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Chapter 19: Problem 12

A gas in a cylinder is held at a constant pressure of 1.80 \(\times\) 10\(^5\) \(Pa\) and is cooled and compressed from 1.70 m\(^3\) to 1.20 m\(^3\). The internal energy of the gas decreases by 1.40 \(\times\) 10\(^5\) J. (a) Find the work done by the gas. (b) Find the absolute value of the heat flow, [\(Q\)] , into or out of the gas, and state the direction of the heat flow. (c) Does it matter whether the gas is ideal? Why or why not?

Short Answer

Expert verified
(a) Work done by the gas is \(-9.0 \times 10^4\) J. (b) Heat flow out of gas is \(|Q| = 2.30 \times 10^5\) J. (c) Gas type (ideal or real) does not matter in this scenario.

Step by step solution

01

Define the variables

We are given the initial and final volumes, the pressure, and the change in internal energy of the gas. Let the initial volume be \( V_i = 1.70 \, \text{m}^3 \), the final volume be \( V_f = 1.20 \, \text{m}^3 \), the pressure \( P = 1.80 \times 10^5 \, \text{Pa} \), and the change in internal energy \( \Delta U = -1.40 \times 10^5 \, \text{J} \).
02

Calculate the work done by the gas

The work done by the gas during compression can be calculated using the formula \( W = P \times (V_f - V_i) \), where \( W \) is the work done, \( P \) is the pressure, and \( V_f \) and \( V_i \) are the final and initial volumes. Substituting the values, we get\[ W = 1.80 \times 10^5 \, \text{Pa} \times (1.20 - 1.70) \, \text{m}^3 \]\[ W = 1.80 \times 10^5 \, \text{Pa} \times (-0.50) \, \text{m}^3 \]\[ W = -9.0 \times 10^4 \, \text{J} \]
03

Use the first law of thermodynamics

According to the first law of thermodynamics, \( \Delta U = Q - W \), where \( \Delta U \) is the change in internal energy, \( Q \) is the heat exchanged, and \( W \) is the work done by the system. We have \( \Delta U = -1.40 \times 10^5 \, \text{J} \) and \( W = -9.0 \times 10^4 \, \text{J} \). Rearrange to find \( Q \):\[ Q = \Delta U + W \]\[ Q = -1.40 \times 10^5 \, \text{J} + (-9.0 \times 10^4 \, \text{J}) \]\[ Q = -2.30 \times 10^5 \, \text{J} \]
04

Determine the direction of heat flow

Since \( Q = -2.30 \times 10^5 \, \text{J} \), heat flows out of the gas as it is a negative value, indicating that the gas lost heat to the surroundings.
05

Analyze the need to consider the type of gas

For this problem, it does not matter whether the gas is ideal because we are dealing with macroscopic properties like pressure, volume, work, and energy that apply universally regardless of the microscopic model of the gas. The first law of thermodynamics holds true for any type of gas, whether ideal or real.

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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 gas in a system undergoes volume change under constant pressure, it either does work or has work done on it. This concept is central to understanding how energy dynamics play out in thermodynamic systems.

In the exercise context, the gas is being compressed, which influences how we view 'work':
  • The work done by the gas during compression is calculated with the formula: \[ W = P \times (V_f - V_i) \]where \( W \) is work, \( P \) is pressure, and \( V_f \), \( V_i \) are final and initial volumes respectively.
  • Here, since the final volume \( V_f \) is smaller than the initial volume \( V_i \), the value of \( V_f - V_i \) is negative, indicating that work is done on the gas, not by the gas.
  • Thus, the calculated work is negative: \[ W = -9.0 \times 10^4 \, \text{J} \]indicating energy being transferred into the gas, forcing it to compress.
Internal energy change
Internal energy is the total energy contained within a system due to its microscopic components. In thermodynamics, changes in internal energy provide insight into energy exchanges within a system.

The First Law of Thermodynamics, stating \( \Delta U = Q - W \), links internal energy change to heat flow and work:
  • In this exercise, the problematic decrease in internal energy is \( \Delta U = -1.40 \times 10^5 \, \text{J} \).
  • Despite the work done on the gas, the total internal energy decreases, signifying that the energy output through heat flow exceeds the work input.
  • Understanding how internal energy changes help depict the net result of energy transactions in a system, informing us about the thermodynamic path and resulting conditions of the system.
Heat flow direction
Heat flow plays a crucial role in understanding energy exchanges in thermodynamic processes. The direction and magnitude of heat flow tell us whether energy is entering or leaving the system.

Here's how this is applied in the exercise:
  • Using the first law formula rearranged to solve for heat, \( Q = \Delta U + W \), informs us of the net heat transfer.\[ Q = -1.40 \times 10^5 \, \text{J} + (-9.0 \times 10^4 \, \text{J}) = -2.30 \times 10^5 \, \text{J} \]
  • The negative \( Q \) value tells us heat is leaving the system, a critical sign of energy distribution as the gas cools and energy disperses into surroundings.
  • This negative heat flow direction denotes that the thermal pathway counteracts internal energy changes, consistent with the physical context of a cooling gas under compression.

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Most popular questions from this chapter

The power output of an automobile engine is directly proportional to the mass of air that can be forced into the volume of the engine's cylinders to react chemically with gasoline. Many cars have a \(turbocharger\), which compresses the air before it enters the engine, giving a greater mass of air per volume. This rapid, essentially adiabatic compression also heats the air. To compress it further, the air then passes through an \(intercooler\) in which the air exchanges heat with its surroundings at essentially constant pressure. The air is then drawn into the cylinders. In a typical installation, air is taken into the turbocharger at atmospheric pressure (1.01 \(\times\) 10\(^5\) Pa), density \(\rho\) = 1.23 kg/m\(^3\), and temperature 15.0\(^\circ\)C. It is compressed adiabatically to 1.45 \(\times\) 10\(^5\) Pa. In the intercooler, the air is cooled to the original temperature of 15.0\(^\circ\)C at a constant pressure of 1.45 \(\times\) 10\(^5\) Pa. (a) Draw a \(pV\)-diagram for this sequence of processes. (b) If the volume of one of the engine's cylinders is 575 cm\(^3\), what mass of air exiting from the intercooler will fill the cylinder at 1.45 \(\times\) 10\(^5\) Pa? Compared to the power output of an engine that takes in air at 1.01 \(\times\) 10\(^5\) Pa at 15.0\(^\circ\)C, what percentage increase in power is obtained by using the turbocharger and intercooler? (c) If the intercooler is not used, what mass of air exiting from the turbocharger will fill the cylinder at 1.45 \(\times\) 10\(^5\) Pa? Compared to the power output of an engine that takes in air at 1.01 \(\times\) 10\(^5\) Pa at 15.0\(^\circ\)C, what percentage increase in power is obtained by using the turbocharger alone?

During an isothermal compression of an ideal gas, 410 J of heat must be removed from the gas to maintain constant temperature. How much work is done by the gas during the process?

On a warm summer day, a large mass of air (atmospheric pressure 1.01 \(\times\) 10\(^5\) Pa) is heated by the ground to 26.0\(^\circ\)C and then begins to rise through the cooler surrounding air. (This can be treated approximately as an adiabatic process; why?) Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 0.850 \(\times\) 10\(^5\) Pa. Assume that air is an ideal gas, with \(\Upsilon\) = 1.40. (This rate of cooling for dry, rising air, corresponding to roughly 1 C\(^\circ\) per 100 m of altitude, is called the dry \(adiabatic\) \(lapse\) \(rate\).)

In a cylinder, 1.20 mol of an ideal monatomic gas, initially at 3.60 \(\times\) 10\(^5\) Pa and 300 K, expands until its volume triples. Compute the work done by the gas if the expansion is (a) isothermal; (b) adiabatic; (c) isobaric. (d) Show each process in a \(pV\)-diagram. In which case is the absolute value of the work done by the gas greatest? Least? (e) In which case is the absolute value of the heat transfer greatest? Least? (f) In which case is the absolute value of the change in internal energy of the gas greatest? Least?

During the time 0.305 mol of an ideal gas undergoes an isothermal compression at 22.0\(^\circ\)C, 392 J of work is done on it by the surroundings. (a) If the final pressure is 1.76 atm, what was the initial pressure? (b) Sketch a \(pV\)-diagram for the process.
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