Question

You are a NASA engineer faced with the task of ensuring that the material on the hull of a spacecraft can withstand puncturing by space debris. The initial cabin air pressure in the craft of 1 atm can drop to 0.7 atm before the safety of the crew is jeopardized. The volume of the cabin is \(100 . \mathrm{m}^{3}\), and the temperature in the cabin is 285 K. Assuming it takes the space shuttle about 8 hours from entry into orbit until landing, what is the largest circular aperture created by a hull puncture that can be safely tolerated assuming that the flow of gas out of the spaceship is effusive? Can the escaping gas from the spaceship be considered as an effusive process? (You can assume that the air is adequately represented by \(\mathrm{N}_{2}\).

Short answer

The largest circular aperture that can be safely tolerated assuming the flow of gas out of the spaceship is effusive is approximately \(D \approx 1.38 \times 10^{-4}\) m or \(138\,\mu \mathrm{m}\). After calculating the mean free path (λ) of the N₂ molecules, if the aperture size is much greater than λ, we can safely assume the process to be effusive; otherwise, it is not a correct assumption.

Step-by-step solution

Calculate the Mass Loss Rate from the Cabin

Using the ideal gas law \(PV = nRT\), we can calculate the moles of gas that escape from the cabin. First, we will find the initial moles (n1) and final moles (n2) in the cabin. Using the ideal gas law: \(n = \frac{PV}{RT}\) Where R is the ideal gas constant (0.0821 L atm/mol K). For initial moles, n1: \(n_1 = \frac{P_1V}{RT}\) Similarly, for final moles, n2: \(n_2 = \frac{P_2V}{RT}\) Now, we can find the mass loss rate: Mass loss rate (molar) = \(\frac{n_1 - n_2}{t}\)

Calculate the Molecule Effusion Rate

We are considering the leak of gas from the cabin as effusive. Using Graham's law, we can find the molecule effusion rate: Molecule effusion rate, \(r\) = \(A\sqrt{\frac{2 \pi RT}{M}}\) Where A is the aperture area, and M is the molar mass of N₂ (28 g/mol).

Calculate the Aperture Area

From step 1 and 2, we can now equate the mass loss rate (step 1) for the escaping gas with the molecule effusion rate (step 2). To find the aperture area, let's call it A. \(A\sqrt{\frac{2 \pi RT}{M}} = \frac{n_1 - n_2}{t}\) Now, solve for A.

Calculate the Aperture Diameter

With the aperture area calculated, we can now determine the diameter of the circular aperture. Let D be the aperture diameter. Then. \(A = \frac{\pi D^2}{4}\) Solve for the aperture diameter, D.

Analyze if the effusive process is correct

We can analyze if considering the process as effusive is suitable by comparing the size of the aperture to the mean free path of the N₂ molecules in the cabin given by: Mean free path (λ) = \(\frac{kT}{\sqrt{2} \pi d^2 P}\) Where k is the Boltzmann constant (1.38 x 10⁻²³ J/K), d is the diameter of N₂ molecules (3.67 x 10⁻¹⁰ m). If the aperture size is much greater than the mean free path (λ) of the N₂ molecules, we can safely assume the process to be effusive. Otherwise, it is not a correct assumption.

Key concepts

Ideal Gas Law

The ideal gas law is a fundamental principle in understanding the behavior of gases under various conditions. It is represented by the equation \( PV = nRT \), where \( P \) is the pressure of the gas, \( V \) is the volume, \( n \) is the number of moles of the gas, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.

When applying this to real-life scenarios, like ensuring the safety of astronauts aboard a spacecraft, an engineer must know the initial and final states of the gas within the cabin to assess risks. The calculation of the number of escaping gas moles, which is essential for solving our exercise, directly uses this law. The idea is to find the changes in pressure and relate it to the volume of gas using constant values such as \( R \) and \( T \) to measure the gas that may escape into space through a puncture in the hull.

In everyday terms, this is like knowing the amount of air inside a balloon that will escape if there is a slight cut and relating it to the size of the cut under consistent room temperature and pressure. The ideal gas law is thus invaluable in predicting the outcome of such situations.

Molecule Effusion Rate

The molecule effusion rate is a measure of how quickly a gas escapes through a tiny hole into a vacuum. It is part of a process known as effusion, which governs how gases move under certain conditions. Effusion rate calculations are particularly useful in the context of space exploration, where maintaining the integrity of a spacecraft's atmosphere is critical.

Graham's law provides a formula to quantify this effusion rate, which can be described by the equation \( r = A\sqrt{\frac{2 \pi RT}{M}} \), with \( A \) representing the area of the aperture, \( R \) the ideal gas constant, \( T \) the temperature, and \( M \) the molar mass of the escaping gas. By determining this rate, engineers can understand the implications of a puncture in a spacecraft's hull and set safety thresholds.

To put it simply, think of the effusion rate as the speed at which air leaks out of a punctured tire. The size of the puncture and the internal pressure of the tire will affect how quickly you need to get to a repair shop! In our NASA engineer's challenge, it's like figuring out how fast the cabin's air is leaking and whether it's fast enough to put the crew in danger.

Mean Free Path

Mean free path refers to the average distance a molecule travels before colliding with another molecule. This is a critical concept in understanding gas behavior, especially in low-pressure environments such as outer space. For gases obeying the ideal gas assumptions, the mean free path \( \lambda \) is given by \( \lambda = \frac{kT}{\sqrt{2} \pi d^2 P} \) where \( k \) is the Boltzmann constant, \( T \) is the temperature, \( d \) is the molecule's diameter, and \( P \) is the pressure.

When a hull puncture occurs on a spacecraft, comparing the size of this puncture with the mean free path can inform whether the effusion assumption is valid. If the aperture is much larger than the mean free path, we can say the process is effusive. This is similar to predicting whether water droplets will pass through a sieve - if the holes are significantly larger than the droplets, the water will flow out unhindered.

In terms of practical applications, understanding the mean free path can help engineers like those at NASA determine how small a puncture can be before it critically affects the environment within a spacecraft. Just as a small enough filter can purify water by trapping contaminants, the integrity of the spacecraft's hull can prevent the loss of precious breathable air.