Connect 1 Semester Access Card for General Chemistry: The Essential Concepts
Connect 1 Semester Access Card for General Chemistry: The Essential Concepts
7th Edition
ISBN: 9781259692543
Author: Raymond Chang Dr.; Kenneth Goldsby Professor
Publisher: McGraw-Hill Education
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Chapter 6, Problem 6.95QP

(a)

Interpretation Introduction

Interpretation:

The mass of Helium in the balloon has to be calculated.

Concept Introduction:

Ideal gas law is applicable to those gases which obey Boyle’s law and Charles’s law. The ideal gas equation can be obtained by combining the equations of Boyle’s law and Charles’s law.

At constant temperature, (Boyle’s law)

1V

At constant volume, (Charles’s law)

PαT

By combining the above equations,

TVPVαTPV=RT

Where R= proportionality constant called as gas constant.

The general equation for ideal gas law is written as,

PV=nRT

Where n= number of moles

(a)

Expert Solution
Check Mark

Answer to Problem 6.95QP

The mass of Helium in the balloon is 3.4×105g.

Explanation of Solution

Given,

Diameter of balloon is 16m

Temperature of inflation is 18°C

The volume of balloon is calculated as,

Volume = 43πr3Volume = 43π(8m)3Volume = (2.1×103m)3×1000L1m3Volume = 2.1×106L

The mass of Helium is calculated as,

nHe = PVRTnHe = (98.7kPa×1atm1.01325×102kPa)(2.1×106L)(0.0821L.atmmoleK)(273+18)KnHe = 8.6×104moleHeMassofHelium = (8.6×104mole)×4.003gHe1moleHeMassofHelium = 3.4×105g

The mass of Helium in the balloon is 3.4×105g.

(b)

Interpretation Introduction

Interpretation:

The work done in joules has to be calculated.

Concept Introduction:

The work done on the system or by the system can be calculated from the equation

w=-PΔV

Where w=work done

P=Pressure

ΔV=Change in volume

(b)

Expert Solution
Check Mark

Answer to Problem 6.95QP

The work done in joules is -2.0×108J.

Explanation of Solution

Given,

Diameter of balloon is 16m

Temperature of inflation is 18°C

Pressure is 98.7kPa

The volume of balloon is calculated as,

Volume = 43πr3Volume = 43π(8m)3Volume = (2.1×103m)3×1000L1m3Volume = 2.1×106L

The work done is calculated as,

w = -PΔVw = -(98.7kPa×1atm1.01325×102kPa)(2.1×106L)(101.3J1L.atm)w = -2.0×108J

The work done in joules is -2.0×108J.

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Chapter 6 Solutions

Connect 1 Semester Access Card for General Chemistry: The Essential Concepts

Ch. 6.5 - Prob. 4PECh. 6.5 - Prob. 1RCCh. 6.6 - Prob. 1PECh. 6.6 - Prob. 2PECh. 6.6 - Prob. 1RCCh. 6 - Prob. 6.1QPCh. 6 - Prob. 6.2QPCh. 6 - Prob. 6.3QPCh. 6 - Prob. 6.4QPCh. 6 - Prob. 6.5QPCh. 6 - Prob. 6.7QPCh. 6 - Prob. 6.8QPCh. 6 - Prob. 6.9QPCh. 6 - Prob. 6.10QPCh. 6 - Prob. 6.11QPCh. 6 - Prob. 6.12QPCh. 6 - 6. 13 The internal energy of an ideal gas depends...Ch. 6 - 6.14 Consider these changes. At constant...Ch. 6 - Prob. 6.15QPCh. 6 - Prob. 6.16QPCh. 6 - Prob. 6.17QPCh. 6 - Prob. 6.18QPCh. 6 - 6.19 Calculate the work done when 50.0 g of tin...Ch. 6 - Prob. 6.20QPCh. 6 - Prob. 6.21QPCh. 6 - Prob. 6.22QPCh. 6 - Prob. 6.23QPCh. 6 - Prob. 6.24QPCh. 6 - Prob. 6.25QPCh. 6 - 6.26 Determine the amount of heat (in kJ) given...Ch. 6 - Prob. 6.27QPCh. 6 - Prob. 6.28QPCh. 6 - Prob. 6.29QPCh. 6 - Prob. 6.30QPCh. 6 - Prob. 6.31QPCh. 6 - Prob. 6.32QPCh. 6 - Prob. 6.33QPCh. 6 - Prob. 6.34QPCh. 6 - Prob. 6.35QPCh. 6 - Prob. 6.36QPCh. 6 - Prob. 6.37QPCh. 6 - Prob. 6.38QPCh. 6 - Prob. 6.39QPCh. 6 - Prob. 6.40QPCh. 6 - Prob. 6.41QPCh. 6 - Prob. 6.42QPCh. 6 - Prob. 6.43QPCh. 6 - Prob. 6.44QPCh. 6 - Prob. 6.45QPCh. 6 - 6.46 The values of the two allotropes of oxygen,...Ch. 6 - 6.47 Which is the more negative quantity at 25°C: ...Ch. 6 - Prob. 6.48QPCh. 6 - Prob. 6.49QPCh. 6 - Prob. 6.50QPCh. 6 - Prob. 6.51QPCh. 6 - Prob. 6.52QPCh. 6 - Prob. 6.53QPCh. 6 - Prob. 6.54QPCh. 6 - Prob. 6.55QPCh. 6 - Prob. 6.56QPCh. 6 - Prob. 6.57QPCh. 6 - 6.58 The first step in the industrial recovery or...Ch. 6 - Prob. 6.59QPCh. 6 - Prob. 6.60QPCh. 6 - Prob. 6.61QPCh. 6 - Prob. 6.62QPCh. 6 - Prob. 6.63QPCh. 6 - Prob. 6.64QPCh. 6 - Prob. 6.65QPCh. 6 - Prob. 6.66QPCh. 6 - Prob. 6.67QPCh. 6 - Prob. 6.68QPCh. 6 - Prob. 6.69QPCh. 6 - Prob. 6.70QPCh. 6 - Prob. 6.71QPCh. 6 - Prob. 6.72QPCh. 6 - Prob. 6.73QPCh. 6 - Prob. 6.74QPCh. 6 - Prob. 6.75QPCh. 6 - Prob. 6.76QPCh. 6 - Prob. 6.77QPCh. 6 - Prob. 6.78QPCh. 6 - Prob. 6.79QPCh. 6 - Prob. 6.80QPCh. 6 - Prob. 6.81QPCh. 6 - Prob. 6.82QPCh. 6 - Prob. 6.83QPCh. 6 - Prob. 6.84QPCh. 6 - Prob. 6.85QPCh. 6 - Prob. 6.86QPCh. 6 - Prob. 6.87QPCh. 6 - Prob. 6.88QPCh. 6 - Prob. 6.89QPCh. 6 - Prob. 6.90QPCh. 6 - Prob. 6.91QPCh. 6 - Prob. 6.92QPCh. 6 - Prob. 6.93QPCh. 6 - Prob. 6.94QPCh. 6 - Prob. 6.95QPCh. 6 - Prob. 6.96QPCh. 6 - Prob. 6.97QPCh. 6 - Prob. 6.98QPCh. 6 - Prob. 6.100QPCh. 6 - Prob. 6.101QPCh. 6 - Prob. 6.102QPCh. 6 - Prob. 6.103QPCh. 6 - Prob. 6.104QPCh. 6 - Prob. 6.105QPCh. 6 - Prob. 6.106SPCh. 6 - Prob. 6.107SPCh. 6 - Prob. 6.109SPCh. 6 - Prob. 6.110SPCh. 6 - Prob. 6.111SPCh. 6 - Prob. 6.112SPCh. 6 - Prob. 6.113SPCh. 6 - Prob. 6.114SPCh. 6 - Prob. 6.115SPCh. 6 - Prob. 6.116SPCh. 6 - Prob. 6.117SPCh. 6 - Prob. 6.118SP
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