Question

Calculate the temperature at which $\text{RMS}$ velocity of oxygen becomes $1.5$ times its value at $\text{NTP}$.

• A. $45{}^{\circ }\text{C}$
• B. $659.25{}^{\circ }\text{C}$
• C. $386.25{}^{\circ }\text{C}$
• D. $387{}^{\circ }\text{K}$

Answer 1

The correct option is C $386.25{}^{\circ }\text{C}$

$v_{rms}=\sqrt{\frac{3RT}{M}}$

Let $T_0$ be the temperature at $\text{NTP}$.

From question,

$\sqrt{\frac{3RT}{M}}=1.5\times \sqrt{\frac{3R{T}_0}{M}}$

$\Rightarrow T=2.25T_0=2.25\times 293$

$\Rightarrow T=659.25{}^{\circ }\text{K}$

$\therefore T=386.25{}^{\circ }\text{C}$

Hence, option $\left(\text{C}\right)$ is correct.