Question

The average velocity of an ideal gas at $0^{o}C$ is $0.4{ms}^{-1}$. If the temperature of the gas is increased to ${546}^{o}C$. The average velocity of the gas will be:

• A. $0.8{ms}^{-1}$
• B. $1.6{ms}^{-1}$
• C. $0.346{ms}^{-1}$
• D. $0.69{ms}^{-1}$

Answer 1

The correct option is D $0.69{ms}^{-1}$

We have,

Average Kinetic Energy ($K{E}_{av}$) of an ideal is directly proportional to temperature ($T$).

That is, $K{E}_{av}\alpha T$

But, we know that $K{E}_{av}=\frac{1}{2}m{V_{av}}^2$ where, $V_{av}$ - average velocity

Therefore, ${V_{av}}^2\alpha T$

$\Rightarrow V_{av}\alpha \sqrt{T}$

Therefore, $\frac{V_{av1}}{V_{av2}}=\sqrt{\frac{T_1}{T_2}}$

Given that,

$T_1=0^{o}C=(0+273)K=273K$

$V_{av1}=0.4m{s}^{-1}$

$T_2={546}^{o}C=(546+273)K=819K$

$V_{av2}=?$

$\frac{V_{av1}}{V_{av2}}=\sqrt{\frac{T_1}{T_2}}=>\frac{0.4}{V_{av2}}=\sqrt{\frac{273}{819}}$

$=>\frac{V_{av2}}{0.4}=\sqrt{\frac{819}{273}}=\sqrt{3}$

$=>V_{av2}=0.4\sqrt{3}=0.69m{s}^{-1}$

So, option D. $0.69m{s}^{-1}$is the correct answer.