Question

The temperature at which the average speed of oxygen molecules is double that of the same molecules at ${0^o}C$ is
A.$546K$
B.$1092K$
C.$277K$
D.$1911K$

Answer 1

The average speed of gas molecules is otherwise known as the mean speed of gases. And here, the average speed of oxygen can be found on the bases of root mean square velocity which is represented as, ${u_{rms}}$. Here, the average speed of oxygen molecules doubles. Hence, we need to take the ratio of average speed and find out the temperature in the second case.

Complete answer:
The temperature at which the average speed of oxygen molecules is double is not equal to $546K$Hence, option (A) is incorrect.
According to the question, the temperature of the oxygen molecule in the first case is equal to ${0^o}C$and it can be converted to kelvin. Hence it is equal to $273K$.
The formula used to find out the average speed of gas is equal to,
$\upsilon = \sqrt {\dfrac{{8RT}}{{\pi {\rm M}}}}$…… (1)
Where, R is equal to universal gas constant,
T is temperature of gas, and M is equal to molar mass.
By substituting the value of temperature in first equation,
$\upsilon = \sqrt {\dfrac{{8R \times 273}}{{\pi {\rm M}}}}$
Consider, temperature in second case is equal to ${T_2}$and the average speed is equal to,
$\upsilon = \sqrt {\dfrac{{8R \times {T_2}}}{{\pi {\rm M}}}}$
And the average speed is double in second case, hence, ${\upsilon _2} = 2{\upsilon _2}$and by substituting the value of average speed,
$\sqrt {\dfrac{{8R \times {T_2}}}{{\pi {\rm M}}}} = 2\sqrt {\dfrac{{8R \times 273}}{{\pi {\rm M}}}}$
Simplify the above equation,
$\sqrt {{T_2}} = 2\sqrt {273}$
Squaring on both sides,
${T_2} = {(2\sqrt {273} )^2}$
Thus, by calculating will get the value of${T_2}$,
${T_2} = 2 \times 273$
${T_2} = 1092K$
Hence, option (B) is correct.
The value of temperature in the second case will not be equal to $277K$. Hence, option (C) is incorrect.
The temperature in the second case, ${T_2}$ is not equal to $1911K$ at which the average speed of oxygen molecules is doubled. Hence, the option (D) is incorrect.

Note:
The root means square velocity, otherwise it is known as rms velocity which is represented as ${u_{rms}}$. And this root mean square velocity is always directly proportional to$\sqrt T$ and which is inversely proportional to the molar mass. Therefore, the average velocity increases with increasing the value of temperature.