Question

At what temperature is the kinetic energy of a gas molecule double that of its value of 27°C

• A. 54°C
• B. 300 K
• C. 327°C
• D. 108°C

Answer 1

Answer: (C)

327°C

Answer 2

$\frac{E_{1}}{E_{2}} = \frac{T_{1}}{T_{2}}$Ž$\frac{E}{2E} = \frac{(273 + 27)}{T_{2}}$Ž$T_{2} = 600K = 273{^\circ}C$