Question

The temperature at which the mean KE of the molecules of gas is one-third of the mean KE of its molecules at ${180}^{o}C$ is

• A. $-{122}^{o}C$
• B. $-{90}^{o}C$
• C. ${60}^{o}C$
• D. ${151}^{o}C$

Answer 1

Answer: (D)

${151}^{o}C$

Kinetic energy of a gas is directly proportional to its temperature
$\therefore \frac{K_1}{K_2}=\frac{T_1}{T_2}$
$\frac{K}{K/3}=\frac{273+180}{T_2}\Rightarrow 3=\frac{453}{T_2}$
$T_2={151}^{o}C$