# Pressure change from one box to another

• heepofajeep
In summary, a .25L box at 1PSI with a leak is inserted into a 1L box at 5psi. The pressure of the 1L box at t=2sec. is 4.8psi. The questions asked were: A) How long will it take for the Pressure in Box 1 to equal the pressure in Box 2? B) What is the size of the leak in Box 2? The relevant equations are PV=nRT, 1/2*density*velocity^2 + P1 = P2 [Bernoulli], and P = Po e^-Ct [generic Pressure Decay]. This is a real-world problem for leak testing a product and a general equation
heepofajeep

## Homework Statement

A .25L box at 1PSI is inserted into a 1L box at 5psi.
The .25L box has a leak.
The pressure of the 1L box at t=2sec. is 4.8psi

A) How long will it take for the Pressure in Box 1 to equal the pressure in Box 2.
B) What is the size of the leak in Box 2?

## Homework Equations

PV=nRT
1/2*density*velocity^2 + P1 = P2 [Bernoulli]
P = Po e^-Ct [generic Pressure Decay] ?

## The Attempt at a Solution

Stuck in a rut... >=(.

Note: this is not actually a true homework problem. In fact, it is a real-world problem for leak testing a product.

I am trying to find a general equation and/or solution to this problem, but I'm not sure how to approach it.

To answer this question, we need to consider the principles of gas laws and fluid dynamics. According to the ideal gas law, PV=nRT, the pressure of a gas is directly proportional to its volume. In this case, the volume of the 1L box is 4 times larger than the .25L box, so we can assume that the pressure in the 1L box will also be 4 times higher.

However, we also need to consider the fact that the .25L box has a leak. This means that the pressure inside the box will decrease over time, following a generic pressure decay equation P = Po e^-Ct. We can use this equation to calculate the pressure in the .25L box at any given time.

For Part A, to find the time it will take for the pressure in Box 1 to equal the pressure in Box 2, we can set up an equation using the ideal gas law for both boxes:

P1V1 = nRT1
P2V2 = nRT2

Where P1 is the pressure in the .25L box, V1 is the volume of the .25L box, P2 is the pressure in the 1L box, V2 is the volume of the 1L box, n is the number of moles of gas, R is the gas constant, and T1 and T2 are the temperatures in the .25L and 1L boxes respectively.

Since we know that P1 decreases over time due to the leak, we can substitute P1 with the generic pressure decay equation: P = Po e^-Ct. This gives us:

Po e^-Ct * V1 = nRT1
P2V2 = nRT2

By setting these two equations equal to each other, we can solve for t, the time it takes for the pressure in Box 1 to equal the pressure in Box 2.

For Part B, we can use Bernoulli's equation, 1/2*density*velocity^2 + P1 = P2, to calculate the velocity of the gas escaping from the leak in Box 2. From there, we can use the equation for volumetric flow rate, Q = A * v, where A is the area of the leak and v is the velocity of the gas, to calculate the size of the leak in Box 2.

In conclusion, solving this problem requires a

## What causes pressure to change between two boxes?

The pressure change between two boxes is caused by a difference in the amount of gas molecules present in each box. If one box has more gas molecules, it will exert more pressure compared to the box with fewer gas molecules.

## How does temperature affect pressure between two boxes?

According to the ideal gas law, pressure is directly proportional to temperature. This means that as the temperature increases, so does the pressure. Therefore, if one box is heated while the other remains at a constant temperature, the pressure in the heated box will increase and cause a pressure difference between the two boxes.

## What is the relationship between volume and pressure in two connected boxes?

According to Boyle's Law, there is an inverse relationship between volume and pressure. This means that as the volume of one box decreases, the pressure inside it will increase, and vice versa. Therefore, if one box has a smaller volume than the other, it will also have a higher pressure.

## Can the pressure between two boxes be equalized?

Yes, the pressure between two boxes can be equalized by opening a valve between them, allowing gas molecules to flow from the high-pressure box to the low-pressure box until the pressures are equal. This is known as pressure equalization.

## How does the shape of the boxes affect pressure change?

The shape of the boxes does not directly affect pressure change. However, if one box has a smaller volume but a larger opening, it will allow for a faster pressure equalization compared to a box with a smaller opening. This is because the larger opening allows for more gas molecules to flow through at once, increasing the rate of pressure equalization.

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