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Chapter 15: Problem 57

An ideal gas at 4.00 atm and \(350 \mathrm{~K}\) is permitted to expand adiabatically to 1.50 times its initial volume. Find the final pressure and temperature if the gas is (a) monatomic with \(C_{p} / C_{V}=\frac{5}{3},\) (b) diatomic with \(C_{p} / C_{V}=\frac{7}{5}\).

Short Answer

Expert verified
(a) Monatomic: \( P_2 \approx 2.19 \text{ atm} \), \( T_2 \approx 277 \text{ K} \); (b) Diatomic: \( P_2 \approx 2.50 \text{ atm} \), \( T_2 \approx 306 \text{ K} \).

Step by step solution

01

Understanding Adiabatic Process

In an adiabatic process, no heat is exchanged with the surroundings. The process follows the equation \( PV^{\gamma} = \text{constant} \), where \( \gamma = \frac{C_p}{C_V} \). For a monatomic gas, \( \gamma = \frac{5}{3} \) and for a diatomic gas, \( \gamma = \frac{7}{5} \).
02

Initial Conditions

The given initial conditions are: Pressure \( P_1 = 4.00 \text{ atm} \), Temperature \( T_1 = 350 \text{ K} \), and the volume expands to \( 1.50 \) times its initial volume, i.e., \( V_2 = 1.50 V_1 \). We need to find the final pressure \( P_2 \) and temperature \( T_2 \).
03

Calculating Final Pressure for Monatomic Gas

Use the adiabatic condition \( P_1 V_1^{\gamma} = P_2 V_2^{\gamma} \). Substitute \( \gamma = \frac{5}{3} \) for a monatomic gas: \[ 4.00 \times V_1^{\frac{5}{3}} = P_2 \times (1.50 V_1)^{\frac{5}{3}} \]Solve for \( P_2 \): \[ P_2 = \frac{4.00}{1.50^{\frac{5}{3}}} \]Calculate \( P_2 \).
04

Calculating Final Temperature for Monatomic Gas

The relation for temperature in an adiabatic process is \( T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \). Substitute \( \gamma = \frac{5}{3} \) for a monatomic gas: \[ 350 \times V_1^{\frac{2}{3}} = T_2 \times (1.50 V_1)^{\frac{2}{3}} \]Solve for \( T_2 \): \[ T_2 = \frac{350}{1.50^{\frac{2}{3}}} \]Calculate \( T_2 \).
05

Calculating Final Pressure for Diatomic Gas

Use the adiabatic condition again, with \( \gamma = \frac{7}{5} \): \[ 4.00 \times V_1^{\frac{7}{5}} = P_2 \times (1.50 V_1)^{\frac{7}{5}} \]Solve for \( P_2 \): \[ P_2 = \frac{4.00}{1.50^{\frac{7}{5}}} \]Calculate \( P_2 \).
06

Calculating Final Temperature for Diatomic Gas

Use the temperature relation for \( \gamma = \frac{7}{5} \): \[ 350 \times V_1^{\frac{2}{5}} = T_2 \times (1.50 V_1)^{\frac{2}{5}} \]Solve for \( T_2 \): \[ T_2 = \frac{350}{1.50^{\frac{2}{5}}} \]Calculate \( T_2 \).

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Key Concepts

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

Ideal Gas
The concept of an ideal gas is central to understanding thermodynamics. An ideal gas is a theoretical gas composed of many randomly moving point particles that interact only through elastic collisions. These gases perfectly obey the ideal gas law, expressed as \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is temperature in Kelvin.
  • Ideal gases assume no intermolecular forces, allowing calculations to simplify.
  • They follow distinct properties making predictions possible about behavior under different conditions.
  • Understanding ideal gases help in approximations related to real gases at high temperatures and low pressures.
In physical chemistry, these assumptions help us explore adiabatic processes more easily, as seen in the original exercise.
Monatomic Gas
A monatomic gas is a type of ideal gas comprised of single atoms. Examples include noble gases like helium and neon. These gases have specific characteristics which make calculations such as those required in adiabatic processes straightforward. For monatomic gases, the ratio of specific heats \( \gamma \) (Cp/Cv) is equal to \( \frac{5}{3} \).
  • This value arises because monatomic gases have fewer degrees of freedom than diatomic or polyatomic gases.
  • With fewer molecular interactions, they behave in a highly predictable manner when undergoing expansion or compression.
This simplicity is noticeable in adiabatic expansions, as demonstrated in solving for pressure and temperature changes in the exercise.
Diatomic Gas
Diatomic gases consist of molecules composed of two atoms, such as oxygen and nitrogen. They have more complex behavior compared to monatomic gases due to additional degrees of freedom. These include vibrational and rotational motions, affecting calculations in thermodynamics. In the adiabatic process, the ratio \( \gamma \) for diatomic gases is \( \frac{7}{5} \).
  • The additional degrees of freedom result in a smaller value of \( \gamma \) compared to monatomic gases.
  • This affects how energy is distributed during a process which, in turn, influences pressure and temperature calculations.
These characteristics are crucial when determining changes in state variables during adiabatic expansions.
Thermodynamics
Thermodynamics is the branch of physics that explores how heat energy converts into other forms of energy and how it affects matter, particularly through awareness of macroscopic variables. Some core principles include the laws of thermodynamics, each addressing different aspects such as energy conservation and heat exchange.
  • The first law relates to the conservation of energy, which serves as a foundation for processes like adiabatic changes.
  • The second law discusses entropy, affecting directionality of energy transformations.
  • Understanding these laws is crucial in engineering applications and natural phenomena analysis.
The exercise underlines the notion of an adiabatic process, a key aspect of thermodynamics without heat exchange with the surroundings.
Pressure and Temperature Calculations
Pressure and temperature calculations are vital when analyzing any process involving gases, like those in the problem at hand. They often dictate the state and behavior of a substance:
  • Pressure refers to the force applied by gas molecules against the walls of their container per unit area.
  • Temperature describes the average kinetic energy of gas molecules and affects pressure measurements.
In adiabatic processes, both variables change without external heat exchange. Formulas derived from laws of thermodynamics allow us to compute changes in pressure and temperature effectively. Calculating these changes accurately assists in deeply understanding gas behavior amidst varying conditions, as shown in the calculations performed for both monatomic and diatomic gases in the exercise.

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Most popular questions from this chapter

Calculate the volume of \(1.00 \mathrm{~mol}\) of liquid water at a temperature of \(20^{\circ} \mathrm{C}\) (at which its density is \(998 \mathrm{~kg} / \mathrm{m}^{3}\) ), and compare this volume with the volume occupied by \(1.00 \mathrm{~mol}\) of steam at \(200^{\circ} \mathrm{C}\). Assume the steam is at atmospheric pressure and can be treated as an ideal gas.

The water pressure at the bottom of the Mariana Trench, which is one of the deepest parts of the Pacific Ocean, is about 1000 atm. How many grams of helium gas would you have to put into a 1 liter volume at room temperature \((300 \mathrm{~K})\) in order to achieve this pressure?

Three moles of an ideal gas are in a rigid cubical box with sides of length \(0.200 \mathrm{~m}\). (a) What is the force that the gas exerts on each of the six sides of the box when the gas temperature is \(20.0^{\circ} \mathrm{C} ?\) (b) What is the force when the temperature of the gas is increased to \(100.0^{\circ} \mathrm{C} ?\)

One way to improve insulation in windows is to fill a sealed space between two glass panes with a gas that has a lower thermal conductivity than that of air. The thermal conductivity \(k\) of a gas depends on its molar heat capacity \(C_{V},\) molar mass \(M,\) and molecular radius \(r\). The dependence on those quantities at a given temperature is approximately \(k \propto C_{V} / r^{2} \sqrt{M}\). The noble gases have properties that make them particularly good choices as insulating gases. Noble gases range from helium (molar mass \(4.0 \mathrm{~g} / \mathrm{mol}\), molecular radius \(0.13 \mathrm{nm}\) ) to xenon (molar mass \(131 \mathrm{~g} / \mathrm{mol}\), molecular radius \(0.22 \mathrm{nm}\) ). (The noble gas radon is heavier than xenon, but radon is radioactive and so is not suitable for this purpose.) Give one reason why the noble gases are preferable to air (which is mostly nitrogen and oxygen) as an insulating material. A. Noble gases are monatomic, so no rotational modes contribute to their molar heat capacity. B. Noble gases are monatomic, so they have lower molecular masses than do nitrogen and oxygen. C. The molecular radii in noble gases are much larger than those of gases that consist of diatomic molecules. D. Because noble gases are monatomic, they have many more degrees of freedom than do diatomic molecules, and their molar heat capacity is reduced by the number of degrees of freedom.

A gas in a cylinder expands from a volume of \(0.110 \mathrm{~m}^{3}\) to \(0.320 \mathrm{~m}^{3} .\) Heat flows into the gas just rapidly enough to keep the pressure constant at \(1.80 \times 10^{5} \mathrm{~Pa}\) during the expansion. The total heat added is \(1.15 \times 10^{5} \mathrm{~J}\). (a) Find the work done by the gas. (b) Find the change in internal energy of the gas.
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