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Five moles of an ideal monatomic gas with an initial temperature of \(127^{\circ} \mathrm{C}\) expand and, in the process, absorb 1200 \(\mathrm{J}\) of heat and do 2100 \(\mathrm{J}\) of work. What is the final temperature of the gas?

Short Answer

Expert verified
The final temperature of the gas is approximately 385.65 K.

Step by step solution

01

Convert the Initial Temperature to Kelvin

The initial temperature given is in degrees Celsius. To convert it to Kelvin, use the formula: \[ T = T_{C} + 273.15 \]Substitute the given temperature:\[ T_{initial} = 127 + 273.15 = 400.15 \, K \].
02

Calculate the Change in Internal Energy

For an ideal monatomic gas, the change in internal energy \( \Delta U \) is given by the first law of thermodynamics: \[ \Delta U = Q - W \]where \( Q = 1200 \) J (heat absorbed) and \( W = 2100 \) J (work done). Therefore,\[ \Delta U = 1200 - 2100 = -900 \, \text{J} \].
03

Use the Formula for Change in Internal Energy

For a monatomic ideal gas, the change in internal energy is also given by:\[ \Delta U = \frac{3}{2} n R \Delta T \]where \( n \) is the number of moles, \( R = 8.314 \, \text{J/mol K} \) is the ideal gas constant, and \( \Delta T \) is the change in temperature. We need to find \( \Delta T \).
04

Solve for Change in Temperature

Substitute \( \Delta U = -900 \, \text{J} \), \( n = 5 \) moles, and \( R = 8.314 \, \text{J/mol K} \) into the formula:\[ -900 = \frac{3}{2} \times 5 \times 8.314 \times \Delta T \]Solve for \( \Delta T \):\[ \Delta T = \frac{-900}{\frac{3}{2} \times 5 \times 8.314} \approx -14.5 \, K \].
05

Find the Final Temperature

Now that we have \( \Delta T \), calculate the final temperature:\[ T_{final} = T_{initial} + \Delta T = 400.15 + (-14.5) = 385.65 \, K \].

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Key Concepts

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

Thermodynamics
Thermodynamics is a branch of physics that explores the relationships between heat, work, and energy. It revolves around the laws guiding these interactions. One of the fundamental concepts in thermodynamics is the First Law, which states that energy can neither be created nor destroyed. Instead, it can only change forms. In the context of the exercise, the First Law of Thermodynamics is expressed with the formula: \[ \Delta U = Q - W \] Here:
  • \(\Delta U\) represents the change in internal energy.
  • \(Q\) is the heat absorbed or released.
  • \(W\) denotes the work done on or by the system.
By knowing \(Q\) and \(W\), you can determine how the internal energy changes and subsequently affect temperature changes in ideal gases.
Monatomic Gas
Monatomic gases are gases made up of single atoms, such as helium or argon. When dealing with thermodynamic equations, monatomic gases have unique properties compared to polyatomic gases. In particular, they have a specific expression relating to their internal energy. For a monatomic ideal gas, the internal energy depends on the number of moles and the temperature. Specifically:\[ \Delta U = \frac{3}{2} n RT \] Where,
  • \(n\) is the number of moles.
  • \(R\) is the ideal gas constant, typically 8.314 J/mol K.
  • \(T\) is the temperature.
In the exercise, this property helps calculate how much the internal energy changes as the gas absorbs heat and does work, ultimately affecting the temperature.
Internal Energy
Internal energy is the total energy contained within a system. It includes kinetic energy from the movement of particles and potential energy from forces among the particles. In the context of an ideal gas, internal energy is primarily a function of temperature. For ideal monatomic gases, the internal energy can be expressed as:\[ U = \frac{3}{2} nRT \] A change in internal energy, \(\Delta U\), links directly to changes in temperature when no phase change occurs. This means that any absorbed heat not spent on doing work results in a temperature increase, while work done by the system on the surroundings might decrease internal energy, as seen in the original exercise.
Temperature Conversion
Temperature conversion is essential when dealing with thermodynamics. Different scales include Celsius, Kelvin, and Fahrenheit. The Kelvin scale is particularly important, as it is used for many scientific calculations involving gas laws. Kelvin temperatures avoid negative numbers and start from absolute zero, the point where particles theoretically stop moving. To convert Celsius to Kelvin, the formula is straightforward:\[ T_{K} = T_{C} + 273.15 \] In the exercise, converting from Celsius to Kelvin was a crucial first step. This allows you to work with the ideal gas law and other thermodynamic formulas more effectively, providing a consistent and absolute measure of temperature which is essential for calculations.

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Most popular questions from this chapter

A player bounces a basketball on the floor, connpressing it to 80.0\(\%\) of its original volume. The air (assume it is essentially N_{2} gas) inside the ball is originally at a temperature of \(20.0^{\circ} \mathrm{C}\) and a pressure of 2.00 \(\mathrm{atm}\) . The ball's diameter is \(23.9 \mathrm{cm} .\) (a) What temperature does the air in the ball reach at its maximum compression? (b) By how much does the internal energy of the air change between the ball's original state and its maximum compression?

Three moles of an ideal monatomic gas expands at a con-stant pressure of 2.50 atm; the volume of the gas changes from \(3.20 \times 10^{-2} \mathrm{m}^{3}\) to \(4.50 \times 10^{-2} \mathrm{m}^{3} .\) (a) Calculate the initial and final temperatures of the gas. (b) Calculate the amount of work the gas does in expanding. (c) Calculate the amount of heat added to the gas. (d) Calculate the change in internal encrgy of the gas.

You close off the nozzle of a bicycle tire pump and very slowly depress the plunger so that the air inside is compressed to half its original volume. Assume the air behaves like an ideal gas. If you do this so slowly that the temperature of the air inside the pump never changes: (a) Is the work done by the air in the pump positive or negative? (b) Is the heat flow to the air positive or negative? (c) What can you say about the relative magnitudes of the heat flow and the work? Explain.

In a cylinder sealed with a piston, you rapidly compress 3.00 \(\mathrm{L}\) of \(\mathrm{N}_{2}\) gas initially at 1.00 atm pressure and \(0.00^{\circ} \mathrm{C}\) to half its original volume. Assume the \(\mathrm{N}_{2}\) behaves like an ideal gas. (a) Calculate the final temperature and pressure of the gas. (b) If you now cool the gas back to \(0.00^{\circ} \mathrm{C}\) without changing the pressure, what is its final volume?

Doughnuts: Breakfast of Champions! A typical doughnut contains \(2.0 \mathrm{~g}\) of protein. \(17.0 \mathrm{~g}\) of carbohydrates, and \(7.0 \mathrm{~g}\) of fat. The average food energy values of these substances are \(4.0 \mathrm{kcal} / \mathrm{g}\) for protein and carbohydrates and \(9.0 \mathrm{kcal} / \mathrm{g}\) for fat. (a) During hcavy exercise, an average person uscs cnergy at a rate of 510 keal \(/\) h. How long would you have to exercise to "work off"one doughnut? (b) If the energy in the doughnut could somehow be converted into the kinetic energy of your body as a whole, how fast could you move after eating the doughnut? Take your mass to be \(60 \mathrm{~kg}_{4}\) and express your answer in \(\mathrm{m} / \mathrm{s}\) and \(\mathrm{km} / \mathrm{h}\)

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