Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 15: Problem 59

An ideal gas is heated at a constant pressure of $2.0 \times 10^{5} \mathrm{Pa}\( from a temperature of \)-73^{\circ} \mathrm{C}$ to a temperature of \(+27^{\circ} \mathrm{C}\). The initial volume of the gas is $0.10 \mathrm{m}^{3} .\( The heat energy supplied to the gas in this process is \)25 \mathrm{kJ} .$ What is the increase in internal energy of the gas?

Short Answer

Expert verified
Answer: The increase in internal energy of the gas is 15 kJ.

Step by step solution

01

Convert temperatures to Kelvin

To do any calculations with temperatures, it's essential to convert them to Kelvin first. To convert a Celsius temperature to Kelvin, we simply add 273.15. Initial temperature in Kelvin: \(T_1 = -73^{\circ}\mathrm{C} + 273.15 = 200.15\ \mathrm{K}\) Final temperature in Kelvin: \(T_2 = 27^{\circ}\mathrm{C} + 273.15 = 300.15\ \mathrm{K}\)
02

Calculate the final volume of the gas

Since we are given the constant pressure, we can use the ideal gas law to find the final volume of the gas. The ideal gas law can be written as \(PV = nRT\) Since the pressure, number of moles n, and the gas constant R are constant, we can write \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\) We are given the initial volume (\(V_1 = 0.10\ \mathrm{m}^3\)) and have found the initial and final temperatures in Kelvin. Now, we can solve for the final volume \(V_2\): \(V_2 = \frac{T_2}{T_1} \times V_1 = \frac{300.15\ \mathrm{K}}{200.15\ \mathrm{K}} \times 0.10\ \mathrm{m}^3 \approx 0.150\ \mathrm{m}^3\)
03

Calculate the work done by the gas

The work done by the gas can be calculated using the constant pressure and the change in volume: \(W = P \Delta V = P(V_2 - V_1) = (2.0 \times 10^5\ \mathrm{Pa}) (0.150\ \mathrm{m}^3 - 0.10\ \mathrm{m}^3) = 10\ \mathrm{kJ}\)
04

Use the First Law of Thermodynamics to find the increase in internal energy

The First Law of Thermodynamics states that \(\Delta U = Q - W\) Where \(\Delta U\) is the change in internal energy, \(Q\) is the heat added, and \(W\) is the work done by the system. We are given the amount of heat energy supplied to the gas \(Q = 25\ \mathrm{kJ}\), and we have calculated the work done by the gas \(W = 10\ \mathrm{kJ}\). Now we can find the increase in internal energy: \(\Delta U = 25\ \mathrm{kJ} - 10\ \mathrm{kJ} = 15\ \mathrm{kJ}\) So, the increase in internal energy of the gas is 15 kJ.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose you mix 4.0 mol of a monatomic gas at \(20.0^{\circ} \mathrm{C}\) and 3.0 mol of another monatomic gas at \(30.0^{\circ} \mathrm{C} .\) If the mixture is allowed to reach equilibrium, what is the final temperature of the mixture? [Hint: Use energy conservation.]
Suppose 1.00 mol of oxygen is heated at constant pressure of 1.00 atm from \(10.0^{\circ} \mathrm{C}\) to \(25.0^{\circ} \mathrm{C} .\) (a) How much heat is absorbed by the gas? (b) Using the ideal gas law, calculate the change of volume of the gas in this process. (c) What is the work done by the gas during this expansion? (d) From the first law, calculate the change of internal energy of the gas in this process.
An ideal gas is in contact with a heat reservoir so that it remains at a constant temperature of \(300.0 \mathrm{K}\). The gas is compressed from a volume of \(24.0 \mathrm{L}\) to a volume of 14.0 L. During the process, the mechanical device pushing the piston to compress the gas is found to expend \(5.00 \mathrm{kJ}\) of energy. How much heat flows between the heat reservoir and the gas and in what direction does the heat flow occur?
A \(0.500-\mathrm{kg}\) block of iron at \(60.0^{\circ} \mathrm{C}\) is placed in contact with a \(0.500-\mathrm{kg}\) block of iron at $20.0^{\circ} \mathrm{C} .\( (a) The blocks soon come to a common temperature of \)40.0^{\circ} \mathrm{C} .$ Estimate the entropy change of the universe when this occurs. [Hint: Assume that all the heat flow occurs at an average temperature for each block.] (b) Estimate the entropy change of the universe if, instead, the temperature of the hotter block increased to \(80.0^{\circ} \mathrm{C}\) while the temperature of the colder block decreased to \(0.0^{\circ} \mathrm{C}\) [Hint: The answer is negative, indicating that the process is impossible. \(]\)
An air conditioner whose coefficient of performance is 2.00 removes $1.73 \times 10^{8} \mathrm{J}$ of heat from a room per day. How much does it cost to run the air conditioning unit per day if electricity costs \(\$ 0.10\) per kilowatt-hour? (Note that 1 kilowatt-hour \(=3.6 \times 10^{6} \mathrm{J} .\) )
See all solutions

Recommended explanations on Physics Textbooks

Medical Physics

Read Explanation

Fields in Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Work Energy and Power

Read Explanation

Geometrical and Physical Optics

Read Explanation

Radiation

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free