Rate of Effusion for Second Gas by Graham's law Calculator

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Gaseous state

Physical spectroscopy

Vapour Pressure

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Important Formulas of Gaseous State

✖The Rate of Effusion of First Gas is the special case of diffusion when the first gas is allowed to escape through the small hole.ⓘ Rate of Effusion of First Gas [r₁]			+10% -10%
✖The Molar Mass of Second Gas is defined as the mass of the gas per mole.ⓘ Molar Mass of Second Gas [M₂]			+10% -10%
✖The Molar Mass of First Gas is defined as the mass of the gas per mole.ⓘ Molar Mass of First Gas [M₁]			+10% -10%

✖The Rate of Effusion of Second Gas is the special case of diffusion when the second gas is allowed to escape through the small hole.ⓘ Rate of Effusion for Second Gas by Graham's law [r₂]

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Formula

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Rate of Effusion for Second Gas by Graham's law

Formula

`"r"_{"2"} = "r"_{"1"}/(sqrt("M"_{"2"}/"M"_{"1"}))`

Example

`"2.772296m³/s"="2.12m³/s"/(sqrt("20.21g/mol"/"34.56g/mol"))`

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Rate of Effusion for Second Gas by Graham's law Solution

STEP 0: Pre-Calculation Summary

Formula Used

Rate of Effusion of Second Gas = Rate of Effusion of First Gas/(sqrt(Molar Mass of Second Gas/Molar Mass of First Gas))
r₂ = r₁/(sqrt(M₂/M₁))
This formula uses 1 Functions, 4 Variables

Functions Used

sqrt - A square root function is a function that takes a non-negative number as an input and returns the square root of the given input number., sqrt(Number)

Variables Used

Rate of Effusion of Second Gas - (Measured in Cubic Meter per Second) - The Rate of Effusion of Second Gas is the special case of diffusion when the second gas is allowed to escape through the small hole.
Rate of Effusion of First Gas - (Measured in Cubic Meter per Second) - The Rate of Effusion of First Gas is the special case of diffusion when the first gas is allowed to escape through the small hole.
Molar Mass of Second Gas - (Measured in Kilogram Per Mole) - The Molar Mass of Second Gas is defined as the mass of the gas per mole.
Molar Mass of First Gas - (Measured in Kilogram Per Mole) - The Molar Mass of First Gas is defined as the mass of the gas per mole.

STEP 1: Convert Input(s) to Base Unit

Rate of Effusion of First Gas: 2.12 Cubic Meter per Second --> 2.12 Cubic Meter per Second No Conversion Required
Molar Mass of Second Gas: 20.21 Gram Per Mole --> 0.02021 Kilogram Per Mole (Check conversion here)
Molar Mass of First Gas: 34.56 Gram Per Mole --> 0.03456 Kilogram Per Mole (Check conversion here)

STEP 2: Evaluate Formula

Substituting Input Values in Formula

r₂ = r₁/(sqrt(M₂/M₁)) --> 2.12/(sqrt(0.02021/0.03456))

Evaluating ... ...

r₂ = 2.77229582592878

STEP 3: Convert Result to Output's Unit

2.77229582592878 Cubic Meter per Second --> No Conversion Required

FINAL ANSWER

2.77229582592878 ≈ 2.772296 Cubic Meter per Second <-- Rate of Effusion of Second Gas

(Calculation completed in 00.021 seconds)

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< 8 Graham's Law Calculators

Rate of Effusion for Second Gas by Graham's law

Go Rate of Effusion of Second Gas = Rate of Effusion of First Gas/(sqrt(Molar Mass of Second Gas/Molar Mass of First Gas))

Rate of Effusion for First Gas by Graham's law

Go Rate of Effusion of First Gas = (sqrt(Molar Mass of Second Gas/Molar Mass of First Gas))*Rate of Effusion of Second Gas

Rate of Effusion for Second Gas given Densities by Graham's law

Go Rate of Effusion of Second Gas = Rate of Effusion of First Gas/(sqrt(Density of Second Gas/Density of First Gas))

Rate of Effusion for First Gas given Densities by Graham's law

Go Rate of Effusion of First Gas = (sqrt(Density of Second Gas/Density of First Gas))*Rate of Effusion of Second Gas

Molar Mass of Second Gas by Graham's law

Go Molar Mass of Second Gas = ((Rate of Effusion of First Gas/Rate of Effusion of Second Gas)^2)*Molar Mass of First Gas

Molar Mass of First Gas by Graham's law

Go Molar Mass of First Gas = Molar Mass of Second Gas/((Rate of Effusion of First Gas/Rate of Effusion of Second Gas)^2)

Density of Second Gas by Graham's law

Go Density of Second Gas = ((Rate of Effusion of First Gas/Rate of Effusion of Second Gas)^2)*Density of First Gas

Density of First Gas by Graham's Law

Go Density of First Gas = Density of Second Gas/((Rate of Effusion of First Gas/Rate of Effusion of Second Gas)^2)

Rate of Effusion for Second Gas by Graham's law Formula

Rate of Effusion of Second Gas = Rate of Effusion of First Gas/(sqrt(Molar Mass of Second Gas/Molar Mass of First Gas))
r₂ = r₁/(sqrt(M₂/M₁))

What is Graham's law?

Graham's law of effusion (also called Graham's law of diffusion) was formulated by Scottish physical chemist Thomas Graham in 1848. Graham found experimentally that the rate of effusion of a gas is inversely proportional to the square root of the molar mass of its particles. Graham's law is most accurate for molecular effusion which involves the movement of one gas at a time through a hole. It is only approximate for diffusion of one gas in another or in air, as these processes involve the movement of more than one gas. In the same conditions of temperature and pressure, the molar mass is proportional to the mass density. Therefore, the rates of diffusion of different gases are inversely proportional to the square roots of their mass densities.

How to Calculate Rate of Effusion for Second Gas by Graham's law?

Rate of Effusion for Second Gas by Graham's law calculator uses Rate of Effusion of Second Gas = Rate of Effusion of First Gas/(sqrt(Molar Mass of Second Gas/Molar Mass of First Gas)) to calculate the Rate of Effusion of Second Gas, The Rate of effusion for second gas by Graham's law formula is defined as rate of diffusion or of effusion of a gas is inversely proportional to the square root of its molecular weight. Rate of Effusion of Second Gas is denoted by r₂ symbol.

How to calculate Rate of Effusion for Second Gas by Graham's law using this online calculator? To use this online calculator for Rate of Effusion for Second Gas by Graham's law, enter Rate of Effusion of First Gas (r₁), Molar Mass of Second Gas (M₂) & Molar Mass of First Gas (M₁) and hit the calculate button. Here is how the Rate of Effusion for Second Gas by Graham's law calculation can be explained with given input values -> 2.772296 = 2.12/(sqrt(0.02021/0.03456)).

FAQ

What is Rate of Effusion for Second Gas by Graham's law?

The Rate of effusion for second gas by Graham's law formula is defined as rate of diffusion or of effusion of a gas is inversely proportional to the square root of its molecular weight and is represented as r₂ = r₁/(sqrt(M₂/M₁)) or Rate of Effusion of Second Gas = Rate of Effusion of First Gas/(sqrt(Molar Mass of Second Gas/Molar Mass of First Gas)). The Rate of Effusion of First Gas is the special case of diffusion when the first gas is allowed to escape through the small hole, The Molar Mass of Second Gas is defined as the mass of the gas per mole & The Molar Mass of First Gas is defined as the mass of the gas per mole.

How to calculate Rate of Effusion for Second Gas by Graham's law?

The Rate of effusion for second gas by Graham's law formula is defined as rate of diffusion or of effusion of a gas is inversely proportional to the square root of its molecular weight is calculated using Rate of Effusion of Second Gas = Rate of Effusion of First Gas/(sqrt(Molar Mass of Second Gas/Molar Mass of First Gas)). To calculate Rate of Effusion for Second Gas by Graham's law, you need Rate of Effusion of First Gas (r₁), Molar Mass of Second Gas (M₂) & Molar Mass of First Gas (M₁). With our tool, you need to enter the respective value for Rate of Effusion of First Gas, Molar Mass of Second Gas & Molar Mass of First Gas and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.

How many ways are there to calculate Rate of Effusion of Second Gas?

In this formula, Rate of Effusion of Second Gas uses Rate of Effusion of First Gas, Molar Mass of Second Gas & Molar Mass of First Gas. We can use 1 other way(s) to calculate the same, which is/are as follows -

Rate of Effusion of Second Gas = Rate of Effusion of First Gas/(sqrt(Density of Second Gas/Density of First Gas))

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