Question

At what temperature will the kinetic energy of gas molecules be double of its value at 27${}^o$C?

• A. ${54}^{o}C$
• B. ${108}^{o}C$
• C. ${300}^{o}C$
• D. ${327}^{o}C$

Answer 1

The correct option is D ${327}^{o}C$

Kinetic energy of gas $\left(E\right)=\frac{3}{2}\times k\times T$
Then
$\frac{E}{T}=k$

where $k$ is Boltzmann constant
$\frac{E_1}{T_1}=\frac{E_2}{T_2}$
Given
$E_1=E$ ; $E_2=2\times E$ ; $T_1={27}^{o}C=300K$ ; $T_2=?$
Then $T_2=E_2\times \frac{T_1}{E_1}=2\times E\times \frac{300}{E}=600K$
$T_2={327}^{o}C$