A gas undergoes two processes. In the first, the volume remains constant at 0.2000.200 m3 and the pressure increases from 2.00×1052.00\$$times10^5 Pa to 5.00×1055.00$$\$$times10^5 Pa. $$The second process is a compression to a volume of 0.1200.120 m3 at a constant pressure of 5.00×1055.00\$$times10^5 Pa. $$Find the total work done by the gas during both processes.

Ch 19: The First Law of Thermodynamics

Young & Freedman Calc - University Physics 14th Edition

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Young & Freedman Calc 14th Edition

Ch 19: The First Law of Thermodynamics

Problem 6b

Chapter 19, Problem 6b

A gas undergoes two processes. In the first, the volume remains constant at $0.200$ m³ and the pressure increases from $2.00 \times 10^{5}$ Pa to $5.00 \times 10^{5}$ Pa. The second process is a compression to a volume of $0.120$ m³ at a constant pressure of $5.00 \times 10^{5}$ Pa. Find the total work done by the gas during both processes.

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Understand that work done by a gas is given by the formula $ W = P \Delta V $, where $ P $ is the pressure and $ \Delta V $ is the change in volume. For a process at constant volume, $ \Delta V = 0 $, so the work done is zero.

In the first process, since the volume remains constant at 0.200 m^3, the work done by the gas is zero because $ \Delta V = 0 $.

In the second process, the gas is compressed at a constant pressure of 5.00 * 10^5 Pa from a volume of 0.200 m^3 to 0.120 m^3. Calculate the change in volume: $ \Delta V = V_{final} - V_{initial} = 0.120 \text{ m}^3 - 0.200 \text{ m}^3 $.

Substitute the values into the work formula for the second process: $ W = P \Delta V = 5.00 \times 10^5 \text{ Pa} \times (0.120 \text{ m}^3 - 0.200 \text{ m}^3) $.

Add the work done in both processes to find the total work done by the gas. Since the work done in the first process is zero, the total work done is just the work from the second process.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Work Done by a Gas

In thermodynamics, the work done by a gas during a process is calculated using the formula W = PΔV, where W is the work, P is the pressure, and ΔV is the change in volume. For processes at constant volume, the work done is zero because there is no change in volume. Understanding this concept is crucial for calculating work in different thermodynamic processes.

Isobaric Process

An isobaric process is a thermodynamic process in which the pressure remains constant. During such a process, the work done by or on the gas can be calculated using the formula W = P(V2 - V1), where V2 and V1 are the final and initial volumes, respectively. This concept is essential for determining the work done during the compression phase of the question.

Isochoric Process

An isochoric process is a thermodynamic process where the volume remains constant. In this type of process, no work is done by the gas because work is a function of volume change (W = PΔV), and ΔV is zero. Understanding this concept helps in recognizing that the first process in the question does not contribute to the total work done.

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Textbook Question

A gas undergoes two processes. In the first, the volume remains constant at $0.200$ m³ and the pressure increases from $2.00$\times$10^5$ Pa to $5.00$\times$10^5$ Pa. The second process is a compression to a volume of $0.120$ m³ at a constant pressure of $5.00$\times$10^5$ Pa. In a $p V$ -diagram, show both processes.