HW5_key - HW5_Chapter 20: 20-5: Q. In this problem, we will...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: HW5_Chapter 20: 20-5: Q. In this problem, we will prove that Equation 20.5 is valid for an arbitrary system. To do this, consider an isolated system made up of two equilibrium subsystems, A and B, which are in thermal contact with each other; in other words, they can exchange energy as heat between themselves. Let subsystem A be an ideal gas and let subsystem B be arbitrary. Suppose now that an infinitesimal reversible process occurs in A accompanied by an exchange of energy as heat (ideal). Simutaneously, another infinitesimal reversible process takes place in B accompanied by an exchange of energy as heat (arbitrary). Because the composite system is isolated, the First Law requires that Now use Equation 20.4 to prove that A. W e use the First Law as suggested in the problem and substitute Equation 20.1 for δqrev (ideal) to write Then is the derivative of a state function. We know that the cyclic integral of a state function is equal to 0. Therefore, we can write and Equation 20.4 holds for any system. 20-6: Q. Calculate and for a reversible cooling of one mole of an ideal gas at a constant volume V1 from P1, V1, T1 to P2, V1, T4 followed by a reversible expansion at constant pressure P2 from P2, V1, T4 to P2, V2, T1 (the final state for all the processes showed in Figure 20.3). Compare your result for with those for paths A, B+B, and D+E in Figure 20.3. A. Step 1. P1, V1, T1 P2, V1, T4 Because there is no change in the volume of the ideal gas, , and we can write Step 2. P2, V1, T4 P2, V2, T1 In this case, we write (by the First Law) For the entire process, P1, V1, T1 P2, V2, T1, we have The value of differs from those found for paths A, B+C, and D+E (section 20-3), but the value of is the same (because entropy is a path-independent function). 20-9: Q. Calculate the value of if one mole of an ideal gas is expanded reversibly and isothermally from 1.00 bar to 0.100 bar. Explain the sign of . A. Because the reaction is isothermal, . For an ideal gas, So we write The value of as is positive because the gas expands. 20-14: Q. Show that If one mole of an ideal gas is taken from T1, V1 to T2, V2, assuming that is independent of temperature. Calculate the value of if one mole of N2(g) is expanded from 20.0 dm3 at 273 K to 300 dm3 at 400 K. Take = 29.4 J∙K-1∙mol-1. A. For the path (T1, V1) (T2, V2), write as Because For N2, and for an ideal gas, we can write this as . W e can then 20-16: Q. In this problem, we will illustrate the condition with a concrete example. Consider the two-component system in Figure 20.8. Each compartment is in equilibrium with a heat reservoir at different temperatures T1 and T2, and the two compartments are separated by a rigid heat-conducting wall. The total change of energy as heat of compartment 1 is where is the energy as heat exchanged with the reservoir and exchanged with compartment 2. Similarly, is the energy as heat Clearly, Show that the entropy change for the two-compartment system is given by where is the entropy exchanged with the reservoirs (surroundings) and is the entropy produced within the two-compartment system. Now show that the condition implies that energy as heat flow spontaneously from a higher temperature to a lower temperature. The value of , however, has no restriction and can be positive, negative or zero. Figure 20.8. A two-compartment system with each compartment in contact with an (essentially infinite) heat reservoir, one at temperature T1 and the other at temperature T2. The two compartments are separated by a rigid heat-conducting wall. A. As stated in the text of the problem, we can write The energy as heat exchanged between compartments 1 and 2 is involved in the entropy transferred between the two compartments. We can therefore express Similarly, the energy as heat exchanged between the compartments and reservoirs is involved in the entropy produced within the two-compartment system, so we can write as These are the only two means of changing the entropy of the system, so we can find be Now take the condition . This is the same as saying that to Arbitrarily, let , then , so and heat is flowing from compartment 1 to compartment 2. If , by the same reasoning compartment 2 to compartment 1. and heat is flowing from 20-18: Q. Vaporization at the normal boiling point ( ) of a substance (the boiling point at one atm) can be regarded as a reversible process because if the temperature is decreased infinitesimal below , all the vapor will condensed to liquid, whereas if it is increased infinitesimal above , all the liquid will vaporize. Calculate the entropy change when two moles of water vaporize at 100.0 . The value of is 40.65 kJ∙mol-1. Comment on the sign of . A. At constant pressure and temperature, isothermal process, (Equation 19.37). And for a reversible As the water becomes more disordered, changing from liquid to vapor, the entropy increases. 20-25: Q. Calculate the change in entropy of the system and of the surroundings and the total change in entropy if one mole of an ideal gas is expanded isothermally and reversibly from a pressure of 10.0 bar to 2.00 bar at 300 K. A. Because this is an isothermal reversible expansion, gas equation to write For an isothermal expansion of an ideal gas, entropy of the gas as For a reversible expansion, , so . W e then use the ideal . We can then write the change of . ...
View Full Document

This note was uploaded on 11/08/2011 for the course CH 353 taught by Professor Bersuker during the Fall '08 term at University of Texas.

Ask a homework question - tutors are online