MCGRAW: CHEMISTRY THE MOLECULAR NATURE
MCGRAW: CHEMISTRY THE MOLECULAR NATURE
8th Edition
ISBN: 9781264330430
Author: VALUE EDITION
Publisher: MCG
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Chapter 6, Problem 6.89P

(a)

Interpretation Introduction

Interpretation:

The initial and final volume of helium bag in early morning and mid-afternoon is to be determined.

Concept introduction:

The ideal gas equation relates volume, moles, temperature, and pressure with each other.

The ideal gas equation is as follows:

PV=nRT (1)

Here,

P is the pressure.

V is the volume.

n is the number of moles of gases.

R is the gas constant.

T is the temperature.

The conversion factor to convert degree Celsius to Kelvin is as follows:

T(K)=T(°C)+273 (2)

(a)

Expert Solution
Check Mark

Answer to Problem 6.89P

The initial and final volume of helium bag is 23.6L and 24.9L respectively.

Explanation of Solution

Substitute 15°C for Tinitial(°C) in the equation (2).

Tinitial(K)=15°C+273=288K

Substitute 30°C for Tfinal(°C) in the equation (2).

Tfinal(K)=30°C+273=303K

Rearrange the equation(1) to calculate the volume.

V=nRTP (3)

Substitute 1mol for n, 0.0821Latm/molK for R, 1atm for P and 288K for Tinitial in the equation(3).

Vinitial=(1mol)(0.0821Latm/molK)(288K)1atm=23.6448L23.6L

Substitute 1mol for n, 0.0821Latm/molK for R, 1atm for P and 303K for Tfinal in the equation(3).

Vfinal=(1mol)(0.0821Latm/molK)(303K)1atm=24.8736L24.9L

Conclusion

The initial and final volume of the helium bag is 23.6L and 24.9L respectively.

(b)

Interpretation Introduction

Interpretation:

The change in internal energy is to be determined.

Concept introduction:

The internal energy of a process is the summation of the kinetic energy and potential energy associated with the process. In the case of a reaction, the change in internal energy (ΔE) is the difference in the energy of the product and reactant.

(b)

Expert Solution
Check Mark

Answer to Problem 6.89P

The change in internal energy is 187J.

Explanation of Solution

The formula to calculate the change in internal energy is,

ΔE=32nR(TfinalTinitial) (4)

Substitute 1mol for n, 0.0821Latm/molK for R, 303K for Tfinal and 288K for Tinitial in the equation (4).

ΔE=32(1mol)(0.0821Latm/molK)(303K288K)=187.065J187J

Conclusion

The change in internal energy is 187J.

(c)

Interpretation Introduction

Interpretation:

Work done by helium is to be calculated.

Concept introduction:

Work (w) is the energy needed to move an object in the opposite direction of the force applied. The work done by the system is negative and the work done on the system is positive. The formula to calculate the work done against an external pressure is:

w=PΔV (5)

Here,

w is the work done by the system or on the system.

P is the external pressure.

ΔV is the change in the volume.

(c)

Expert Solution
Check Mark

Answer to Problem 6.89P

Work done by helium is 1.2×102J.

Explanation of Solution

The initial volume of gas is 23.6448L.

The final volume of gas is 24.8736L.

The formula to calculate ΔV of the system is:

ΔV=VFinalVInitial (6)

Substitute 23.6448L for VInitial and 24.8736L for VFinal in the equation (6).

ΔV=16.3 L10.5 L=5.8L

Substitute 1atm for P and 5.8L for ΔV in the equation (5).

w=(1atm)(5.8L)=(5.8atmL)(101.3J1atmL)=124.75J1.2×102J

Conclusion

Work done by helium is 1.2×102J.

(d)

Interpretation Introduction

Interpretation:

Heat transferred is to be calculated.

Concept introduction:

The first law of thermodynamics gives the relation between the work, heat and internal energy. According to this law “the total energy of the system plus surrounding remains constant”. The expression to the first law of thermodynamics is:

ΔE=q+w (7)

Here,

ΔE is the internal energy of the system.

q is the heat released or absorbed.

w is the work done by the system or on the system.

(d)

Expert Solution
Check Mark

Answer to Problem 6.89P

The heat transferred is 3.1×102J.

Explanation of Solution

Rearange equation (7) to calculate q.

q=ΔEw (8)

Substitute 187.065J for ΔE and 124.75J for w in the equation (7).

q=187.065J(124.75J)=311.815J3.1×102J

Conclusion

The heat transferred is 3.1×102J.

(e)

Interpretation Introduction

Interpretation:

ΔH of the process is to be calculated.

Concept introduction:

The internal energy of a process is the summation of the kinetic energy and potential energy associated with the process. In the case of a reaction, the change in internal energy (ΔE) is the difference in the energy of the product and reactant but at constant pressure, the PΔV work gets eliminated and the change in enthalpy (ΔH) is measured.

In the case of a reaction, the change in enthalpy (ΔH) is the difference in the energy of the product and reactant. The general expression to calculate ΔH is,

ΔH=HProductHReactant (9)

Here,

ΔH is the change in enthalpy of the system.

HProduct is the enthalpy of the products.

HReactant is the enthalpy of the reactants.

The heat flow at constant pressure (qp) is equal to the change in enthalpy change.

(e)

Expert Solution
Check Mark

Answer to Problem 6.89P

ΔH of the process is 3.1×102J.

Explanation of Solution

The heat transferred in the process is 3.1×102J.

The heat flow at constant pressure (qp) is equal to the change in enthalpy change. Therefore, the value of the change in the enthalpy (ΔH) is 3.1×102J.

Conclusion

ΔH of the process is 3.1×102J.

(f)

Interpretation Introduction

Interpretation:

The relationship between ΔH and q of the process is to be determined.

Concept introduction:

The internal energy of a process is the summation of the kinetic energy and potential energy associated with the process. In the case of a reaction, the change in internal energy (ΔE) is the difference in the energy of the product and reactant but at constant pressure, the PΔV work gets eliminated and the change in enthalpy (ΔH) is measured.

In the case of a reaction, the change in enthalpy (ΔH) is the difference in the energy of the product and reactant. The general expression to calculate ΔH is,

ΔH=HProductHReactant (9)

Here,

ΔH is the change in enthalpy of the system.

HProduct is the enthalpy of the products.

HReactant is the enthalpy of the reactants.

Work (w) is the energy needed to move an object in the opposite direction of the force applied. The work done by the system is negative and the work done on the system is positive. The formula to calculate the work done against an external pressure is:

w=PΔV (5)

Here,

w is the work done by the system or on the system.

P is the external pressure.

ΔV is the change in the volume.

The first law of thermodynamics gives the relation between the work, heat and internal energy. According to this law “the total energy of the system plus surrounding remains constant”. The expression to first law of thermodynamics is:

ΔE=q+w (7)

Here,

ΔE is the internal energy of the system.

q is the heat released or absorbed.

w is the work done by the system or on the system.

(f)

Expert Solution
Check Mark

Answer to Problem 6.89P

The heat flow at constant pressure (qp) is equal to the change in enthalpy change.

Explanation of Solution

The formula to calculate ΔH is as follows:

ΔH=ΔE+PΔV (10)

Substitute w for PΔV and q+w for ΔE in the equation (10).

ΔH=q+w+(w)=qp

Conclusion

The heat flow at constant pressure (qp) is equal to the change in enthalpy change.

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Chapter 6 Solutions

MCGRAW: CHEMISTRY THE MOLECULAR NATURE

Ch. 6.3 - When 25.0 mL of 2.00 M HNO3 and 50.0 mL of 1.00 M...Ch. 6.3 - Prob. 6.6BFPCh. 6.3 - Prob. 6.7AFPCh. 6.3 - Prob. 6.7BFPCh. 6.4 - Prob. 6.8AFPCh. 6.4 - Prob. 6.8BFPCh. 6.5 - Prob. 6.9AFPCh. 6.5 - Prob. 6.9BFPCh. 6.6 - Prob. 6.10AFPCh. 6.6 - Prob. 6.10BFPCh. 6.6 - Prob. 6.11AFPCh. 6.6 - Prob. 6.11BFPCh. 6.6 - Prob. B6.1PCh. 6.6 - Prob. B6.2PCh. 6 - Prob. 6.1PCh. 6 - Prob. 6.2PCh. 6 - Prob. 6.3PCh. 6 - Prob. 6.4PCh. 6 - Prob. 6.5PCh. 6 - Prob. 6.6PCh. 6 - Prob. 6.7PCh. 6 - Prob. 6.8PCh. 6 - Prob. 6.9PCh. 6 - A system releases 255 cal of heat to the...Ch. 6 - What is the change in internal energy (in J) of a...Ch. 6 - Prob. 6.12PCh. 6 - Prob. 6.13PCh. 6 - Thermal decomposition of 5.0 metric tons of...Ch. 6 - Prob. 6.15PCh. 6 - The external pressure on a gas sample is 2660...Ch. 6 - The nutritional calorie (Calorie) is equivalent to...Ch. 6 - If an athlete expends 1950 kJ/h, how long does it...Ch. 6 - Prob. 6.19PCh. 6 - Hot packs used by skiers produce heat via the...Ch. 6 - Prob. 6.21PCh. 6 - Prob. 6.22PCh. 6 - For each process, state whether ΔH is less than...Ch. 6 - Prob. 6.24PCh. 6 - Prob. 6.25PCh. 6 - Prob. 6.26PCh. 6 - Prob. 6.27PCh. 6 - Prob. 6.28PCh. 6 - Prob. 6.29PCh. 6 - Prob. 6.30PCh. 6 - Prob. 6.31PCh. 6 - Prob. 6.32PCh. 6 - What data do you need to determine the specific...Ch. 6 - Is the specific heat capacity of a substance an...Ch. 6 - Prob. 6.35PCh. 6 - Both a coffee-cup calorimeter and a bomb...Ch. 6 - Find q when 22.0 g of water is heated from 25.0°C...Ch. 6 - Calculate q when 0.10 g of ice is cooled from...Ch. 6 - A 295-g aluminum engine part at an initial...Ch. 6 - Prob. 6.40PCh. 6 - Two iron bolts of equal mass—one at 100.°C, the...Ch. 6 - Prob. 6.42PCh. 6 - Prob. 6.43PCh. 6 - Prob. 6.44PCh. 6 - Prob. 6.45PCh. 6 - A 30.5-g sample of an alloy at 93.0°C is placed...Ch. 6 - When 25.0 mL of 0.500 M H2SO4 is added to 25.0 mL...Ch. 6 - Prob. 6.48PCh. 6 - Prob. 6.49PCh. 6 - A chemist places 1.750 g of ethanol, C2H6O, in a...Ch. 6 - High-purity benzoic acid (C6H5COOH; ΔH for...Ch. 6 - Two aircraft rivets, one iron and the other...Ch. 6 - A chemical engineer burned 1.520 g of a...Ch. 6 - Prob. 6.54PCh. 6 - Prob. 6.55PCh. 6 - Prob. 6.56PCh. 6 - Consider the following balanced thermochemical...Ch. 6 - Prob. 6.58PCh. 6 - Prob. 6.59PCh. 6 - When 1 mol of KBr(s) decomposes to its elements,...Ch. 6 - Prob. 6.61PCh. 6 - Compounds of boron and hydrogen are remarkable for...Ch. 6 - Prob. 6.63PCh. 6 - Prob. 6.64PCh. 6 - Prob. 6.65PCh. 6 - Prob. 6.66PCh. 6 - Prob. 6.67PCh. 6 - Prob. 6.68PCh. 6 - Prob. 6.69PCh. 6 - Prob. 6.70PCh. 6 - Prob. 6.71PCh. 6 - Write the balanced overall equation (equation 3)...Ch. 6 - Prob. 6.73PCh. 6 - Prob. 6.74PCh. 6 - Prob. 6.75PCh. 6 - Prob. 6.76PCh. 6 - Prob. 6.77PCh. 6 - Prob. 6.78PCh. 6 - Prob. 6.79PCh. 6 - Prob. 6.80PCh. 6 - Prob. 6.81PCh. 6 - Prob. 6.82PCh. 6 - Calculatefor each of the following: SiO2(s) +...Ch. 6 - Prob. 6.84PCh. 6 - Prob. 6.85PCh. 6 - The common lead-acid car battery produces a large...Ch. 6 - Prob. 6.87PCh. 6 - Prob. 6.88PCh. 6 - Prob. 6.89PCh. 6 - Prob. 6.90PCh. 6 - Prob. 6.91PCh. 6 - Prob. 6.92PCh. 6 - The following scenes represent a gaseous reaction...Ch. 6 - Prob. 6.94PCh. 6 - Prob. 6.95PCh. 6 - Prob. 6.96PCh. 6 - Prob. 6.97PCh. 6 - Prob. 6.98PCh. 6 - Prob. 6.99PCh. 6 - Prob. 6.100PCh. 6 - Prob. 6.101PCh. 6 - Prob. 6.102PCh. 6 - Prob. 6.103PCh. 6 - Prob. 6.104PCh. 6 - Prob. 6.105PCh. 6 - Prob. 6.106PCh. 6 - Liquid methanol (CH3OH) canbe used as an...Ch. 6 - Prob. 6.108P
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