According to Graham's law at a given temperature, the ratio of the rates of diffusion \[\;\gamma A/\gamma B\] of gases A and B is given by
A.\[\mathop {\dfrac{{PA}}{{PB}}.(\dfrac{{MA}}{{MB}})}\nolimits^{1/2} \]
B.\[\mathop {\dfrac{{MA}}{{MB}}.(\dfrac{{PA}}{{PB}})}\nolimits^{1/2} \]
C.\[\mathop {\dfrac{{PA}}{{PB}}.(\dfrac{{MB}}{{MA}})}\nolimits^{1/2} \]
D.\[\sqrt {\dfrac{1}{M}} \]
Answer
621.6k+ views
Hint: We must know that Graham's law is a gas law that makes narration of the rate of diffusion or effusion of a gas to its molar mass. We can use Graham's law generally to find how much faster one gas efferves than another.
Complete step by step answer:
We must understand that Graham's law states that the rate at which a gas effuses or diffuses is inversely proportional to the square root of the molar masses of that gas.
So, this law states that lighter the gas, faster and quicker will be the effusion or diffusion and vice versa.
According to Graham's law at a given temperature
\[\gamma \propto P\sqrt {\dfrac{1}{M}} \]
Where,
P = the pressure of gas
M = molar mass of the gas
Now, consider two gases, A and B
So, their rate of effusion will be \[\gamma A\] and \[\gamma B\]respectively
∴The ratio of the rates of diffusion of gases A and B is given by,
\[\dfrac{{\gamma A}}{{\gamma B}} = \dfrac{{PA}}{{PB}}\sqrt {\dfrac{{MB}}{{MA}}} \]
Therefore,
$\dfrac{{\gamma A}}{{\gamma B}} = \dfrac{{PA}}{{PB}}{\left( {\dfrac{{MB}}{{MA}}} \right)^{1/2}}$
Hence, option C is the correct option.
In Graham's Law, generally we look more often to the rate of effusion than we will look at its speed. So, by considering this law, we will see about the amount of gas that diffuses or moves per unit time and not how fast the individual particles are moving.
Note:
We can use Graham's law only to evaluate the rate of diffusion or effusion of various gases at a constant temperature. At very high concentrations of gas molecules, Graham’s law breaks down. And we must know that the Diffusion is the process of slowly mixing two molecules of gases together. Effusion is the process that happens when a gas is allowed to escape through a small opening of its container.
Complete step by step answer:
We must understand that Graham's law states that the rate at which a gas effuses or diffuses is inversely proportional to the square root of the molar masses of that gas.
So, this law states that lighter the gas, faster and quicker will be the effusion or diffusion and vice versa.
According to Graham's law at a given temperature
\[\gamma \propto P\sqrt {\dfrac{1}{M}} \]
Where,
P = the pressure of gas
M = molar mass of the gas
Now, consider two gases, A and B
So, their rate of effusion will be \[\gamma A\] and \[\gamma B\]respectively
∴The ratio of the rates of diffusion of gases A and B is given by,
\[\dfrac{{\gamma A}}{{\gamma B}} = \dfrac{{PA}}{{PB}}\sqrt {\dfrac{{MB}}{{MA}}} \]
Therefore,
$\dfrac{{\gamma A}}{{\gamma B}} = \dfrac{{PA}}{{PB}}{\left( {\dfrac{{MB}}{{MA}}} \right)^{1/2}}$
Hence, option C is the correct option.
In Graham's Law, generally we look more often to the rate of effusion than we will look at its speed. So, by considering this law, we will see about the amount of gas that diffuses or moves per unit time and not how fast the individual particles are moving.
Note:
We can use Graham's law only to evaluate the rate of diffusion or effusion of various gases at a constant temperature. At very high concentrations of gas molecules, Graham’s law breaks down. And we must know that the Diffusion is the process of slowly mixing two molecules of gases together. Effusion is the process that happens when a gas is allowed to escape through a small opening of its container.
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