At what temperature the kinetic energy of a gas molecule is half of the value at $27^{\circ} \mathrm

Question

At what temperature the kinetic energy of a gas molecule is half of the value at $27^{\circ} \mathrm{C}$ ?\

Professor PC Thomas Classes · Answer

Kinetic energy of a gas molecule $\mathrm{K}=\frac{3}{2} \mathrm{kT}$ where k is the Boltzmann's constant and T is the temperature in kelvin.\ $	herefore \frac{\mathrm{K}}{\mathrm{K}^{\prime}}=\frac{\mathrm{T}}{\mathrm{T}^{\prime}}$ or $\frac{\mathrm{K}}{\mathrm{K} / 2}=\frac{300}{\mathrm{~T}^{\prime}}$\ or $2=\frac{300}{\mathrm{~T}^{\prime}}$ or $\mathrm{T}^{\prime}=150 \mathrm{~K}=-123^{0} \mathrm{C}$.\