# Gas effusing through hole, working out time dependence

• physconomic
In summary, the conversation discusses a thermally insulated container with a small hole containing a gas with known properties at time t = 0. The question is how to determine the time dependence of density and temperature as the gas effuses out through the hole. The participant suggests using the equation ##N_L = \frac{1}{4} n A = \frac{1}{4} \frac{N_0}{v} A \sqrt{\frac{8kT}{\pi m}}##, but is unsure how to find n(t) and T(t). The solution involves using the open-system version of the 1st law of thermodynamics.
physconomic
Homework Statement
Consider instead a thermally insulated container of volume V with a
small hole of area A, containing a gas with molecular mass m. At time t = 0, the density is ##n_0## and temperature is ##T_0##. As gas effuses out through a small hole, both density and temperature inside the container will drop. Work out their time dependence, n(t) and T(t) in terms of the quantities given above.
Relevant Equations
##N_L = \frac{1}{4} n A <u> = \frac{1}{4} \frac{N_0}{v} A \sqrt{\frac{8kT}{\pi m}}##
Consider instead a thermally insulated container of volume V with a
small hole of area A, containing a gas with molecular mass m. At time t = 0, the density is ##n_0## and temperature is ##T_0##. As gas effuses out through a small hole, both density and temperature inside the container will drop. Work out their time dependence, n(t) and T(t) in terms of the quantities given above.

I know that ##N_L = \frac{1}{4} n A <u> = \frac{1}{4} \frac{N_0}{v} A \sqrt{\frac{8kT}{\pi m}}##, but not sure how to use this to find n(t) and T(t). I think I need to find the flux of the energy to know if the temperature is decreasing and then find n and t from that but not sure exactly how to do this. Any help greatly appreciated.

This can be solved using the open-system (control volume) version of the 1st law. Are you familiar with that?

## 1. What is gas effusion?

Gas effusion is the process by which gas molecules escape through a small hole or opening in a container. This can occur due to a difference in pressure or temperature between the inside and outside of the container.

## 2. How does gas effusion through a hole work?

Gas effusion through a hole is governed by Graham's Law, which states that the rate of effusion is inversely proportional to the square root of the molar mass of the gas. This means that lighter gases will effuse more quickly than heavier gases.

## 3. What factors affect the rate of gas effusion?

The rate of gas effusion is affected by the size of the hole, the pressure and temperature of the gas, and the molar mass of the gas. A smaller hole, higher pressure and temperature, and lower molar mass will result in a faster rate of effusion.

## 4. How can I calculate the time dependence of gas effusion through a hole?

The time dependence of gas effusion can be calculated using Graham's Law and the ideal gas law. By knowing the initial conditions (pressure, temperature, and molar mass) and the size of the hole, you can determine the rate of effusion and how it changes over time.

## 5. What are the practical applications of studying gas effusion through holes?

Studying gas effusion through holes can have many practical applications, such as in the design of gas separation processes, leak detection systems, and gas flow meters. It can also help in understanding the behavior of gases in different environments and predicting their movement and diffusion.

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