Question

The temperature of a gas is increased from ${27}^{o}C$ to such an extent that its $rms$ speed be double the speed at ${27}^{o}C$.
The final temperature will be

• A. ${927}^{o}C$
• B. ${250}^{o}C$
• C. ${650}^{o}C$
• D. ${1200}^{o}C$

Answer 1

The correct option is A ${927}^{o}C$

Let temperature is $T$ at which rms speed of gas is doubled.

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

$\Rightarrow v=\sqrt{\frac{3R\times 300}{M}}$ ...1

$\Rightarrow 2v=\sqrt{\frac{3R\times T}{M}}$ ....2

Solving equation 1 and 2, we get $\Rightarrow T=1200K$

Hence, $T=1200-273={927}^{o}C$