Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 15: Problem 9

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.

Short Answer

Expert verified
Question: A 1.00 mol oxygen gas (O2) is heated from 10.0°C to 25.0°C under constant pressure of 1.00 atm. The heat capacity of O2 under constant pressure (Cp) is 29.4 J/mol.K. Calculate (a) the heat absorbed by the gas, (b) the change in volume, (c) the work done during expansion, and (d) the change in internal energy. Answer: (a) The heat absorbed by the gas is 441 J. (b) The change in volume is 0.00206 m^3. (c) The work done during expansion is -208.73 J. (d) The change in internal energy is 232.27 J.

Step by step solution

01

Part (a): Calculate the heat absorbed by the gas

To find the heat absorbed (q) by the gas, use the formula: q = n * Cp * ΔT where n is the number of moles, Cp is the heat capacity at constant pressure, and ΔT is the change in temperature (T2 - T1). ΔT = T2 - T1 = 25.0 - 10.0 = 15.0 K q = (1.00 mol) * (29.4 J/mol.K) * (15.0 K) q = 441 J So, the gas absorbs 441 J of heat.
02

Part (b): Calculate the change in volume

We can use the ideal gas law to find the change in volume (ΔV): PV = nRT First, convert pressure from atm to Pa by multiplying it with the conversion factor (1 atm = 101325 Pa). Then, calculate the initial and final volumes using the ideal gas law: P = 1.00 atm * 101325 Pa/atm = 101325 Pa R = 8.314 J/mol.K Initial volume (V1): V1 = nRT1 / P V1 = (1.00 mol)(8.314 J/mol.K)(283.15 K) / 101325 Pa V1 = 0.02241 m^3 Final volume (V2): V2 = nRT2 / P V2 = (1.00 mol)(8.314 J/mol.K)(298.15 K) / 101325 Pa V2 = 0.02447 m^3 Change in volume (ΔV): ΔV = V2 - V1 ΔV = 0.02447 - 0.02241 ΔV = 0.00206 m^3 The change in volume is 0.00206 m^3.
03

Part (c): Calculate the work done

To find the work done (w) by the gas during expansion, we use the formula: w = -P(ΔV) w = -101325 Pa * 0.00206 m^3 w = -208.73 J The work done by the gas during expansion is -208.73 J.
04

Part (d): Calculate the change in internal energy

To find the change in internal energy (ΔU), we use the first law of thermodynamics: ΔU = q + w ΔU = 441 J + (-208.73 J) ΔU = 232.27 J The change in internal energy of the gas is 232.27 J.

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

A reversible refrigerator has a coefficient of performance of \(3.0 .\) How much work must be done to freeze \(1.0 \mathrm{kg}\) of liquid water initially at \(0^{\circ} \mathrm{C} ?\)
Show that the coefficient of performance for a reversible heat pump is $1 /\left(1-T_{\mathrm{C}} / T_{\mathrm{H}}\right)$.
(a) How much heat does an engine with an efficiency of \(33.3 \%\) absorb in order to deliver \(1.00 \mathrm{kJ}\) of work? (b) How much heat is exhausted by the engine?
A \(0.50-\mathrm{kg}\) block of iron $[c=0.44 \mathrm{kJ} /(\mathrm{kg} \cdot \mathrm{K})]\( at \)20.0^{\circ} \mathrm{C}$ is in contact with a \(0.50-\mathrm{kg}\) block of aluminum \([c=\) $0.900 \mathrm{kJ} /(\mathrm{kg} \cdot \mathrm{K})]$ at a temperature of \(20.0^{\circ} \mathrm{C} .\) The system is completely isolated from the rest of the universe. Suppose heat flows from the iron into the aluminum until the temperature of the aluminum is \(22.0^{\circ} \mathrm{C}\) (a) From the first law, calculate the final temperature of the iron. (b) Estimate the entropy change of the system. (c) Explain how the result of part (b) shows that this process is impossible. [Hint: since the system is isolated, $\left.\Delta S_{\text {System }}=\Delta S_{\text {Universe }} .\right]$
A reversible heat engine has an efficiency of \(33.3 \%\) removing heat from a hot reservoir and rejecting heat to a cold reservoir at $0^{\circ} \mathrm{C} .$ If the engine now operates in reverse, how long would it take to freeze \(1.0 \mathrm{kg}\) of water at \(0^{\circ} \mathrm{C},\) if it operates on a power of \(186 \mathrm{W} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Torque and Rotational Motion

Read Explanation

Mechanics and Materials

Read Explanation

Waves Physics

Read Explanation

Physical Quantities And Units

Read Explanation

Atoms and Radioactivity

Read Explanation

Measurements

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