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\(\bullet\) 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 Gas: Final Pressure = 2.22 atm, Final Temperature = 274.8 K. (b) Diatomic Gas: Final Pressure = 2.62 atm, Final Temperature = 303.2 K.

Step by step solution

01

Understand Adiabatic Processes

In an adiabatic process, no heat is exchanged with the environment. For an ideal gas, the relationship between pressure, volume, and temperature is given by the equation \( PV^\gamma = \text{constant} \), where \( \gamma = \frac{C_p}{C_V} \). Also, the equation \( TV^{\gamma-1} = \text{constant} \) is used.
02

Establish Initial Conditions

The initial conditions are: \( P_1 = 4.00 \) atm, \( T_1 = 350 \) K, and \( V_2 = 1.50V_1 \) (the volume after expansion is 1.5 times the initial volume).
03

Calculate Final Pressure for Monatomic Gas

For a monatomic gas, \( \gamma = \frac{5}{3} \). Use the equation \( P_1 V_1^\gamma = P_2 V_2^\gamma \). Substituting the values: \[ (4.00)(V_1)^{5/3} = P_2 (1.50V_1)^{5/3} \]Simplifying, \[ P_2 = \frac{4.00}{1.50^{5/3}} \approx 2.22 \text{ atm} \]
04

Calculate Final Temperature for Monatomic Gas

For temperature, use the equation \( T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1} \). For \( \gamma = \frac{5}{3} \),\[ 350 \times V_1^{2/3} = T_2 \times (1.50V_1)^{2/3} \]Simplifying, \[ T_2 = \frac{350}{1.50^{2/3}} \approx 274.8 \text{ K} \]
05

Calculate Final Pressure for Diatomic Gas

For a diatomic gas, \( \gamma = \frac{7}{5} \). Use the equation \( P_1 V_1^\gamma = P_2 V_2^\gamma \). Substituting the values: \[ (4.00)(V_1)^{7/5} = P_2 (1.50V_1)^{7/5} \]Simplifying, \[ P_2 = \frac{4.00}{1.50^{7/5}} \approx 2.62 \text{ atm} \]
06

Calculate Final Temperature for Diatomic Gas

For temperature, use the equation \( T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1} \). For \( \gamma = \frac{7}{5} \),\[ 350 \times V_1^{2/5} = T_2 \times (1.50V_1)^{2/5} \]Simplifying, \[ T_2 = \frac{350}{1.50^{2/5}} \approx 303.2 \text{ K} \]

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Key Concepts

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

Monatomic Gas
A monatomic gas is a simple type of gas where each molecule consists of a single atom. Common examples include the noble gases such as helium (He), neon (Ne), and argon (Ar). These gases have a specific heat ratio, represented by \( \gamma \), of \( \frac{5}{3} \). This value is derived from the degrees of freedom of the atoms, as they can move in three independent directions (translational freedom) but have no rotational or vibrational freedom since they are single atoms.

In thermodynamic processes, like the adiabatic expansion in the given exercise, this ratio \( \gamma \) is critical in determining the changes in pressure and temperature. Specifically, for a monatomic ideal gas, the adiabatic process can be described by the equation \( PV^\gamma = \text{constant} \), where \( P \) is pressure and \( V \) is volume. This relationship allows us to calculate how the gas will behave as it expands without exchanging heat with its surroundings.
Diatomic Gas
A diatomic gas consists of molecules that have two atoms. These can be either of the same element, such as hydrogen \((\text{H}_2)\), nitrogen \((\text{N}_2)\), or oxygen \((\text{O}_2)\), or different elements like in carbon monoxide \((\text{CO})\). Diatomic gases have a specific heat ratio \( \gamma \) of \( \frac{7}{5} \) because their atoms can move translationally and also rotate.

Rotation adds more degrees of freedom, which slightly changes the energy dynamics of the gas when compared to monatomic gases. In the case of diatomic gases undergoing adiabatic processes, the equation \( PV^\gamma = \text{constant} \) still holds, but \( \gamma \) is reflective of these additional degrees of freedom available to diatomic molecules. This altered \( \gamma \) value affects calculations for pressure and temperature during the adiabatic expansion of the gas.
Ideal Gas Law
The Ideal Gas Law is a fundamental equation in thermodynamics that relates the pressure, volume, temperature, and number of moles of a gas. It is expressed as \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.

This law assumes that gas molecules are point particles that do not exhibit intermolecular forces and occupy no space. While this might not represent reality perfectly, it provides a useful approximation, especially under conditions of high temperature and low pressure.
  • For monatomic gases like helium, the interactions are minimal, making them excellent real-world approximations to ideal gases.
  • Diatomic gases, although slightly more complex, generally behave similarly under standard conditions.
The Ideal Gas Law is often a starting point before considering additional specifics like adiabatic processes, when calculating changes under different thermodynamic conditions.
Thermodynamics
Thermodynamics is the branch of physics that deals with heat, work, and energy transfer. It provides the rules governing interactions and transformations of energy between systems. It is built on four fundamental laws: the zeroth, first, second, and third laws of thermodynamics.

In the context of gases, thermodynamics explores how systems exchange energy with their surroundings and involve concepts like pressure, volume, and temperature. An adiabatic process, as seen in the exercise, is a specific thermodynamic process where the system undergoes expansion or compression without heat transfer. This leads to changes in the internal energy and temperature of the gas.
  • When working with gases, these principles are critical as they outline how temperature and pressure change when no heat is added or removed.
  • Moreover, understanding thermodynamics allows one to predict and analyze complex systems beyond just gases, including engines and refrigerators.
By applying these principles, you can solve various problems involving energy transfer and transformation efficiently.

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Most popular questions from this chapter

\(\bullet$$\bullet\) We have two equal-size boxes, \(\mathrm{A}\) and \(\mathrm{B}\) . Each box contains gas that behaves as an ideal gas. We insert a thermometer into each box and find that the gas in box \(A\) is at a temperature of \(50^{\circ} \mathrm{C}\) while the gas in box \(B\) is at \(10^{\circ} \mathrm{C} .\) This is all we know about the gas in the boxes. Which of the following statements must be true? Which could be true? (a) The pressure in \(A\) is higher than in \(B\) . (b) There are more molecules in \(A\) than in \(B\) . (c) \(A\) and \(B\) cannot contain the same type of gas. (d) The molecules in \(A\) have more average kinetic energy per molecule than those in \(B\) (e) The molecules in \(A\) are moving faster than those in \(B\) . Explain the reasoning behind your answers.

\(\bullet\) Helium gas with a volume of 2.60 \(\mathrm{L}\) under a pressure of 1.30 atm and at a temperature of \(41.0^{\circ} \mathrm{C}\) is warmed until both the pressure and volume of the gare are doubled. (a) What is the final temperature? (b) How many grams of helium are there? The molar mass of helium is 4.00 \(\mathrm{g} / \mathrm{mol}.\)

\(\bullet\) Lung volume. The total lung volume for a typical person is 6.00 L. A person fills her lungs with air at an absolute pressure of 1.00 atm. Then, holding her breath, she compresses her chest cavity, decreasing her lung volume to 5.70 L. What is the pressure of the air in her compressed lungs, assuming that the temperature of the air remains constant?

\(\bullet$$\bullet\) Heat \(Q\) flows into a monatomic ideal gas, and the volume increases while the pressure is kept constant. What fraction of the heat energy is used to do the expansion work of the gas?

\(\bullet\) A Jaguar XK8 convertible has an eight-cylinder engine. At the beginning of its compression stroke, one of the cylinders contains 499 \(\mathrm{cm}^{3}\) of air at atmospheric pressure \(\left(1.01 \times 10^{5} \mathrm{Pa}\right)\) and a temperature of \(27.0^{\circ} \mathrm{C}\) At the end of the stroke, the air has been compressed to a volume of 46.2 \(\mathrm{cm}^{3}\) and the gauge pressure has increased to \(2.72 \times 10^{6} \mathrm{Pa} .\) Compute the final temperature.

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