JEE MAIN - Physics (2023 - 8th April Evening Shift - No. 13)
The temperature at which the kinetic energy of oxygen molecules becomes double than its value at $$27^{\circ} \mathrm{C}$$ is
$$627^{\circ} \mathrm{C}$$
$$927^{\circ} \mathrm{C}$$
$$327^{\circ} \mathrm{C}$$
$$1227^{\circ} \mathrm{C}$$
توضیح
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:
$T_f = 600K - 273 = 327 ^\circ C$