/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 109 A rigid tank whose volume is \(1... [FREE SOLUTION] | 91Ó°ÊÓ

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A rigid tank whose volume is \(10 \mathrm{~L}\) is initially evacuated. A pinhole develops in the wall, and air from the surroundings at 1 bar, \(25^{\circ} \mathrm{C}\) enters until the pressure in the tank becomes 1 bar. No significant heat transfer between the contents of the tank and the surroundings occurs. Assuming the ideal gas model with \(k=1.4\) for the air, determine (a) the final temperature in the tank, in \({ }^{\circ} \mathrm{C}\), and (b) the amount of air that leaks into the tank, in \(g\).

Short Answer

Expert verified
The final temperature is 25°C. The mass of air that leaks in is 11.72 g.

Step by step solution

01

Understand the Problem

Identify that air is leaking into the tank at constant pressure and that the ideal gas model applies.
02

Apply the Ideal Gas Law

Recall the Ideal Gas Law: \[ PV = nRT \] where: - \( P \) is pressure (1 bar), - \( V \) is volume (10 L), - \( n \) is the amount of gas in moles, - \( R \) is the gas constant, - \( T \) is temperature in Kelvin.
03

Determine Initial Conditions

The initial state has no gas in the tank, so initially, \( n_{initial} = 0 \).
04

Final Conditions

In the final state, the pressure inside the tank equals the surrounding pressure (1 bar), and no heat transfer occurs. Thus, the process is isothermal.
05

Isothermal Process

Since the process is isothermal, the final temperature is the same as the surrounding temperature. \[ T_{final} = T_{surroundings} = 25^{\textdegree} \text{C} = 298.15 \text{ K} \]
06

Calculate the Amount of Air (moles)

Using the ideal gas law to find the final amount of air: \[ n_{final} = \frac{PV}{RT} \] where: - \( P = 1 \text{ bar} = 100000 \text{ Pa} \) - \( V = 10 \text{ L} = 0.01 \text{ m}^3 \) - \( R = 8.314 \text{ J/mol·K} \) - \( T = 298.15 \text{ K} \) \[ n_{final} = \frac{100000 \times 0.01}{8.314 \times 298.15} \approx 0.404 \text{ mol} \]
07

Convert Moles to Mass

To find the mass of air that leaks into the tank, multiply the moles of air by the molar mass of air (approximately 29 g/mol):\[ \text{Mass} = n_{final} \times M = 0.404 \text{ mol} \times 29 \text{ g/mol} = 11.72 \text{ g} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Isothermal Process
Understanding an isothermal process is crucial for solving problems like this one. An isothermal process means the temperature remains constant throughout the process. So, in our example, no significant heat transfer occurs between the tank and surroundings.
An isothermal process can be described by Boyle's Law, which states that for a given mass of an ideal gas, the product of pressure (P) and volume (V) is constant when the temperature is kept constant: \[ PV = \text{constant} \]
Since temperature doesn't change in an isothermal process, the final temperature in the tank will be the same as the surroundings, which is 25°C or 298.15 K in this problem.
Molar Mass of Air
The molar mass of air is the average mass of the air's molecules. Air is primarily composed of nitrogen (N_2), oxygen (O_2), argon (Ar), and trace amounts of other gases.
The molar mass of air is calculated using the weighted average based on the abundance of these gases. For simplification in our calculations, the molar mass of air is often approximated as 29 g/mol.
This molar mass is used to convert the number of moles of air to the mass of air. For example, if we have 0.404 moles of air (as calculated using the ideal gas law), the mass can be found by multiplying the number of moles by the molar mass:\[ \text{Mass} = n \times \text{Molar Mass} \] This approach helps us find that 0.404 moles multiplies by 29 g/mol results in approximately 11.72 grams of air.
Rigid Tank Problem
A rigid tank problem involves finding out how gases behave within a fixed volume container when subjected to changes in pressure or temperature. In our problem, the tank’s volume is constant at 10 liters (0.01 m³).
Since the tank doesn't change in size (hence 'rigid'), any change in the quantity of gas or its state parameters (like pressure or temperature) can be directly assessed using the ideal gas law. Given, the process is isothermal (constant temperature) and the pressure inside the tank eventually balances with the surroundings at 1 bar, we can simplify our calculations significantly.
The ideal gas law formula, \( PV = nRT \) allows us to solve for the unknowns (such as the number of moles of air that leaks in) since the volume (V) and the pressure (P) are known, and the gas constant (R) and temperature (T) are standard values (R = 8.314 J/mol·K and T = 298.15 K).
This results in the final mole count for air which we can then convert to mass using the air's molar mass. All these steps must consider the rigid nature of the tank, ensuring no volume change impacts our calculations.

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Most popular questions from this chapter

Air enters a nozzle operating at steady state at \(720^{\circ} \mathrm{R}\) with negligible velocity and exits the nozzle at \(500^{\circ} \mathrm{R}\) with a velocity of \(1450 \mathrm{ft} / \mathrm{s}\). Assuming ideal gas behavior and neglecting potential energy effects, determine the heat transfer in Btu per lb of air flowing.

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