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Chapter 15: Problem 44

\(\bullet\) 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.

Short Answer

Expert verified
(a) Work done by the gas is \(3.78 \times 10^{4} \ \mathrm{J}\). (b) Change in internal energy is \(7.72 \times 10^{4} \ \mathrm{J}\).

Step by step solution

01

Understand the problem

We need to find the work done by the gas during constant pressure expansion and the change in internal energy using the first law of thermodynamics.
02

Calculate the work done by the gas

The work done by the gas during an isobaric process (constant pressure) is given by the formula: \[ W = P imes \Delta V \]where \( P = 1.80 \times 10^{5} \ \mathrm{Pa} \) and \( \Delta V = V_{f} - V_{i} = 0.320 \ \mathrm{m}^{3} - 0.110 \ \mathrm{m}^{3} = 0.210 \ \mathrm{m}^{3} \).Substitute the values:\[ W = 1.80 \times 10^{5} \ \mathrm{Pa} \times 0.210 \ \mathrm{m}^{3} = 3.78 \times 10^{4} \ \mathrm{J} \].
03

Use First Law of Thermodynamics

The change in internal energy \( \Delta U \) of the system is given by the first law of thermodynamics:\[ \Delta U = Q - W \]where \( Q = 1.15 \times 10^{5} \ \mathrm{J} \) is the total heat added to the system and \( W = 3.78 \times 10^{4} \ \mathrm{J} \) is the work done calculated previously.
04

Calculate the change in internal energy

Substitute the values of \( Q \) and \( W \) into the equation from the previous step:\[ \Delta U = 1.15 \times 10^{5} \ \mathrm{J} - 3.78 \times 10^{4} \ \mathrm{J} = 7.72 \times 10^{4} \ \mathrm{J} \].

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Key Concepts

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

Work Done by Gas
When a gas expands inside a cylinder, it can do work on the surroundings. Work done by the gas depends on how and under what conditions the gas expands. In the case of our exercise, the expansion happens under constant pressure, which is known as an isobaric process.

To find out how much work the gas does, we can use the formula:
  • The formula for work done by the gas (\(W\) is given by: \( W = P \times \Delta V \)
  • Here, \(P\) is the pressure, which is constant at \(1.80 \times 10^{5} \, \mathrm{Pa}\), and \(\Delta V\) is the change in volume \(0.210 \, \mathrm{m}^{3}\)
Plugging the values, we've found that work done is \(3.78 \times 10^{4} \, \mathrm{J}\). This tells us how much energy was transferred from the gas to do work as it expanded, keeping pressure steady.
Internal Energy Change
The internal energy of a gas changes when heat is added or removed, or when the gas does work. In our problem, heat was added to keep the pressure constant. The magnitude of this internal energy change can be calculated by using the first law of thermodynamics.
  • The first law states: \(\Delta U = Q - W\).
  • Here, \(\Delta U\) is the change in internal energy, \(Q\) is the heat added to the gas, and \(W\) is the work done by the gas.
In this exercise, we know:
  • The heat added \(Q = 1.15 \times 10^{5} \, \mathrm{J}\).
  • The work done \(W = 3.78 \times 10^{4} \, \mathrm{J}\).
By substituting these values into the equation, we find the change in internal energy to be \(7.72 \times 10^{4} \, \mathrm{J}\).
This change reflects how much energy the gas acquired due to heating, minus what was used for work.
Isobaric Process
An isobaric process occurs when the pressure of the system remains constant as other conditions such as volume and temperature change. In our scenario, the gas undergoes expansion at a steady pressure of \(1.80 \times 10^{5} \, \mathrm{Pa}\).

This type of process is common in real-life applications involving gas under cylinders or pistons, where maintaining a constant pressure is necessary.
In such processes, the relationship of pressure to work and volume change is simplified, allowing us to use the formula for work done directly. While the pressure remains constant, for an ideal gas, changes in other state properties like volume and temperature allow us to draw valuable conclusions.

These principles help in understanding practical applications of thermodynamics and the behavior of gases in controlled environments.

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Most popular questions from this chapter

\(\bullet$$\bullet\) A cylinder with a piston contains 0.250 mol of ideal oxygen at a pressure of \(2.40 \times 10^{5}\) 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.

\(\bullet$$\bullet\) The effect of altitude on the lungs. (a) Calculate the change in air pressure you will experience if you climb a 1000 \(\mathrm{m}\) moun- tain, assuming that the temperature and air density do not change over this distance and that they were \(22^{\circ} \mathrm{Cand} 1.2 \mathrm{kg} / \mathrm{m}^{3},\) respectively, at the bottom of the mountain. (b) If you took a 0.50 \(\mathrm{L}\) breath at the foot of the mountain and managed to hold it until you reached the top, what would be the volume of this breath when you exhaled it there?

\(\bullet\) Suppose you do 457 \(\mathrm{J}\) of work on 1.18 moles of ideal He gas in a perfectly insulated container. By how much does the internal energy of this gas change? Does it increase or decrease?

\(\bullet$$\bullet\) Starting with 2.50 \(\mathrm{mol}\) of \(\mathrm{N}_{2}\) gas (assumed to be ideal) in a cylinder at 1.00 atm and \(20.0^{\circ} \mathrm{C},\) a chemist first heats the gas at constant volume, adding \(1.52 \times 10^{4} \mathrm{J}\) of heat, then continues heating and allows the gas to expand at constant pressure to twice its original volume. (a) Calculate the final temperature of the gas. (b) Calculate the amount of work done by the gas. (c) Calculate the amount of heat added to the gas while it was expanding. (d) Calculate the change in internal energy of the gas for the whole process.

\(\bullet$$\cdot\) 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 \(C_{p} / C_{V}=\frac{7}{5} .\)
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