Ch 19: The First Law of Thermodynamics

Young & Freedman Calc - University Physics 14th Edition

Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook

Young & Freedman Calc 14th Edition

Ch 19: The First Law of Thermodynamics

Problem 16

Chapter 19, Problem 16

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?

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Understand that during an isothermal process for an ideal gas, the temperature remains constant. According to the first law of thermodynamics, the change in internal energy (ΔU) is zero for an isothermal process.

Recall the first law of thermodynamics, which is given by the equation: ΔU = Q - W, where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system.

Since the process is isothermal, ΔU = 0. Therefore, the equation simplifies to 0 = Q - W, which can be rearranged to W = Q.

In this problem, 410 J of heat is removed from the gas, which means Q = -410 J (since heat is removed, it is negative).

Substitute Q = -410 J into the equation W = Q to find the work done by the gas. This will give you the value of work done by the gas during the isothermal compression.

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Key Concepts

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

Isothermal Process

An isothermal process is a thermodynamic process in which the temperature of the system remains constant. For an ideal gas, this implies that any heat added to the system is used to do work, and any heat removed is equivalent to the work done on the system. This is described by the equation Q = W, where Q is the heat exchanged and W is the work done.

First Law of Thermodynamics

The First Law of Thermodynamics states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system on its surroundings. In an isothermal process for an ideal gas, the internal energy change is zero, so the heat exchanged is equal to the work done, i.e., ΔU = 0, thus Q = W.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation that relates the pressure, volume, and temperature of an ideal gas through the equation PV = nRT, where P is pressure, V is volume, T is temperature, n is the number of moles, and R is the ideal gas constant. In an isothermal process, the temperature remains constant, so the product of pressure and volume remains constant as well.

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Related Practice

Textbook Question

When water is boiled at a pressure of $2.00$ atm, the heat of vaporization is $2.20 \times 10^{6}$ J/kg and the boiling point is $120$ °C. At this pressure, $1.00$ kg of water has a volume of $1.00 \times 10^{- 3}$ m³, and $1.00$ kg of steam has a volume of $0.824$ m³. Compute the work done when $1.00$ kg of steam is formed at this temperature.

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Textbook Question

When water is boiled at a pressure of $2.00$ atm, the heat of vaporization is $2.20\times10^6$ J/kg and the boiling point is $120$ °C. At this pressure, $1.00$ kg of water has a volume of $1.00\times10^{-3}$ m³, and $1.00$ kg of steam has a volume of $0.824$ m³. Compute the increase in internal energy of the water.

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Textbook Question

An ideal gas is taken from $a$ to $b$ on the $p V$ -diagram shown in Fig. E $19.15$ . During this process, $700$ J of heat is added and the pressure doubles. How does the internal energy of the gas at $a$ compare to the internal energy at $b$ ? Be specific and explain.

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Textbook Question

A cylinder contains $0.0100$ mol of helium at $T equals 27.0$ °C. How much heat is needed to raise the temperature to $67.0$ °C while keeping the volume constant? Draw a $p V$ -diagram for this process.

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Textbook Question

A cylinder contains $0.0100$ mol of helium at $T = 27.0$ °C. If instead the pressure of the helium is kept constant, how much heat is needed to raise the temperature from $27.0$ °C to $67.0$ °C? Draw a $p V$ -diagram for this process.

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Textbook Question

A cylinder contains $0.0100$ mol of helium at $T = 27.0$ °C. What accounts for the difference between your answers to parts (a) and (b)? In which case is more heat required? What becomes of the additional heat?

(a) How much heat is needed to raise the temperature to $67.0$ °C while keeping the volume constant? Draw a $p V$ -diagram for this process.

(b) If instead the pressure of the helium is kept constant, how much heat is needed to raise the temperature from $27.0$ °C to $67.0$ °C? Draw a $p V$ -diagram for this process.

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During an isothermal compression of an ideal gas, 410410 J of heat must be removed from the gas to maintain constant temperature. How much work is done by the gas during the process?

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Key Concepts

Isothermal Process

First Law of Thermodynamics

Ideal Gas Law

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?