Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 40

One mole of \(\mathrm{H}_{2} \mathrm{O}(g)\) at 1.00 atm and \(100 .^{\circ} \mathrm{C}\) occupies a volume of 30.6 L. When 1 mole of \(\mathrm{H}_{2} \mathrm{O}(g)\) is condensed to 1 mole of \(\mathrm{H}_{2} \mathrm{O}(l)\) at 1.00 atm and \(100 .^{\circ} \mathrm{C}, 40.66 \mathrm{kJ}\) of heat is released. If the density of \(\mathrm{H}_{2} \mathrm{O}(l)\) at this temperature and pressure is \(0.996 \mathrm{g} / \mathrm{cm}^{3},\) calculate \(\Delta E\) for the condensation of 1 mole of water at 1.00 atm and \(100 .^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The change in internal energy, \(\Delta E\), for the condensation of 1 mole of water at 1.00 atm and \(100^{\circ}\)C is -37.562 kJ.

Step by step solution

01

1. Calculate the number of moles of water vapor

We are given that 1 mole of water vapor is being condensed. So, n = 1 moles of \(\mathrm{H}_{2} \mathrm{O}(g)\).
02

2. Determine the heat, q, released during condensation

We are given that 40.66 kJ of heat is released during the condensation process. Since it is released, it will be a negative value. So, q = -40.66 kJ.
03

3. Calculate the initial and final volume of the system

The initial volume of the system is given as 30.6 L, which is the volume of 1 mole of water vapor. To find the final volume, we can use the given density of liquid water at this temperature and pressure, which is 0.996 g/cm³. First, find the mass of 1 mole of water: mass = (1 mole) × (18.015 g/mol) = 18.015 g Now, we can calculate the volume of liquid water using the density formula: volume = mass/density Convert the density to g/L: density = 0.996 g/cm³ × \(\frac{1000 \mathrm{cm}^{3}}{\mathrm{L}}\) = 996 g/L The final volume, V₂, can be calculated as: V₂ = \(\frac{18.015 \mathrm{g}}{996 \mathrm{g}/\mathrm{L}}\) = 0.0181 L
04

4. Calculate the work, w, done by the system

The work done by the system can be calculated using the formula for work done by a gas at constant pressure: \[w = -P(V_{2}-V_{1})\] We are given that the pressure, P, is 1.00 atm. First, let's convert pressure to J/L units by using the conversion factor: 1 atm = 101.325 J/L P = 1.00 atm × \(\frac{101.325 \mathrm{J}}{\mathrm{L}}\) = 101.325 J/L Now, we can substitute the given values into the formula for work: w = -101.325 J/L × (0.0181 L - 30.6 L) = -101.325 J/L × (-30.582 L) = 3098.00 J
05

5. Calculate the change in internal energy, \(\Delta E\)

Now we apply the first law of thermodynamics, which states that the change in internal energy, \(\Delta E\), can be calculated as a sum of heat, q, and work, w: \(\Delta E = q + w\) Plugging in the values obtained in Steps 2 and 4: \(\Delta E = -40.66 \mathrm{kJ} + 3098 \mathrm{J}\) Since the energy units are different, we must convert one of them to be consistent. Let's convert 3098 J to kJ: 3098 J = 3.098 kJ Now we find \(\Delta E\): \(\Delta E = -40.66 \mathrm{kJ} + 3.098 \mathrm{kJ} = -37.562 \mathrm{kJ}\) Hence, the change in internal energy, \(\Delta E\), for the condensation of 1 mole of water at 1.00 atm and \(100^{\circ}\)C is -37.562 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

For the reaction \(\mathrm{HgO}(s) \rightarrow \mathrm{Hg}(l)+\frac{1}{2} \mathrm{O}_{2}(g), \Delta H=+90.7 \mathrm{kJ}\). a. What quantity of heat is required to produce 1 mole of mercury by this reaction? b. What quantity of heat is required to produce 1 mole of oxygen gas by this reaction? c. What quantity of heat would be released in the following reaction as written? $$2 \mathrm{Hg}(l)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{HgO}(s)$$

Consider the reaction $$\mathrm{B}_{2} \mathrm{H}_{6}(g)+3 \mathrm{O}_{2}(g) \longrightarrow \mathrm{B}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2} \mathrm{O}(g) \quad \Delta H=-2035 \mathrm{kJ}$$ Calculate the amount of heat released when \(54.0 \mathrm{g}\) of diborane is combusted.

Consider the dissolution of \(\mathrm{CaCl}_{2}\) : $$\mathrm{CaCl}_{2}(s) \longrightarrow \mathrm{Ca}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q) \quad \Delta H=-81.5 \mathrm{kJ}$$ An \(11.0-\mathrm{g}\) sample of \(\mathrm{CaCl}_{2}\) is dissolved in 125 g water, with both substances at \(25.0^{\circ} \mathrm{C}\). Calculate the final temperature of the solution assuming no heat loss to the surroundings and assuming the solution has a specific heat capacity of \(4.18 \mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g}\).

Given the following data $$\begin{array}{cl}\mathrm{P}_{4}(s)+6 \mathrm{Cl}_{2}(g) \longrightarrow 4 \mathrm{PCl}_{3}(g) & \Delta H=-1225.6 \mathrm{kJ} \\ \mathrm{P}_{4}(s)+5 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) & \Delta H=-2967.3 \mathrm{kJ} \\\ \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{PCl}_{5}(g) & \Delta H=-84.2 \mathrm{kJ} \\ \mathrm{PCl}_{3}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{Cl}_{3} \mathrm{PO}(g) & \Delta H=-285.7 \mathrm{kJ} \end{array}$$ calculate \(\Delta H\) for the reaction $$\mathrm{P}_{4} \mathrm{O}_{10}(s)+6 \mathrm{PCl}_{5}(g) \longrightarrow 10 \mathrm{Cl}_{3} \mathrm{PO}(g)$$

In a coffee-cup calorimeter, \(50.0 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{AgNO}_{3}\) and \(50.0 \mathrm{mL}\) of \(0.100 \mathrm{M}\) HCl are mixed to yield the following reaction: $$\mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{AgCl}(s)$$ The two solutions were initially at \(22.60^{\circ} \mathrm{C},\) and the final temperature is \(23.40^{\circ} \mathrm{C}\). Calculate the heat that accompanies this reaction in \(\mathrm{kJ} / \mathrm{mol}\) of \(\mathrm{AgCl}\) formed. Assume that the combined solution has a mass of \(100.0 \mathrm{g}\) and a specific heat capacity of \(4.18 \mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g}\).
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Making Measurements

Read Explanation

Kinetics

Read Explanation

Organic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Nuclear Chemistry

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