Calculate the Ratio of Effusion Rates of Oxygen to Hydrogen

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Effusion Rate Ratio Calculator (Graham's Law)

Calculation Results:

function calculateEffusionRatio() { var m1 = parseFloat(document.getElementById('gas1_mass').value); var m2 = parseFloat(document.getElementById('gas2_mass').value); var resultArea = document.getElementById('result_area'); var ratioText = document.getElementById('ratio_text'); var inverseText = document.getElementById('inverse_text'); if (isNaN(m1) || isNaN(m2) || m1 <= 0 || m2 <= 0) { alert("Please enter valid positive molar masses."); return; } // Graham's Law: Rate1 / Rate2 = sqrt(M2 / M1) // Ratio of Oxygen (Rate1) to Hydrogen (Rate2) var ratio = Math.sqrt(m2 / m1); var inverseRatio = 1 / ratio; resultArea.style.display = "block"; ratioText.innerHTML = "The ratio of the effusion rate of Oxygen (O₂) to Hydrogen (H₂) is " + ratio.toFixed(4) + "."; inverseText.innerHTML = "This means that Hydrogen effuses approximately " + inverseRatio.toFixed(2) + " times faster than Oxygen under identical conditions of temperature and pressure."; }

Understanding the Effusion Rate Ratio of Oxygen to Hydrogen

In the world of physical chemistry, Graham's Law of Effusion provides a fascinating look at how gases move through small openings. When comparing oxygen (O₂) and hydrogen (H₂), the difference in their effusion rates is stark, driven primarily by their vastly different molar masses.

What is Graham's Law?

Graham's Law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, it is expressed as:

Rate₁ / Rate₂ = √(Molar Mass₂ / Molar Mass₁)

This means that lighter gases will travel through a pinhole or porous membrane much faster than heavier gases. Because hydrogen is the lightest element on the periodic table, it effuses significantly faster than oxygen.

Step-by-Step Calculation

To calculate the ratio of the effusion rate of oxygen to hydrogen, we use the following standard molar masses:

  • Oxygen (O₂): ~32.00 g/mol
  • Hydrogen (H₂): ~2.02 g/mol

Plugging these into the formula:

  1. Identify Gas 1 as Oxygen and Gas 2 as Hydrogen.
  2. Set up the ratio: Rate(O₂) / Rate(H₂) = √(Molar Mass H₂ / Molar Mass O₂)
  3. Calculate: √(2.02 / 32.00) = √0.063125
  4. Final Result: ~0.251

This calculation shows that for every molecule of oxygen that passes through a barrier, approximately four molecules of hydrogen will have passed through in the same amount of time.

Real-World Implications

This principle is not just a theoretical exercise. Understanding effusion rates is critical in several industrial and scientific applications:

  • Isotope Separation: Graham's Law was historically used to separate uranium isotopes for nuclear applications.
  • Gas Leakage: Because hydrogen is so small and light, it is notoriously difficult to contain. It can leak through seals and materials that would easily hold heavier gases like oxygen or nitrogen.
  • Respiratory Therapy: Understanding how different gas mixtures diffuse in the lungs is vital for medical professionals dealing with respiratory mechanics.

Why Molar Mass Matters

The kinetic molecular theory of gases tells us that at a constant temperature, all gas particles have the same average kinetic energy. However, because kinetic energy is defined as 1/2 * mass * velocity², a lighter particle (like Hydrogen) must move at a higher velocity than a heavier particle (like Oxygen) to maintain the same energy level. This higher velocity directly leads to a faster rate of effusion.

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