Question

Calculate the temperature at which the root mean square velocity of chlorine gas molecules is twice the root mean square velocity of carbon dioxide gas molecules at ${40}^{0}C$.

• A. $500K$
• B. $1000K$
• C. $1500K$
• D. $2000K$

Answer 1

The correct option is D $2000K$

The expression for the root mean square velocity is $u=\sqrt{\frac{3RT}{M}}$

For carbon dioxide gas

$u=\sqrt{\frac{3R\times (40+273)}{44}}$

$u=\sqrt{3R\times 7.114}$ ...... (1)

For chlorine gas

$u=\sqrt{\frac{3RT}{71}}$ ..... (2)

The root mean square velocity of chlorine gas molecules is twice The root mean square velocity of carbon dioxide gas molecules.

From (1) and (2)

$\sqrt{\frac{3RT}{71}}=2\times \sqrt{3R\times 7.114}$

$\frac{3RT}{71}=4\times 3R\times 7.114$

$\frac{T}{71}$ $=4\times 7.114$

$T=71\times 4\times 7.114$

$T=2020K\simeq 2000$ K