Question

find the ratio of RMS velocity of Methane gas at 27°C and oxygen gas at 32°F​

Answer 1

Answer:

3:4 is your answer

Answer 2

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

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}}$

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

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

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

$T_2=[\text]{Temperature at which }.</p><p> [tex]O_2\text]{ is present}=27\°C=(27+273)K=300K[tex]$

$T_1=?K$

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.