Question

the temperature at which the root mean square velocity of SO2 molecules is the same as that of O2 at 27C is

Answer 1

Answer: At 327K, the root mean square velocity of both the gases will be equal.

Explanation: Root mean square velocity is related to the temperature and molar mass of the gas. It's expression is given by:

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

Where,

R = Gas constant

T = Temperature (in Kelvin)

M = Molar mass of gas

Now, we need to find the temperature at which root mean square velocity of $SO_2$ and $O_2$ is same

$V_{rms}_1=V_{rms}_2$

Squaring and cancelling the terms on both the sides.

$\sqrt{\not{3}\not{R}\frac{T_1}{M_1}}=\sqrt{\not{3}\not{R}\frac{T_2}{M_2}}$

$\frac{T_1}{M_1}=\frac{T_2}{M_2}$

$M_1=\text{Molar mass of }SO_2=64g/mol$

$M_2=\text{Molar mass of }O_2=32g/mol$

$T_2=\text{Temperature at which }O_2\text{ is present}=27\°C=(27+273)K=300K$

$T_1=?K$

Putting values in above equation:

$\frac{T_1}{64g/mol}=\frac{300K}{32g/mol}$

$T_1=600K$

$T_1=327 \°C$

This is the temperature of $SO_2$ at which root mean square vales of both the gases will be equal.

Answer 2

Answer:

Explanation:

See this image below to get the answer