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A \(1.1-\mathrm{m}^{3}\) volume of air at a pressure of \(101 \mathrm{kPa}\) is compressed to a volume of \(0.45 \mathrm{~m}^{3} .\) (a) What is the pressure in the compressed volume if the compression process is isentropic? (b) What is the pressure if the compression process is isothermal?

Short Answer

Expert verified
(a) Isentropic pressure: Calculate using \(P_2 = P_1 (V_1/V_2)^\gamma\). (b) Isothermal pressure: \(P_2 \approx 246.89 \, \text{kPa}\).

Step by step solution

01

Understand the Problem

We have a volume of air being compressed. The initial volume is \(1.1 \, \text{m}^3\) with an initial pressure of \(101 \, \text{kPa}\). The final volume is \(0.45 \, \text{m}^3\). We need to find the final pressure for two scenarios: isentropic and isothermal compression.
02

Apply Isentropic Compression Formula

For isentropic compression, we use the relation \( P_1 V_1^\gamma = P_2 V_2^\gamma \) where \( \gamma \) (the adiabatic index) for air is approximately \(1.4\). We are given \( P_1 = 101 \, \text{kPa}\), \( V_1 = 1.1 \, \text{m}^3\), \( V_2 = 0.45 \, \text{m}^3\), and need to find \( P_2 \).
03

Calculate for Isentropic Compression

Rearrange the formula to find \( P_2 \):\[ P_2 = P_1 \left( \frac{V_1}{V_2} \right)^\gamma = 101 \, \text{kPa} \left( \frac{1.1}{0.45} \right)^{1.4} \]Compute the pressure.
04

Apply Isothermal Compression Formula

In isothermal processes, the relation is \( P_1 V_1 = P_2 V_2 \). Given \( P_1 = 101 \, \text{kPa}\), \( V_1 = 1.1 \, \text{m}^3\), \( V_2 = 0.45 \, \text{m}^3\), find \( P_2 \) using:\[ P_2 = \frac{P_1 V_1}{V_2} = \frac{101 \, \text{kPa} \times 1.1}{0.45} \]Compute this value.
05

Calculate for Isothermal Compression

Substitute the given values into the formula:\[ P_2 = \frac{101 \, \text{kPa} \times 1.1}{0.45} \approx 246.89 \, \text{kPa} \]The pressure during isothermal compression is approximately \(246.89 \, \text{kPa}\).

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Key Concepts

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

Compression
Compression is a fundamental concept in thermodynamics. It involves reducing the volume of a gas, resulting in changes to its pressure and, depending on the process, potentially its temperature. In our exercise, we consider two types of compression: isentropic and isothermal.

During isentropic compression, the process is adiabatic, meaning no heat is transferred to or from the gas. This makes the gas work done purely on its internal energy changes. Conversely, isothermal compression keeps the temperature constant throughout the process. Heat must be exchanged to maintain temperature as the gas compresses.

Understanding these two methods provides insight into how gases behave differently under varying conditions of pressure and volume.
Adiabatic Index
The adiabatic index, often represented with the Greek letter \( \gamma \), is a crucial parameter in describing adiabatic processes, which are a component of isentropic processes. It is the ratio of specific heat at constant pressure \( (C_p) \) to specific heat at constant volume \( (C_v) \). For air, \( \gamma \) is approximately 1.4.

This index indicates how compressible a substance is under adiabatical conditions. In our calculation for isentropic compression, the adiabatic index allows us to relate initial and final states of pressure and volume without involving heat exchange.

The use of \( \gamma \) in equations like \( P_1 V_1^\gamma = P_2 V_2^\gamma \) enables predictions about the pressure or volume of a gas after compression.
Pressure Calculation
Pressure calculation during compression depends on whether the process is isentropic or isothermal. In the isentropic scenario, the equation \( P_1 V_1^\gamma = P_2 V_2^\gamma \) adjusts for changes without heat transfer. Here, the key task is to solve for \( P_2 \) using given \( P_1, V_1, V_2 \), and \( \gamma \).

For an isothermal process, the calculation follows the formula \( P_1 V_1 = P_2 V_2 \). This relation assumes temperature constancy and thus directly links pressure to inverse volume changes.

Successfully calculating the pressure requires attention to these relations and accurate substitution of known values. Complexities like expansions or contractions are crucial in determining final results based on initial conditions.
Volume Change
Volume change is intrinsic to understanding both isentropic and isothermal compressions. It is the transformation of a gas's occupied space during these processes.

In our example, the gas starts at 1.1 m³ and is compressed to 0.45 m³. This substantial reduction necessitates calculation of subsequent pressure change.

Whether isentropic or isothermal, acknowledging volume change helps link the initial and final states of the gas. The calculations address how a smaller volume can increase pressure or, under isothermal conditions, remain stable thermally despite compressive forces.

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Most popular questions from this chapter

A viscometer is constructed with two \(30-\mathrm{cm}\) -long concentric cylinders, one \(20.0 \mathrm{~cm}\) in diameter and the other \(20.2 \mathrm{~cm}\) in diameter. A torque of \(0.13 \mathrm{~N} \cdot \mathrm{m}\) is required to rotate the inner cylinder at \(400 \mathrm{rpm} .\) Calculate the viscosity of the fluid.

Consider the case of an air bubble released from the bottom of a \(12-\mathrm{m}\) -deep lake as shown in Figure 1.21 . The bubble is filled with air, it has an initial diameter of \(6 \mathrm{~mm},\) no air is lost or gained in the bubble as it rises, and the air in the bubble and the surrounding lake water maintain a constant temperature of \(20^{\circ} \mathrm{C}\). Atmospheric pressure, \(p_{\mathrm{atm}},\) on the surface of the lake is \(101.3 \mathrm{kPa},\) and the pressure, \(p,\) at any depth, \(z\), below the surface of the lake is given by \(p=p_{\text {atm }}+\gamma z\), where \(\gamma\) is the specific weight of the water in the lake. Estimate the diameter of the bubble when it surfaces.

The mass of air in a tank is estimated as \(12 \mathrm{~kg},\) and the temperature and pressure of the air in the tank are measured as \(67^{\circ} \mathrm{C}\) and \(210 \mathrm{kPa}\), respectively. Estimate the volume of air in the tank.

Remote measurements of the properties of a mystery gas find that the molecular weight of the gas is \(35 \mathrm{~g} / \mathrm{mol}\) and the specific heat of the gas at constant pressure is \(1025 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). Estimate the speed of sound in this gas at a temperature of \(22{ }^{\circ} \mathrm{C}\). Assume that the behavior of the gas can be approximated by that of an ideal gas.

Use prefixes to express the following quantities with magnitudes in the range of \(0.01-1000:(\) a \() 6.27 \times 10^{7} \mathrm{~N},\) (b) \(7.28 \times 10^{5} \mathrm{~Pa},\) (c) \(4.76 \times 10^{-4} \mathrm{~m}^{2},\) and (d) \(8.56 \times\) \(10^{5} \mathrm{~m}\).

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