Explanation The pressure is 1.10 atm and internal energy change is 71.5 kJ. Also, volume decrease of 13.6 L. The PV work can be calculated as follows: w = − P Δ V Putting the values, w = − ( 1.10 atm ) ( − 13.6 L ) = + 14.96 L atm Since, 1 L atm = 101 .325 J Thus, w = + 14.96 L atm ( 101 .325 J 1 L atm ) = + 1516 J The internal energy change, Δ E is 71

8th Edition

ISBN: 9780135204634

Author: Robinson

Publisher: PEARSON

See similar textbooks

Concept explainers

Thermochemistry

Thermochemistry can be considered as a branch of thermodynamics that deals with the connections between warmth, work, and various types of energy, formed because of different synthetic and actual cycles. Thermochemistry describes the energy changes that …

Exergonic Reaction

The term exergonic is derived from the Greek word in which ‘ergon’ means work and exergonic means ‘work outside’. Exergonic reactions releases work energy. Exergonic reactions are different from exothermic reactions, the one that releases only heat energ…

Videos

Question

Chapter 9, Problem 9.65SP

Interpretation Introduction

Interpretation:

The change in enthalpy for the reaction needs to be determined.

Concept introduction:

The PV work can be calculated as follows:

w=−PΔV

Here, P is pressure and ΔV is change in volume.

It is related to the internal energy and enthalpy as follows:

ΔH=ΔE+PΔV

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 9, Problem 9.64SP

Chapter 9, Problem 9.66SP

Students have asked these similar questions

we were assigned to dilute 900ppm in to 18ppm by using only 250ml vol flask. firstly we did calc and convert 900ppm to 0.9 ppm to dilute in 1 liter. to begin the experiment we took 0,225g of kmno4 and dissolved in to 250 vol flask. then further we took 10 ml sample sol and dissolved in to 100 ml vol flask and put it in to a spectrometer and got value of 0.145A . upon further calc we got v2 as 50ml . need to find DF, % error (expval and accptVal), molarity, molality. please write the whole report. thank you The format, tables, introduction, procedure and observation, result, calculations, discussion and conclusion

Q5. Predict the organic product(s) for the following transformations. If no reaction will take place (or the reaction is not synthetically useful), write "N.R.". Determine what type of transition state is present for each reaction (think Hammond Postulate). I Br₂ CH3 F2, light CH3 Heat CH3 F₂ Heat Br2, light 12, light CH3 Cl2, light No

None

Chapter 9 Solutions

CHEMISTRY-MASTERINGCHEMISTRY W/ETEXT

Ch. 9 - Prob. 9.1P Ch. 9 - Conceptual APPLY 9.2 How much work is done in...Ch. 9 - PRACTICE 9.3 The reaction between hydrogen and...Ch. 9 - Conceptual APPLY 9.4 The following reaction has E...Ch. 9 - Use the following thermo chemical equation to...Ch. 9 - APPLY 9.6 Approximately, 1.8106 kJ of energy is...Ch. 9 - Classify the following reaction as end o- or ex...Ch. 9 - APPLY 9.8 Instant hot packs and cold packs contain...Ch. 9 - PRACTICE 9.9 What is the specific heat of lead in...Ch. 9 - APPLY 9.10 Calculate the heat capacity (C) of a...

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

The change in enthalpy for the reaction needs to be determined. Concept introduction: The PV work can be calculated as follows: w = − P Δ V Here, P is pressure and Δ V is change in volume. It is related to the internal energy and enthalpy as follows: Δ H = Δ E + P Δ V

Solution Summary: The author explains that the change in enthalpy for the reaction needs to be determined.

Concept explainers

Videos

Interpretation:

The change in enthalpy for the reaction needs to be determined.

Concept introduction:

The PV work can be calculated as follows:

w=−PΔV

Here, P is pressure and ΔV is change in volume.

It is related to the internal energy and enthalpy as follows:

ΔH=ΔE+PΔV

Want to see the full answer?

Chapter 9 Solutions

The change in enthalpy for the reaction needs to be determined. Concept introduction: The PV work can be calculated as follows: w = − P Δ V Here, P is pressure and Δ V is change in volume. It is related to the internal energy and enthalpy as follows: Δ H = Δ E + P Δ V

Solution Summary: The author explains that the change in enthalpy for the reaction needs to be determined.

Concept explainers

Videos

Interpretation: The change in enthalpy for the reaction needs to be determined. Concept introduction: The PV work can be calculated as follows: w=−PΔV Here, P is pressure and ΔV is change in volume. It is related to the internal energy and enthalpy as follows: ΔH=ΔE+PΔV

Want to see the full answer?

Chapter 9 Solutions

Interpretation:

The change in enthalpy for the reaction needs to be determined.

Concept introduction:

The PV work can be calculated as follows:

w=−PΔV

Here, P is pressure and ΔV is change in volume.

It is related to the internal energy and enthalpy as follows:

ΔH=ΔE+PΔV