Skip to main content
Ch 19: The First Law of Thermodynamics
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Non è quello che usi tu?Cambia libro di testo
Capitolo 19, Problema 16

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?

Guida verificata passo dopo passo
1
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.

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.
Durata del video:
2m

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

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.
Video consigliato:
Percorso guidato
06:13
Entropy & Ideal Gas Processes

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.
Video consigliato:
Percorso guidato
08:04
The First Law of Thermodynamics

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.
Video consigliato:
Percorso guidato
07:21
Ideal Gases and the Ideal Gas Law
Pratica correlata
Domanda del libro di testo

A cylinder contains 0.01000.0100 mol of helium at T=27.0T = 27.0°C. If instead the pressure of the helium is kept constant, how much heat is needed to raise the temperature from 27.027.0°C to 67.067.0°C? Draw a pVpV-diagram for this process.

1335
views
Domanda del libro di testo

A cylinder contains 0.01000.0100 mol of helium at T=27.0T = 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 67.067.0°C while keeping the volume constant? Draw a pVpV-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 27.027.0°C to 67.067.0°C? Draw a pVpV-diagram for this process.

1265
views
Domanda del libro di testo

When water is boiled at a pressure of 2.002.00 atm, the heat of vaporization is 2.20×1062.20\(\times\)10^6 J/kg and the boiling point is 120120°C. At this pressure, 1.001.00 kg of water has a volume of 1.00×10−31.00\(\times\)10^{-3} m3, and 1.001.00 kg of steam has a volume of 0.8240.824 m3. Compute the increase in internal energy of the water.

1798
views
Domanda del libro di testo

An ideal gas is taken from aa to bb on the pVpV-diagram shown in Fig. E19.1519.15. During this process, 700700 J of heat is added and the pressure doubles. How does the internal energy of the gas at aa compare to the internal energy at bb? Be specific and explain.

2321
views
Domanda del libro di testo

A cylinder contains 0.01000.0100 mol of helium at T=27.0T = 27.0°C. How much heat is needed to raise the temperature to 67.067.0°C while keeping the volume constant? Draw a pVpV-diagram for this process.

1378
views
Domanda del libro di testo

When water is boiled at a pressure of 2.002.00 atm, the heat of vaporization is 2.20×1062.20\(\times\)10^6 J/kg and the boiling point is 120120°C. At this pressure, 1.001.00 kg of water has a volume of 1.00×10−31.00\(\times\)10^{-3} m3, and 1.001.00 kg of steam has a volume of 0.8240.824 m3. Compute the work done when 1.001.00 kg of steam is formed at this temperature.

4116
views