Question

The average velocity of an ideal gas at $0^{o}C$ is $0.4$ m/s. If the temperature of gas is increased to ${546}^{o}C$, its average velocity will be:

• A. 0.8 m/s
• B. 1.6m/s
• C. 0.346m/s
• D. 0.69m/s

Answer 1

Answer: (D)

0.69m/s

Solution:- (D) $0.69m/s$

As we know that,

$V_{av.}=\sqrt{\frac{8RT}{\pi m}}$

For same gas at different temperature,

$V_{av.}\propto \sqrt{T}$

Given:-

$V$ at $0℃=0.4m/s$

$V$ at $546℃=V=?$

$T_1=0℃=273K$

$T_2=546℃=(546+273)K=819$

Therefore,

$\frac{0.4}{V}=\sqrt{\frac{273}{819}}$

$\Rightarrow V=\sqrt{3}\times 0.4=0.69m/s$

Hence the average velocity at $546℃$ is $0.69m/s$.