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An Introduction to Thermal Physics
Found in: Page 156
An Introduction to Thermal Physics

An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279

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Short Answer

A muscle can be thought of as a fuel cell, producing work from the metabolism of glucose:

C6H12O6+6O26CO2+6H2O

(a) Use the data at the back of this book to determine the values of ΔHand ΔG for this reaction, for one mole of glucose. Assume that the reaction takes place at room temperature and atmospheric pressure.

(b) What is maximum amount of work that a muscle can perform , for each mole of glucose consumed, assuming ideal operation?

(c) Still assuming ideal operation, how much heat is absorbed or expelled by the chemicals during the metabolism of a mole of glucose?

(d) Use the concept of entropy to explain why the heat flows in the direction it does?

(e) How would your answers to parts (a) and (b) change, if the operation of the muscle is not ideal?

(a) The value of ΔH is -2803.04 kJ and the value of ΔG is -2878.94 kJ.

(b) The maximum amount of workdone is 2878.94 KJ.

(c) The amount of heat absorbed is 75.9 KJ.

(d) Here, the heat is positive and entropy is positive. So, the heat flows into the system.

(e) Here less amount of energy will leave the system, but Gibbs free energy and enthalpy will be same.

See the step by step solution

Step by Step Solution

Step 1: Explanation

The chemical reaction for the fuel cell of the muscle is

C6H12O6+6O26CO2+6H2O

Formula used:

Write the expression for the enthalpy change for the reaction.

ΔH=6HCo2+6HH2O-HC6H12O6-6HO2 (1)

Here, HCO2 is the enthalpy for CO2,HH2O is the enthalpy of H2O,HC4H2O4 is the enthalpy for C6H12O6 and Ho2 is the enthalpy of O2.

Write the expression for the Gibbs energy change for the reaction.

ΔG=6GCO2+6GH2O-GCbH12O4-6GO2(2)

Step 2: Calculation

Calculation:

Refer table at the back of the book.

Substitute-285.83 kJfor HCO4,-393.51 kJ for HH2O,0 for HCnH2O6 and -1273 kJ for Ho2in expression (1).

ΔH=-6(285.83 kJ+393.51 kJ)-6(0)+1273 kJ=-2803.04 kJ

Substitute-237.13 kJ for GCO2,-394.36 kJ for GH2O,0 for GC6H2O3 and -910 kJ for GO2in expression (2).

ΔG=-6(237.13 kJ+394.36 kJ)-6(0)+910 kJ=-2878.94 kJ

Thus, the value of ΔH is -2803.04 kJ and the value of ΔG is -2878.94 kJ.

Step 3. (b) GIven information

The chemical reaction for the fuel cell of the muscle is

C6H12O6+6O2→6CO2+6H2O.

Workdone for ideal operation,

W=GwhereW=workdoneG=Gibbs energy

Step 4. Calculation

As, G=2878.94 KJ.

So, W=2878.94 KJ

Step 5. Conclusion

The maximum amount of workdone is 2878.94 KJ.

Step 5. Given information

The chemical reaction for the fuel cell of the muscle is

C6H12O6+6O2→6CO2+6H2O.

The enthalpy is less than the amount of the work extracted.

The expression for the heat absorbed.

Q=W-∆H

where

Q=heat absorbed.

Step 7. Calculation

Here

W=2878.94 KJH=2803.04 KJ

So, Q=2878.94 KJ-2803.04 KJ=75.9 KJ

Step 8. Conclusion

The amount of heat absorbed is 75.9 KJ.

Step 9. Given information

The chemical reaction for the furl cell of the muscle is

C6H12O6+6O2→6CO2+6H2O.

The expression for the entropy change for the reaction.

S=6SCO2+6SH2O-SC6H12O6-6SO2·······(3)whereSCO2=Entropy for CO2SH2O=Entropy of H2OSC6H12O6=Entropy of C6H12O6SO2=Entropy of O2

Entropy in terms of heat absorbed

S=QT········(4)whereQ=heat absorbedT=absolute temperature

Step 10. Calculation

From the table

SCO2=69.91 J·K-1SH2O=213.74 J·K-1SC6H12O6=205.14 J·K-1SO2=212 J·K-1

So,

S=669.91+213.74=6205.14-212 J·K-1=259.06 J·K-1

Hence,

Q=259.06 J·K-1298 K=77.2 KJ

Step 11. Conclusion

Thus, the heat is positive and entropy is also positive. So, the heat flows into the system.

Step 12. Given information

The chemical reaction for the fuel cell of the muscle is

C6H12O6+6O2→6CO2+6H2O.

In the ideal reaction, the work equals the change in the Gibbs energy. But when the reaction is not ideal the amount of work is less than the Gibbs energy. Here, as the change of entropy is same, therefore less heat flow enters the system and less energy leaves the system. But, the Gibbs free energy and enthalpy are same whether or not the operation is ideal or not.

Step 13. Conclusion

Thus, less amount of energy will leave the system, but Gibbs free energy and enthalpy will be same.

Most popular questions for Physics Textbooks

Problem 5.58. In this problem you will model the mixing energy of a mixture in a relatively simple way, in order to relate the existence of a solubility gap to molecular behaviour. Consider a mixture of A and B molecules that is ideal in every way but one: The potential energy due to the interaction of neighbouring molecules depends upon whether the molecules are like or unlike. Let n be the average number of nearest neighbours of any given molecule (perhaps 6 or 8 or 10). Let n be the average potential energy associated with the interaction between neighbouring molecules that are the same (4-A or B-B), and let uAB be the potential energy associated with the interaction of a neighbouring unlike pair (4-B). There are no interactions beyond the range of the nearest neighbours; the values of μo and μAB are independent of the amounts of A and B; and the entropy of mixing is the same as for an ideal solution.

(a) Show that when the system is unmixed, the total potential energy due to neighbor-neighbor interactions is 12Nnu0. (Hint: Be sure to count each neighbouring pair only once.)

(b) Find a formula for the total potential energy when the system is mixed, in terms of x, the fraction of B.

(c) Subtract the results of parts (a) and (b) to obtain the change in energy upon mixing. Simplify the result as much as possible; you should obtain an expression proportional to x(1-x). Sketch this function vs. x, for both possible signs of uAB-u0.

(d) Show that the slope of the mixing energy function is finite at both end- points, unlike the slope of the mixing entropy function.

(e) For the case uAB>u0, plot a graph of the Gibbs free energy of this system

vs. x at several temperatures. Discuss the implications.

(f) Find an expression for the maximum temperature at which this system has

a solubility gap.

(g) Make a very rough estimate of uAB-u0for a liquid mixture that has a

solubility gap below 100°C.

(h) Use a computer to plot the phase diagram (T vs. x) for this system.

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