Equal volumes of two gases \(A\) and \(B\) diffuse through a porous pot in 20 and 10 seconds respectively. If the molar mass of \(A\) be 80, find the molar mass of \(B\)

The effusion rate of a gas refers to how quickly a gas passes through a small opening or porous barrier. It is a key concept in understanding the behavior of gases in various conditions and has practical applications in fields such as materials science, environmental science, and engineering.

According to Graham's Law of Effusion, the rate at which a gas effuses is inversely proportional to the square root of its molar mass. In simpler terms, lighter gases effuse faster than heavier gases. This relationship can be quantitatively expressed as \( r \propto \frac{1}{\sqrt{M}} \), where \( r \) is the rate of effusion and \( M \) is the molar mass of the gas. When comparing two gases, therefore, we can use the formula \( \frac{r_A}{r_B} = \sqrt{ \frac{M_B}{M_A} } \).

For students tackling physical chemistry problems that involve effusion rates, understanding this law is essential. It allows for the prediction of how different gases will behave under the same conditions. It's important to remember that effusion is different from diffusion, the latter being the spread of gas molecules throughout a space, which we will discuss in another section.

Molar mass calculation is an integral part of solving many physical chemistry problems, including those pertaining to gas effusion. The molar mass of a substance is the mass in grams of one mole of the substance, measured in grams per mole (g/mol).

To calculate the molar mass of a gas when given its effusion rate, we can manipulate Graham's Law: \( \frac{r_A}{r_B} = \sqrt{ \frac{M_B}{M_A} } \). By rearranging the equation and substituting known values, we can find the unknown molar mass.

In the exercise provided, we determine the molar mass of gas \(B\) using the given effusion times \(t_A\) and \(t_B\) for gases \(A\) and \(B\) respectively. This process, shown in the step-by-step solution, relies on converting the effusion times to rates, and then using those rates to calculate the unknown molar mass. It is a prime example of the type of conversion and manipulation skills needed when dealing with physical chemistry equations.

Understanding how to perform molar mass calculations is pivotal for any student studying chemistry, as it is a fundamental concept that recurs in various applications, from chemical reactions to the study of gas laws.

Gas diffusion is the process by which gas molecules spread from an area of higher concentration to an area of lower concentration, often resulting in an even distribution of the gas within the space. This natural spreading or mingling of gas molecules is driven by their kinetic energy and does not require an external force.

While diffusion is similar to effusion in that they both involve the movement of gas molecules, they take place under different conditions. Diffusion occurs within a container or an area where the gas is free to spread out, while effusion refers to the escape of the gas through a small opening into another space, such as the outside environment.

In educational practice, helping students visualize gas diffusion can be enhanced by using analogies or simulations. For instance, the spread of a perfume scent in a room is a relatable example of diffusion. Understanding gas diffusion is necessary when discussing topics such as respiratory processes, material sciences (in terms of gas permeability), and even weather patterns.

Solving physical chemistry problems requires a clear understanding of the concepts involved and the ability to perform precise calculations. These problems can range from simple to complex, often involving the application of various scientific laws, such as Graham's Law of Effusion in the given exercise.

To effectively solve these problems, students should first grasp the fundamental principles, such as the nature of gases and their behaviors under different physical conditions. Once the principles are understood, systematically approaching the problem can make the process smoother. This means identifying given information, understanding what is being asked, determining which equations or laws are applicable, and then methodically solving for the unknown variables.

In situations where molar mass needs to be determined from effusion rates, breaking the problem down into steps, as shown in the solution provided, can greatly assist in comprehension and execution. Moreover, practicing a variety of problems enhances problem-solving skills and helps students gain confidence in their abilities to tackle challenging physical chemistry questions.

• Understand the principles
• Approach the problem systematically
• Determine applicable equations
• Solve for unknown variables
• Practice with a variety of problems