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Chapter 15: Problem 85

A gas, while expanding under isobaric conditions, does 480 \(\mathrm{J}\) of work. The pressure of the gas is \(1.6 \times 10^{5} \mathrm{Pa}\) , and its initial volume is \(1.5 \times 10^{-3} \mathrm{m}^{3} .\) What is the final volume of the gas?

Short Answer

Expert verified
The final volume of the gas is \(4.5 \times 10^{-3} \mathrm{m}^{3}\).

Step by step solution

01

Identify the Known Parameters

We know that the work done by the gas, \( W \), is 480 \( \mathrm{J} \). The pressure, \( P \), is \( 1.6 \times 10^5 \mathrm{Pa} \), and the initial volume, \( V_i \), is \( 1.5 \times 10^{-3} \mathrm{m}^3 \). We need to find the final volume, \( V_f \).
02

Apply the Isobaric Work Formula

For isobaric (constant pressure) processes, the work done by the gas is given by the formula: \( W = P \Delta V \), where \( \Delta V \) is the change in volume. \( \Delta V = V_f - V_i \).
03

Rearrange the Formula to Solve for Final Volume

Rearrange the work formula to find \( V_f \):\[ W = P(V_f - V_i) \]\[ V_f = \frac{W}{P} + V_i \]
04

Substitute the Values into the Equation

Substitute the given values into the rearranged equation:\[ V_f = \frac{480 \mathrm{J}}{1.6 \times 10^5 \mathrm{Pa}} + 1.5 \times 10^{-3} \mathrm{m}^3 \]
05

Perform the Calculation

Calculate the change in volume:\[ \Delta V = \frac{480}{1.6 \times 10^5} = 3 \times 10^{-3} \mathrm{m}^3 \]Then add the initial volume:\[ V_f = 3 \times 10^{-3} \mathrm{m}^3 + 1.5 \times 10^{-3} \mathrm{m}^3 = 4.5 \times 10^{-3} \mathrm{m}^3 \]
06

Conclusion

The final volume of the gas after expansion is \( 4.5 \times 10^{-3} \mathrm{m}^3 \).

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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 deals with heat, work, and the laws governing the conversion between these energy forms in different systems. When we explore thermodynamic processes, we not only delve into the transfer of energy but also the concepts of temperature, volume, and pressure.
These parameters describe the state of a substance. In our specific example, we're investigating the behavior of a gas under an isobaric process, which is an important part of thermodynamics.
  • Internal Energy: This involves looking at the energy contained within the system, due to the motions and interactions of molecules.
  • First Law of Thermodynamics: This states that energy cannot be created or destroyed, only transferred or converted.
Understanding these aspects helps in explaining how and why energy changes occur in physical systems.
Work Done by Gas
When a gas expands, it can do work on its surroundings by pushing against external forces, like a piston. Sometimes, this work can manifest as an observable movement, similar to how a balloon expands.
For an isobaric process where the pressure remains constant, the work done, denoted by the letter (W), is calculated through the relationship:\[ W = P \Delta V \]Here, (P) stands for pressure and (\Delta V) is the change in volume, which is the final volume minus the initial volume.
This principle helps us understand how work and energy transfer occur when the volume of a gas changes.
Volume Expansion
Volume expansion is the increase in volume of a substance, in this case, an expanding gas. The phenomenon occurs as a gas absorbs heat and does work against external pressures.
In the realm of thermodynamics, we're often interested in how the change in volume (\Delta V) takes place. For the problem at hand:
  • Initial Volume, \(V_i\): The starting volume of the system. For this case, it was given as \(1.5 \times 10^{-3}\) m鲁.
  • Change in Volume, \(\Delta V\): This is derived by rearranging the work formula, leading to the calculation \(\Delta V = 3 \times 10^{-3}\) m鲁.
  • Final Volume, \(V_f\): After volume expansion, this is calculated as the initial volume plus the change in volume, resulting in \(4.5 \times 10^{-3}\) m鲁.
Understanding these values helps in visualizing how gases behave under specified conditions.
Constant Pressure
Constant pressure, or isobaric conditions, imply that during the process, the pressure exerted by the gas remains unchanged. This is typical of many practical processes, like inflating a tire or heating a liquid.
Under constant pressure, changes in a gas's volume directly correlate with work being done by or on the gas. This is especially useful because:
  • Predictability: It allows us to predict how the gas will behave when heat is added or removed.
  • Simplicity: The calculations become straightforward, as one parameter (pressure) remains the same throughout.
  • Real-world Applications: Many engines and industrial processes operate under these conditions, offering practical insights.
By understanding constant pressure scenarios, we can better analyze and balance energies within a system.

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Most popular questions from this chapter

A power plant taps steam superheated by geothermal energy to 505 \(\mathrm{K}\) (the temperature of the hot reservoir) and uses the steam to do work in turning the turbine of an electric generator. The steam is then converted back into water in a condenser at 323 \(\mathrm{K}\) (the temperature of the cold reservoir), after which the water is pumped back down into the earth where it is heated again. The output power (work per unit time) of the plant is 84000 kilowatts. Determine (a) the maximum efficiency at which this plant can operate and \((\mathbf{b})\) the minimum amount of rejected heat that must be removed from the condenser every twenty-four hours.

The work done by one mole of a monatomic ideal gas \(\left(\gamma=\frac{5}{3}\right)\) in expanding adiabatically is 825 \(\mathrm{J}\) . The initial temperature and volume of the gas are 393 \(\mathrm{K}\) and 0.100 \(\mathrm{m}^{3} .\) Obtain \(\quad\) (a) the final temperature and (b) the final volume of the gas.

The temperatures indoors and outdoors are 299 and \(312 \mathrm{K},\) respectively. A Carnot air conditioner deposits \(6.12 \times 10^{5} \mathrm{J}\) of heat outdoors. How much heat is removed from the house?

Heat engines take input energy in the form of heat, use some of that energy to do work, and exhaust the remainder. Similarly, a person can be viewed as a heat engine that takes an input of internal energy, uses some of it to do work, and gives off the rest as heat. Suppose that a trained athlete can function as a heat engine with an efficiency of 0.11 . (a) What is the magnitude of the internal energy that the athlete uses in order to do \(5.1 \times 10^{4} \mathrm{J}\) of work? (b) Determine the magnitude of the heat the athlete gives off.

Multiple-Concept Example 6 deals with the same concepts as this problem does. What is the efficiency of a heat engine that uses an input heat of \(5.6 \times 10^{4} \mathrm{J}\) and rejects \(1.8 \times 10^{4} \mathrm{J}\) of heat?
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