/*! 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 81 A rigid tank of volume \(0.75 \m... [FREE SOLUTION] | 91Ó°ÊÓ

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A rigid tank of volume \(0.75 \mathrm{~m}^{3}\) is initially evacuated. A hole develops in the wall, and air from the surroundings at 1 bar, \(25^{\circ} \mathrm{C}\) flows in until the pressure in the tank reaches 1 bar. Heat transfer between the contents of the tank and the surroundings is negligible. Determine the final temperature in the tank, in \({ }^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The final temperature in the tank is 25°C.

Step by step solution

01

- Understand the Problem

The tank starts empty and air flows in from the surroundings (1 bar, 25°C) until the pressure in the tank equals the surrounding pressure (1 bar). Heat transfer is negligible.
02

- Identify Assumptions

Assume the ideal gas behavior for air and note the process is isobaric (constant pressure).
03

- Apply the Ideal Gas Law

Use the ideal gas law, \[ PV = nRT \]. Given that pressure (P) remains constant and initial conditions are known, identify the relationship for final temperature.
04

- Initial State

The tank is initially evacuated, so we only consider the final state.
05

- Calculate Final Temperature

Under isobaric conditions and since heat transfer is negligible, the final temperature must equal the initial temperature of the entering air. Thus, the final temperature is simply 25°C.

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Key Concepts

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

rigid tank thermodynamics
A rigid tank is a crucial concept in thermodynamics.
It means that the tank's volume does not change regardless of the process occurring inside.
For this exercise, the rigid tank starts empty (evacuated).
When air from the surroundings at 1 bar and 25°C enters the tank until the pressure inside matches the outside pressure, the volume remains fixed at 0.75 m³.
Because the volume remains constant, it simplifies calculations involving gas behavior and makes it easier to apply thermodynamic principles like the ideal gas law.
isobaric process
An isobaric process is one where the pressure remains constant.
In this exercise, the pressure in the tank rises from zero to 1 bar, matching the surrounding pressure.
It's important to note that heat exchange with the surroundings is negligible.
The air enters and fills the tank until the pressures equalize, making this an isobaric process.
Understanding this concept is key to knowing that despite changes in other state properties, the pressure remains steady during the entry of air into the tank.
final temperature calculation
To calculate the final temperature in the tank, we rely on the principles of the ideal gas law and the conditions described.
Given that the process is isobaric and the tank volume is constant (rigid), we start by recognizing that the tank pressure (1 bar) remains constant.
Utilizing the ideal gas law, : : PV = nRT, where: V is volume, n is the amount of gas, R is the gas constant, and T is the temperature. - We realize since volume and pressure are constant, temperature must also remain unchanged if heat transfer is negligible. Simply put, the temperature of the air entering the tank (25°C) is the same as the final temperature in the tank due to isobaric conditions and no heat transfer. Thus, the final temperature is 25°C.
This calculation showcases a fundamental usage of thermodynamic principles in analyzing gas behavior in a closed, constant volume, and an isobaric environment.

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Most popular questions from this chapter

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