Skip to main content
Ch 19: The First Law of Thermodynamics
Chapter 19, Problem 19

Five moles of an ideal monatomic gas with an initial temperature of 127°C expand and, in the process, absorb 1500 J of heat and do 2100 J of work. What is the final temperature of the gas?

Verified step by step guidance
1
Convert the initial temperature from Celsius to Kelvin by using the formula: T(K) = T(°C) + 273.15.
Use the first law of thermodynamics, which states that the change in internal energy of the system (\(\Delta U\)) is equal to the heat added to the system (Q) minus the work done by the system (W): \(\Delta U = Q - W\).
Calculate the change in internal energy (\(\Delta U\)) using the values of heat absorbed and work done: \(\Delta U = 1500 \, \text{J} - 2100 \, \text{J}\).
For an ideal monatomic gas, the change in internal energy can also be expressed in terms of the number of moles (n), the gas constant (R), and the change in temperature (\(\Delta T\)): \(\Delta U = n \cdot \frac{3}{2} \cdot R \cdot \Delta T\).
Solve for the final temperature (T_f) by rearranging the equation: \(T_f = T_i + \Delta T\), where \(\Delta T = \frac{\Delta U}{n \cdot \frac{3}{2} \cdot R}\) and substituting the initial temperature (T_i) converted to Kelvin.

Verified Solution

Video duration:
5m
This video solution was recommended by our tutors as helpful for the problem above.
Was this helpful?

Key Concepts

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

Ideal Gas Law

The Ideal Gas Law relates the pressure, volume, temperature, and number of moles of an ideal gas through the equation PV = nRT. This law is fundamental in understanding the behavior of gases under various conditions and is essential for calculating changes in temperature when heat and work are involved.
Recommended video:
Guided course
07:21
Ideal Gases and the Ideal Gas Law

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. This principle is crucial for analyzing energy transfers in thermodynamic processes, such as the one described in the question.
Recommended video:
Guided course
08:04
The First Law of Thermodynamics

Monatomic Ideal Gas

A monatomic ideal gas consists of single atoms and follows the ideal gas behavior closely. The specific heat capacities for monatomic gases are well-defined, which allows for straightforward calculations of temperature changes when heat is added or work is done, making it easier to determine the final temperature in this scenario.
Recommended video:
Guided course
07:21
Ideal Gases and the Ideal Gas Law
Related Practice
Textbook Question
A cylinder contains 0.100 mol of an ideal monatomic gas. Initially the gas is at 1.00 * 10^5 Pa and occupies a volume of 2.50 * 10^-3 m^3. (b) If the gas is allowed to expand to twice the initial volume, find the final temperature (in kelvins) and pressure of the gas if the expansion is (i) isothermal; (ii) isobaric; (iii) adiabatic.
775
views
Textbook Question
A cylinder contains 0.100 mol of an ideal monatomic gas. Initially the gas is at 1.00 * 10^5 Pa and occupies a volume of 2.50 * 10^-3 m^3. (b) If the gas is allowed to expand to twice the initial volume, find the final temperature (in kelvins) and pressure of the gas if the expansion is (i) isothermal; (ii) isobaric; (iii) adiabatic.
506
views
Textbook Question
Work Done in a Cyclic Process. (a) In Fig. 19.7a, consider the closed loop 1 → 3 → 2 → 4 → 1. This is a cyclic process in which the initial and final states are the same. Find the total work done by the system in this cyclic process, and show that it is equal to the area enclosed by the loop.

1832
views
Textbook Question
A gas in a cylinder is held at a constant pressure of 1.80 * 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 * 10^5 J. (c) Does it matter whether the gas is ideal? Why or why not?
435
views
Textbook Question
Boiling Water at High Pressure. When water is boiled at a pressure of 2.00 atm, the heat of vaporization is 2.20 * 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 * 10^-3 m^3 , and 1.00 kg of steam has a volume of 0.824 m^3. (b) Compute the increase in internal energy of the water.
493
views
Textbook Question
An experimenter adds 970 J of heat to 1.75 mol of an ideal gas to heat it from 10.0°C to 25.0°C at constant pressure. The gas does +223 J of work during the expansion. (b) Calculate γ for the gas.
480
views