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A quantity of \(\mathrm{CH}_{4}(\mathrm{g})\) at \(298 \mathrm{K}\) is compressed reversibly and adiabatically from \(50.0 \mathrm{bar}\) to 200 bar. Assuming ideal behavior, calculate the final temperature of the \(\mathrm{CH}_{4}(\mathrm{g})\). Take \(\bar{C}_{v}=3 R\)

Short Answer

Expert verified
The final temperature of \(\mathrm{CH}_4(\mathrm{g})\) is approximately 422 K.

Step by step solution

01

Use Adiabatic Process Formula

For an ideal gas undergoing a reversible adiabatic process, the relation between the initial and final states is given by:\[T_2 = T_1 \left( \frac{P_2}{P_1} \right)^{\frac{\gamma - 1}{\gamma}} \]where \( T_1 = 298 \ K \), \( P_1 = 50.0 \ bar \), \( P_2 = 200 \ bar \), and \( \gamma = \frac{C_p}{C_v} \).
02

Calculate Specific Heat Ratio \(\gamma\)

Since \( \bar{C}_v = 3R \), and for a monoatomic ideal gas \( \bar{C}_p = \bar{C}_v + R \), we have:\[\bar{C}_p = 4R\]Thus, the specific heat ratio \( \gamma \) is:\[\gamma = \frac{\bar{C}_p}{\bar{C}_v} = \frac{4R}{3R} = \frac{4}{3}\]
03

Substitute Values into Adiabatic Equation

Substitute the values into the adiabatic process formula:\[T_2 = 298 \left( \frac{200}{50} \right)^{\frac{4/3 - 1}{4/3}}\]\[T_2 = 298 \times 4^{\frac{1}{4}}\]Calculate \( 4^{1/4} \).
04

Calculate Final Temperature

Calculate \( 4^{1/4} \) which approximates to \( 1.414 \), then:\[T_2 = 298 \times 1.414\]\[T_2 \approx 421.772 \ K\]

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental concept in thermodynamics. It relates the pressure, volume, and temperature of an ideal gas. An ideal gas is a hypothetical gas that perfectly follows the law under all conditions of temperature and pressure.
The formula for the Ideal Gas Law is:
  • \( PV = nRT \)
where:
  • \( P \) is the pressure of the gas,
  • \( V \) is the volume,
  • \( n \) is the amount of gas in moles,
  • \( R \) is the universal gas constant,
  • \( T \) is the temperature of the gas in Kelvin.
This equation is useful for making predictions about gas conditions when one of the variables changes. Although in our exercise, the assumptions made are based on ideal gas behavior, it's crucial to check if real-world gases approximate this model closely under given conditions.
Specific Heat Capacity
Specific Heat Capacity is essentially a measure of how much heat energy is required to change the temperature of a given amount of a substance by one-degree Celsius. For gases, we typically use molar heat capacities, which tell us how much energy is required for one mole of substance.
In our exercise, we used molar specific heat capacities like:
  • \( \bar{C}_v = 3R \): This is the heat capacity at constant volume.
  • \( \bar{C}_p = 4R \): This is the heat capacity at constant pressure.
Knowing \( \bar{C}_v \) helped us determine \( \gamma \), the ratio of \( \bar{C}_p \) to \( \bar{C}_v \), which is vital for calculations in adiabatic processes. This ratio determines how a gas behaves when undergoing processes without heat transfer.
Thermodynamics
Thermodynamics is the branch of physics that deals with heat, work, and the forms of energy involved in chemical reactions. It provides the fundamental principles that describe changes in energy and thermodynamic quantities. Understanding thermodynamics is crucial for analyzing the behavior of gases in processes like the one in our exercise. The principles of energy conservation and enthalpy changes guide us in predicting how gas properties will change under different process conditions, such as temperature and pressure changes.
When working with thermodynamic problems, remember:
  • Energy cannot be created or destroyed (First Law of Thermodynamics).
  • Entropy, or disorder, always increases in isolated systems (Second Law of Thermodynamics).
  • Perfect energy conversions are impossible (Second Law of Thermodynamics).
These insights form the basis for calculating various parameters in problem-solving, including changes in temperature.
Reversible Processes
Reversible processes are idealized processes that never actually occur in nature but give us important insights into the thermodynamic world. They are processes that proceed infinitely slowly, allowing the system to remain in thermodynamic equilibrium throughout. In the real world, all processes are irreversible because they occur over finite times, leading to a loss of useful energy in the form of entropy changes. However, when calculating theoretical scenarios like our exercise, reversible processes help simplify the calculations.
  • In a reversible adiabatic process, no heat exchange occurs.
  • Energy changes only occur through work done on or by the system.
These conditions make reversible processes a crucial model when evaluating the limits of efficiency or work output of a thermodynamic engine or machine.

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Most popular questions from this chapter

Suppose that a 10 -kg mass of iron at \(20^{\circ} \mathrm{C}\) is dropped from a height of 100 meters. What is the kinetic energy of the mass -just before it hits the ground? What is its speed? What would be the final temperature of the mass if all its kinetic energy at impact is transformed into internal energy? Take the molar heat capacity of iron to be \(\bar{C}_{P}=25.1 \mathrm{J} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\) and the gravitational acceleration constant to be \(9.80 \mathrm{m} \cdot \mathrm{s}^{-2}\)

The \(\Delta_{\mathrm{r}} H^{\circ}\) values for the following equations are \\[ \begin{array}{ll} 2 \mathrm{Fe}(\mathrm{s})+\frac{3}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s}) & \Delta_{\mathrm{r}} H^{\circ}=-206 \mathrm{kJ} \cdot \mathrm{mol}^{-1} \\ 3 \mathrm{Fe}(\mathrm{s})+2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{s}) & \Delta_{\mathrm{r}} H^{\circ}=-136 \mathrm{kJ} \cdot \mathrm{mol}^{-1} \end{array} \\] Use these data to calculate the value of \(\Delta_{\mathrm{r}} H\) for the reaction described by \\[ 4 \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+\mathrm{Fe}(\mathrm{s}) \longrightarrow 3 \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{s}) \\]

One mole of a monatomic ideal gas initially at a pressure of 2.00 bar and a temperature of \(273 \mathrm{K}\) is taken to a final pressure of 4.00 bar by the reversible path defined by \(P / V=\) constant. Calculate the values of \(\Delta U, \Delta H, q,\) and \(w\) for this process. Take \(\bar{C}_{V}\) to be equal to \(12.5 \mathrm{J} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\)

In this problem, we will discuss a famous experiment called the Joule-Thomson experiment. In the first half of the 19 th century, Joule tried to measure the temperature change when a gas is expanded into a vacuum. The experimental setup was not sensitive enough, however, and he found that there was no temperature change, within the limits of his error. Soon afterward, Joule and Thomson devised a much more sensitive method for measuring the temperature change upon expansion. In their experiments (see Figure 19.11 ), a constant applied pressure \(P_{1}\) causes a quantity of gas to flow slowly from one chamber to another through a porous plug of silk or cotton. If a volume, \(V_{1},\) of gas is pushed through the porous plug, the work done on the gas is \(P_{1} V_{1}\). The pressure on the other side of the plug is maintained at \(P_{2},\) so if a volume \(V_{2}\) enters the right-side chamber, then the net work is given by \\[ w=P_{1} V_{1}-P_{2} V_{2} \\] The apparatus is constructed so that the entire process is adiabatic, so \(q=0 .\) Use the First Law of Thermodynamics to show that \\[ U_{2}+P_{2} V_{2}=U_{1}+P_{1} V_{1} \\] or that \(\Delta H=0\) for a Joule-Thomson expansion. Starting with \\[ d H=\left(\frac{\partial H}{\partial P}\right)_{T} d P+\left(\frac{\partial H}{\partial T}\right)_{P} d T \\] show that \\[ \left(\frac{\partial T}{\partial P}\right)_{H}=-\frac{1}{C_{P}}\left(\frac{\partial H}{\partial P}\right)_{T} \\] Interpret physically the derivative on the left side of this equation. This quantity is called the Joule-Thomson coefficient and is denoted by \(\mu_{\mathrm{JT}} .\) In Problem 19-54 you will show that it equals zero for an ideal gas. Nonzero values of \((\partial T / \partial P)_{H}\) directly reflect intermolecular interactions. Most gases cool upon expansion [a positive value of \(\left.(\partial T / \partial P)_{H}\right]\) and a JouleThomson expansion is used to liquefy gases.

Derive an expression for the reversible isothermal work of an expansion of a gas that obeys the Peng-Robinson equation of state.

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