Question

\(\mathrm{X} \mathrm{mL}\) of \(\mathrm{H}_{2}\) has effused through a hole in a container in 5 seconds. The time taken for the effusion of the same volume of the gas specified below under identical conditions is (a) 10 seconds : \(\mathrm{He}\) (b) 20 seconds : \(\mathrm{O}_{2}\) (c) 25 seconds : \(\mathrm{CO}\) (d) 55 seconds : \(\mathrm{CO}_{2}\)

Short answer

(b) 20 seconds for O\textsubscript{2} fits Graham's law.

Step-by-step solution

Understand Graham's Law of Effusion

Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, this is expressed as:\[\text{Rate of effusion of gas 1} / \text{Rate of effusion of gas 2} = \sqrt{M_2 / M_1}\]where \(M_1\) and \(M_2\) are the molar masses of the gases.

Apply the Law for Hydrogen and Helium

The rate of effusion is proportional to the reciprocal of the time taken if the same volume of gas effuses. For hydrogen (H extsubscript{2}) and helium (He), with times 5 seconds and 10 seconds respectively:\[\frac{X}{t_{\text{H}}} \propto \frac{1}{t_{\text{H}}} \text{ and } \frac{X}{t_{\text{He}}} \propto \frac{1}{t_{\text{He}}}\]\[\frac{1/t_{\text{H}}}{1/t_{\text{He}}} = \sqrt{\frac{M_{\text{He}}}{M_{\text{H}}}}\]Given: \(t_{\text{H}} = 5\, \text{seconds}\), \(t_{\text{He}} = 10\, \text{seconds}\), compare to determine if this fits the relationship.

Calculate the Molar Mass Ratios

Calculate the molar masses:- \(M_{\text{H}_2} = 2.02\, \text{g/mol}\)- \(M_{\text{He}} = 4.00\, \text{g/mol}\)Using the times for the hydrogen and helium:\[\frac{5}{10} = \sqrt{\frac{4.00}{2.02}} \]Check if the experimental ratio equals the calculated ratio after solving the square root.

Evaluate for Oxygen and Carbon Monoxide

For other gases like O extsubscript{2} and CO:- \(t_{\text{O}_2} = 20\, \text{seconds}\), \(M_{\text{O}_2} = 32.00\, \text{g/mol}\)- \(t_{\text{CO}} = 25\, \text{seconds}\), \(M_{\text{CO}} = 28.01\, \text{g/mol}\)Use Graham's law to see which gas aligns.\[\frac{1/5}{1/t} = \sqrt{\frac{M}{2.02}} \]Compare calculated time with given times.

Validate for Carbon Dioxide

Finally, validate for CO extsubscript{2} and determine consistency.- \(t_{\text{CO}_2} = 55\, \text{seconds}\), \(M_{\text{CO}_2} = 44.01\, \text{g/mol}\)Calculate if the time taken is consistent with:\[\frac{1/5}{1/55} = \sqrt{\frac{44.01}{2.02}}\]Compare given time with calculated result from the formula.

Key concepts

Molar Mass

The molar mass of a substance is simply the mass of one mole of that substance. It's usually measured in grams per mole (g/mol). When we're talking about gases, the molar mass is a key factor in how they behave, especially in processes like effusion. Consider hydrogen ( hydrogen has a molar mass of 2.02 g/mol) and helium (helium has a molar mass of 4.00 g/mol). As you can see, these numbers tell us that a mole of helium is heavier than a mole of hydrogen. Understanding molar mass is essential because it relates to how quickly or slowly a gas will effuse, which is dictated by Graham's Law. So, the lighter the molar mass, the faster the gas effuses through a small opening, and vice versa.

Effusion Rate

Effusion refers to the process by which gas molecules escape from a container through a tiny opening. The rate at which this happens is known as the effusion rate. According to Graham's Law, the effusion rate is inversely proportional to the square root of the gas's molar mass. This means if you have two gases and want to compare how fast each will effuse, you can use their molar masses for this calculation:

• The lighter the gas (in terms of molar mass), the faster its effusion rate. Think of a gas like hydrogen effusing much faster than a heavier gas like carbon dioxide.
• Mathematically, if you know the effusion rates of two gases, and their molar masses, you can predict how these gases will behave under the same conditions.

This is crucial in understanding why hydrogen, with its small molar mass, effuses faster compared to heavier gases such as oxygen or carbon dioxide.

Gas Laws

Gas laws describe how gases behave under various conditions, involving relationships between pressure, volume, and temperature. Graham's Law of Effusion is specifically concerned with how gases move through small holes, relating this movement to their molar masses. In Graham's Law:

• The rate of effusion is inversely related to the square root of the gas's molar mass. If one gas has a much lower molar mass than another, it will effuse much more rapidly.
• Graham's Law helps predict the relative rates of effusion for different gases under identical conditions, such as the same temperature and pressure.

In broader terms, gas laws help us understand and predict gas behavior in different scenarios, making them a cornerstone of chemistry and physics.