Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 122

Consider a balloon filled with helium at the following conditions. $$ \begin{array}{l}{313 \mathrm{g} \mathrm{He}} \\ {1.00 \mathrm{atm}} \\ {1910 . \mathrm{L}} \\ {\text { Molar Heat Capacity }=20.8 \mathrm{J} / \mathrm{C} \cdot \mathrm{mol}}\end{array} $$ The temperature of this balloon is decreased by \(41.6^{\circ} \mathrm{C}\) as the volume decreases to \(1643 \mathrm{L},\) with the pressure remaining constant. Determine \(q, w,\) and $\Delta E(\text { in } \mathrm{kJ} \text { ) for the compression of }$ the balloon.

Short Answer

Expert verified
In the compression process of a helium-filled balloon with constant pressure, the heat (q) is -67 kJ, the work (w) is 10.9 kJ, and the change in internal energy (ΔE) is -56.1 kJ.

Step by step solution

01

Calculate the number of moles

To calculate the number of moles, we use the Ideal Gas Law. \[PV = nRT\] We know the mass m of helium is given (313 g) and the molar mass of helium is 4 g/mol. Therefore, the number of moles (n) can be calculated as: \[n = \frac{m}{M}\] \[n = \frac{313 \text{ g}}{4 \text{ g/mol}}\] \[n = 78.25 \text{ mol}\]
02

Calculate the initial temperature T1

Use the Ideal Gas Law again and solve for the initial temperature T1. \[PV = nRT\] Rearranging the equation for T1: \[T1 = \frac{PV}{nR}\] We know P = 1.00 atm, V = 1910 L, n = 78.25 mol, and R ≈ 0.0821 L·atm/K·mol. Now, we can plug the values into the equation. \[T1 = \frac{(1.00 \text{ atm})(1910 \text{ L})}{(78.25 \text{ mol})(0.0821 \text{ L·atm/K·mol})}\] \[T1 ≈ 298.6 \text{K}\]
03

Calculate the final temperature T2

We know that the temperature is decreased by 41.6°C, which is equivalent to 41.6 K. Therefore, we can find the final temperature T2: \[T2 = T1 - \Delta T\] \[T2 = 298.6 \text{ K} - 41.6 \text{K}\] \[T2 ≈ 257 \text{K}\]
04

Calculate heat q

Now, we can use the equation for heat to calculate q: \[q = nC\Delta T\] We have n = 78.25 mol, C = 20.8 J/C·mol, and ΔT = T2 - T1 = -41.6 K. \[q = (78.25 \text{ mol})(20.8 \text{ J/C·mol})(-41.6 \text{ K})\] \[q ≈ -6.7 \times 10^{4} \text{ J}\] Convert the value of q to kJ: \[q ≈ -67 \text{ kJ}\]
05

Calculate work w

To calculate work (w), we use the equation: \[w = -P\Delta V\] We have P = 1.00 atm, and we need to find ΔV. We are given the initial volume V1 = 1910 L, and the final volume V2 = 1643 L. \[\Delta V = V2 - V1\] \[\Delta V = 1643 \text{ L} - 1910 \text{ L}\] \[\Delta V = -267 \text{ L}\] Now, we can calculate w: \[w = -(1.00 \text{ atm})(-267 \text{ L})\] \[w ≈ 267 \text{ L·atm}\] Convert the value of w to kJ: \[w ≈ 10.9 \text{ kJ}\]
06

Calculate the change in internal energy ΔE

Using the First Law of Thermodynamics, we can now calculate the change in internal energy ΔE: \[\Delta E = q + w\] \[\Delta E = -67 \text{ kJ} + 10.9 \text{ kJ}\] \[\Delta E ≈ -56.1 \text{ kJ}\]
07

Final Answers

1. Heat (q): -67 kJ 2. Work (w): 10.9 kJ 3. Change in Internal Energy (ΔE): -56.1 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

What is the difference between \(\Delta H\) and \(\Delta E ?\)

In a coffee-cup calorimeter, 50.0 \(\mathrm{mL}\) of 0.100\(M \mathrm{AgNO}_{3}\) and 50.0 \(\mathrm{mL}\) of 0.100 \(\mathrm{M} \mathrm{HCl}\) 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 reacture in kJ/mol of AgCl formed. Assume that the combined solution has a mass of 100.0 \(\mathrm{g}\) and a specific heat capacity of 4.18 \(\mathrm{J} / \rho \mathrm{C} \cdot \mathrm{g} .\)

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 -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 $\mathrm{J} /^{\prime} \mathrm{C} \cdot \mathrm{g} .$

The bomb calorimeter in Exercise 112 is filled with 987 \(\mathrm{g}\) water. The initial temperature of the calorimeter contents is $23.32^{\circ} \mathrm{C} .\( A \)1.056-\mathrm{g}\( sample of benzoic acid \)\left(\Delta E_{\mathrm{comb}}=\right.\( \)-26.42 \mathrm{kJ} / \mathrm{g}$ ) is combusted in the calorimeter. What is the final temperature of the calorimeter contents?

How is average bond strength related to relative potential energies of the reactants and the products?
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Nuclear Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Reactions

Read Explanation

Chemical Analysis

Read Explanation

Physical 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