules …

The kinetic energy of an ideal gas is given by the equation:

$KE = \frac{3}{2} kT$

where (k) is Boltzmann's constant and (T) is the absolute temperature in kelvins. Therefore, the kinetic energy of a gas is directly proportional to its temperature.

If the kinetic energy doubles, the temperature must also double. The original temperature is given as ( $27^\circ C$ ), which is equal to (300 K) in absolute terms. Therefore, the final temperature ( $T_f$ ) in kelvins is:

$T_f = 2 \cdot 300 K = 600 K$

Converting this back to degrees Celsius gives:

$T_f = 600K - 273 = 327 ^\circ C$