At what temperature the velocity of O2 molecules have the same velocity as SO2 as 47C
Answers
Answer:
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}}V
rms
=
M
3RT
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_2SO
2
and O_2O
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}}
3
R
M
1
T
1
=
3
R
M
2
T
2
\frac{T_1}{M_1}=\frac{T_2}{M_2}
M
1
T
1
=
M
2
T
2
M_1=\text{Molar mass of }SO_2=64g/molM
1
=Molar mass of SO
2
=64g/mol
M_2=\text{Molar mass of }O_2=32g/molM
2
=Molar mass of O
2
=32g/mol
T_2=\text{Temperature at which }O_2\text{ is present}=27\°C=(27+273)K=300KT
2
=Temperature at which O
2
is present=27\°C=(27+273)K=300K
T_1=?KT
1
=?K
Putting values in above equation:
\frac{T_1}{64g/mol}=\frac{300K}{32g/mol}
64g/mol
T
1
=
32g/mol
300K
T_1=600KT
1
=600K
T_1=327 \°CT
1
=327\°C
This is the temperature of SO_2SO
2
at which root mean square vales of both the gases will be equal.