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Chapter 15: Problem 38

Three moles of an ideal monatomic gas expand at a constant 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 energy of the gas.

Short Answer

Expert verified
Initial Temp: ~326.7 K, Final Temp: ~459.3 K, Work done: ~3299 J, Heat added: ~8268 J, Internal energy change: ~4969 J.

Step by step solution

01

Identify the Given Quantities

We are given the following:- Number of moles, \( n = 3 \)- Constant pressure, \( P = 2.50 \text{ atm} = 2.50 \times 101325 \text{ Pa} \)- Initial volume, \( V_i = 3.20 \times 10^{-2} \text{ m}^3 \)- Final volume, \( V_f = 4.50 \times 10^{-2} \text{ m}^3 \)- The gas is ideal and monatomic.
02

Convert Pressure to SI Units

Convert the given pressure from atm to Pa (Pascals):\[ P = 2.50 \times 101325 \text{ Pa} = 253312.5 \text{ Pa} \]
03

Use Ideal Gas Law to Find Initial Temperature

Use the ideal gas law formula: \[ PV = nRT \]Solving for the initial temperature \( T_i \):\[ T_i = \frac{PV_i}{nR} \]Substitute the values:\[ T_i = \frac{(253312.5)(3.20 \times 10^{-2})}{3 \times 8.314} \approx 326.69 \text{ K} \]
04

Use Ideal Gas Law to Find Final Temperature

Similarly, calculate the final temperature \( T_f \) using the final volume:\[ T_f = \frac{PV_f}{nR} \]Substitute the values:\[ T_f = \frac{(253312.5)(4.50 \times 10^{-2})}{3 \times 8.314} \approx 459.31 \text{ K} \]
05

Calculate Work Done by the Gas

The work done by the gas during expansion at constant pressure is given by:\[ W = P (V_f - V_i) \]Substitute the given values:\[ W = 253312.5 (4.50 \times 10^{-2} - 3.20 \times 10^{-2}) \approx 3299.06 \text{ J} \]
06

Calculate the Heat Added Using First Law of Thermodynamics

The first law of thermodynamics states:\[ \Delta U = Q - W \]For an ideal monatomic gas, the change in internal energy \( \Delta U \) can also be calculated as:\[ \Delta U = \frac{3}{2}nR\Delta T \]Now calculate \( \Delta U \):\[ \Delta T = T_f - T_i = 459.31 - 326.69 = 132.62 \text{ K} \]\[ \Delta U = \frac{3}{2} \times 3 \times 8.314 \times 132.62 \approx 4968.93 \text{ J} \]Rearrange the first law equation for heat \( Q \):\[ Q = \Delta U + W = 4968.93 + 3299.06 \approx 8267.99 \text{ J} \]
07

Summarize the Results

- Initial Temperature \( T_i \approx 326.69 \text{ K} \)- Final Temperature \( T_f \approx 459.31 \text{ K} \)- Work Done \( W \approx 3299.06 \text{ J} \)- Heat Added \( Q \approx 8267.99 \text{ J} \)- Change in Internal Energy \( \Delta U \approx 4968.93 \text{ J} \)

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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 equation in thermodynamics, used to relate the pressure, volume, and temperature of a gas. The formula is expressed as \(PV = nRT\), where:
  • \(P\) is the pressure of the gas (in Pascals when using SI units)
  • \(V\) is the volume of the gas (in cubic meters)
  • \(n\) is the number of moles of the gas
  • \(R\) is the universal gas constant, approximately 8.314 \(\text{J/(mol} \cdot \text{K)}\)
  • \(T\) is the temperature in Kelvin

If you know any three of these quantities, you can easily solve for the fourth. In the original exercise, this law was used to determine the initial and final temperatures of the gas as it expanded under constant pressure. By plugging in the values for pressure, volume, and the number of moles, the temperatures were calculated each time the gas was at a different volume.
Internal Energy
Internal energy refers to the total energy stored within a system. It includes kinetic and potential energies of the molecules that make up the system. For an ideal monatomic gas, like the one in the exercise, the internal energy is purely translational kinetic energy.

The change in internal energy, \(\Delta U\), can be calculated using the formula:
  • \(\Delta U = \frac{3}{2}nR\Delta T\)
  • Here, \(\Delta T\) represents the change in temperature (final temperature minus initial temperature)

As temperature rises, so does the internal energy, because the molecular movement speeds up, increasing kinetic energy. In the exercise, calculating this change allowed us to determine how much energy entered or left the system as heat was added or removed.
Work Done by Gas
When a gas expands, it does work on its surroundings. The work done by a gas at constant pressure is computed by the equation: \(W = P (V_f - V_i)\).

This formula indicates that the work done depends directly on the pressure and the change in volume, \(V_f - V_i\).
  • \(V_f\) is the final volume, while \(V_i\) is the initial volume

During the expansion in the exercise, work was done by the gas, causing an increase in volume. It's essential in thermodynamics to understand that work is a way that energy can be transferred from a system to its environment or vice versa.
First Law of Thermodynamics
The first law of thermodynamics is essentially the law of energy conservation adapted to systems involving thermal energy. It is expressed as \(\Delta U = Q - W\), where:
  • \(\Delta U\) is the change in internal energy of the gas
  • \(Q\) is the heat added to the system
  • \(W\) is the work done by the system

This equation states that the energy added to the gas as heat, minus the work done by the gas, will result in a change in the internal energy of the gas.

In the exercise, the total energy change in the system was balanced by the heat added and the work done, perfectly illustrating this principle. Understanding this law is crucial, as it ties together heat, work, and internal energy in thermodynamics.

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Most popular questions from this chapter

A cylinder with a piston contains \(0.250 \mathrm{~mol}\) of ideal oxygen at a pressure of \(2.40 \times 10^{5} \mathrm{~Pa}\) and a temperature of \(355 \mathrm{~K}\). The gas first expands isobarically to twice its original volume. It is then compressed isothermally back to its original volume, and finally it is cooled isochorically to its original pressure. (a) Show the series of processes on a \(p V\) diagram. (b) Compute the temperature during the isothermal compression. (c) Compute the maximum pressure. (d) Compute the total work done by the piston on the gas during the series of processes.

IA flask contains a mixture of neon (Ne), krypton (Kr), and radon (Rn) gases. Compare (a) the average kinetic energies of the three types of atoms; (b) their root-mean-square speeds. (Hint: The periodic table in Appendix C shows the molar mass (in g/mol) of each element.)

A gas under a constant pressure of \(1.50 \times 10^{5} \mathrm{~Pa}\) and with an initial volume of \(0.0900 \mathrm{~m}^{3}\) is cooled until its volume becomes \(0.0600 \mathrm{~m}^{3}\). (a) Draw a \(p V\) diagram of this process. (b) Calculate the work done by the gas.

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.) Estimate the ratio of the thermal conductivity of Xe to that of He. A. 0.015 B. 0.061 C. 0.10 D. 0.17

A cylindrical tank has a tight-fitting piston that allows the volume of the tank to be changed. The tank originally contains \(0.110 \mathrm{~m}^{3}\) of air at a pressure of 3.40 atm. The piston is slowly pulled out until the volume of the gas is increased to \(0.390 \mathrm{~m}^{3}\). If the temperature remains constant, what is the final value of the pressure?
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