### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# Physical Chemistry II CH 433

NCS

GPA 3.54

### View Full Document

## 19

## 0

## Popular in Course

## Popular in Chemistry

This 6 page Class Notes was uploaded by Sienna Shields on Thursday October 15, 2015. The Class Notes belongs to CH 433 at North Carolina State University taught by Stefan Franzen in Fall. Since its upload, it has received 19 views. For similar materials see /class/223997/ch-433-north-carolina-state-university in Chemistry at North Carolina State University.

## Reviews for Physical Chemistry II

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 10/15/15

Chemistry 433 Lecture 9 Entropy and the Second Law NC State University 9232008 Spontaneity of Chemical Reactions One might be tempted based on the results ofthermo chemistry to predict that all exothermic reactions would be spontaneous The corollary this would be the statement endothermic reactions that are s ontaneous Of course heat must be taken up 39om the surroundings in order for such processes to occur Nonetheless the enthalpy ofthe reaction does not determine whether or not the reaction will occur only how much heat will be required or generated by the reaction The observation that gases expand to ll a vacuum and that different substances spontaneously mix when introduced into the same vessel are thher examples that require quantitative explanation Spontaneity of Chemical Reactions As you might guess by now we are going to de ne a new state Jnction that will explain all of these observations and de ne the direction ofspontaneous processes This state Jnction is the entropy Entropy is related to heat and heat ow and yet heat is not a state function Recall that q is a path Jnction It turns out that the state Jnction needed to describe spontaneous change is the heat divided by the temperature Here we simply state this result We will prove that entropy is a state Jnction in this lecture Engines Historically people were interested in understanding the ef ciency with which heat is converted into work This was a very important question at the dawn ofthe industrial revolution since it was easy to conceive ofan engine powered by steam but it turned out to be quite dif cult to build one that was ef cient enough to get anything done In an engine there is a cycle in which fuel is burned to heat 0 ora steam engine there Is a net release 0 eat Therefore we all understand that Sq 2 0 forthe cycle A cyclic heat engine common I I Ill Amman The work is WWi WiiWmWiv H q q qm 39 Wl 39 Wm Forthe adiabatic steps qll qlV Forthe isothermal steps AU 0 III Work and Heat for the Cycle Neither the work nor the heat is a state function Neither one is zero forthe cycle as should be the case for a state function The work is VquotiWiiWiiiWiv n RTmInNQIvigt CVltTmrTmo nRTmInwAAa cv w n RTmlnV2N1 nRTm danANa since w nRTmlnV2N1 nRTm dlnV1N2 since v4 V1N2 w n RTmlnV2N1 n RTmmlnV2N1 property of logarithms The heat is q q qm since qH qW 0 for adiabatic processes M wm since dU 0 for isothermal steps q nRTmInltv2N1gt nRTmInmAa q nRThmlnV2N1 nRTm dlnV1N2 since VANa V1N2 q nRThmlnV2N1 nRTm danzVO property of logarithms Tm mm le 5 A new state function Entropy The heat is not a state Jnction The sum q qm is notzero From this point on we will make the following de nitions Cli qhm Clm qculd V V q qm 70 nRTmln nRTm n I 0 1 1 am qm am V2 V2 T nRn VJ an 0 Tm However the heat divided by temperature is a state function This reasoning leads to the idea of a state function called the entropy We can write qrev AS T 9232008 Thermodynamics of an Engine The cycle just described could be the cycle for a pl in a steam engine or in an internal combustion engI The hot gas that expands following combustion ofa small quantity offossil fuel drives the cycle If you think about the fact that the piston is connected to the cranksha you will realize that the external ressure on the iston is ton he that the expansion is perfectly reversible we can apply the above reasoning to your car The formalism above for the entropy can be used to tell us the thermodynamic ef ciency ofthe engine Thermodynamic Efficiency We de ne the ef ciency as the work extracted divided by the total heat input n IWtotaI IHRTcold Tnoz n V2 VOI Tmz Tm qwr quotRTmln V2 V1 Tm The ef ciency de ned here is the ideal best case It assumes a reversible process with no losses due to friction The exhaust TEE d cannot be less than the temperature of the surroundings Question Your car has an operating temperature of 400 K lfthe ambient temperature is 300 K what is the thermodynamic ef ciency of the the engine A 75 Question Your car has an operating temperature of 400 K lfthe ambient temperature is 300 K what is the thermodynamic ef ciency ofthe the engine A 75 B 50 c 25 D 5 IWzozaI Tmz T d Tm 300K T39 qmz m 1 Tm 400K 03925 Question The thermodynamic cycle was derived for reversible expansions What are the consequences if the cycle is not perfectly reversible A The work of expansion will decrease B The work of compression will decrease C There will be no adiabatic expansion D There can be no cycle 9232008 Question The thermodynamic cycle was derived for reversible expansions What are the consequences ifthe cycle is not perfectly reversible A The work of expansion will decrease B The work of compression will decrease C There will be no adiabatic expansion D There can be no cycle The Thermodynamic Temperature Scale The de nition of entropy is qml39l39hut qmmTmm 0 We can write this as tamTm qmmrrmm Since qmd is negative we can combine the minus sign with qmd and write the expression as anmlnnut Iqmurrmm and nally IqmlIqmml TimTm The ratio ofthe heats is equal to the ratio of temperatures for two steps in a thermodynamic cycle This de nes a temperature scale and allows one to measure temperature as well is this scheme represents a thermometer Both this expression and the thermodynamic ef ciency further imply that there is an absolute zero oftemperature Heat Transfer To examine the function that we have just de ned let us imagine that we place to identical metal bricks in contact with one another If one of the bricks is at equilibrium at 300 K and the other at 500 K what will the new equilibrium temperature be lntuitively you would say 400 K and you would imagine that heat ows s ontaneousl 39om the warmer brick to the colder brick The entropy function makes these ideas quantitative q q 2 l ql 4 AS 72 f T1 500 K Using this de nition M by the temperature Heat Transfer Let s assume that heat ows 39om the hot body to the cold body Then q is negative the ow from the hot body and q2 is positive the ow into the cold body Moreover q2 39Cll Cl This means that we can substitute in q to obtain ll ll 4 AS T2 E qT2 E 300 500 00013 q For this calculation it does not matter how big q is but only that it is a postive number so that AS is positive T1 500K Using this de nition A 0 which says that the process is spontaneous System and surroundings Up to this point we have considered the system but we have not concerned ourselves with the relationship between the system and the surroundings When we consider heat ow In the state function entropy we need to carefully account for how the heat ow takes place in order to determine the effect We will show that for reversible processes the entropy change is zero However for irreversible processes the entropy change can be positive spontaneous or negative reverse process is spontaneous For isolated systems the ang e Is also zero However when there is heat be nonzero In the case that the heat ow is irreversible the entropy change will be nonzero Calculating reversible and irreversible paths It is important to reiterate that the calculation ofthe entropy ofthe system always follows a reversible pa You might ask well what happens ifthe process is not reversib e To consider this let us the example of expansion ofgas in a cylinder The process can occur along different paths a constant ressure expans39on w PEXAV PBXIf V b reversible isothermal expansion the work w nRTlnVW For both a irreversible and b reversible we will calculate the entropy of the system along a reversible path A W w v i For the reversible path we can use the fact that ASSW ASSVS to obtain ASSW annVW 9232008 System and surroundings The heat transfer example shows us that we always must consider the surroundings Any time the system releases heat it goes into the surroundings and this contributes to the overall entropy change Thus the total entropy is Astutal Assurr A sys For a reversible process the total entropy is zero lfa The rule is always calculate the entropy ofthe system along a reversible path lfthe process is truly reversible then tutal 0 and Assurr 39 Assys Understanding the irreversible path For the irreversible path we use the actual work of the constant pressure expansion w PEXAV PEXKVf V to calculate ASSW qlT wT PMV VT where T is the same temperature we used for the isothermal expansion Note that the sign is opposite since the heat is owing into the surround ings and out of the system We see that in this case the entropy change ASSW is smaller in magnitude than ASSVS know this from the rst law where we saw that the irreversible work of expansion is always less than the reversible work of expansion Thus Astmal gt 0 for the irreversible process Statements of the second law Asa qmrr 0 or A8 M gt 0 The entropy of an irreversible process is greater than zero lfthe process is spontaneous Clausius the entropy of the universe tends towards a maximum Famous words Die Entropie des Universums strebt ein Maximum zuquot The dependence ofthe entropy on volume For a constant temperature isothermalexpansion we have Sq wE The logic behind this statement is that the internal energy change is zero for a constant temperature process and so SqE aww To calculate the reversible work we simply plug in 6wEV V According to the ideal gas law P nRTN so d8 anVN 2 v L d8 17 V V The result ofthis equation is that AS annV2V1 at constant tempera ure The dependence ofthe entropy on temperature The entropy change as a function ofthe temperature is derived at constant volume using the fact that dU Sq nC dT The reversible heat in this case qw is a constant volume heat and so it can be replaced by d8 aqmT nCVdTT at constant volume To obtain AS we need to integrate both sides L zdS CJd TT We obtain s s2 s1 ncvlnmm Exactly the same reasoning applies at constant pressure so that AS nCplnT2T History of Refrigeration In prehistoric times game was stored in a cave or packed n now to presene the game for times when food was not available Later ice was hanested in the winter to be used in the summer Ice is still used for cooling and storing food The intermediate stage in the history of cooling foods was to add chemicals like sodium nitrate or potassium nitrate to water causing the temperature to fall Cooling wine via this method was recorded in 1550 as were the words quotto refrigeratequot The evolution to mechanical refrigeration a compressor with refrigerant was a long slow procss and was introduced in the last quarter ofthe 19th century Source US Department ongriculture Importance of Refrigeration quotDanger ZoneII some doubling in number in as little as 20 minuts A refrigerator set at 40 F or below will protect most foods 9232008 Cyclic Refrigeration This consiss ofa refrigeration cycle where heat is removed from a lowtemperature space or source and rejected to a hightemperature sink with the help of external work and is inverse the thermodynamic power cycle In the power cycle heat Is supplied from a high temperature source to the engine part of the heat being used to produce work and the rst being rejected to a lowtemperature sink This satis es the second law of thermodynamics Coefficient of Performance The coefficient of performance or COP sometimes CP ofa heat pump is the ratio of the output heat to the supplied rk or COP qm leal where is the useful heat su lied b the condenser and w is the work consumed by the compressor Note COP has no units therefore heat and work must be expressed in the e units qcod W and W qmz qmdv and qm is the heat given off by the hot heat reservoir Coef cient of Performance Therefore by substituting for w qm COPMEW ha qcoa Therefore Tm 003mm 39 m temperature Thut is the temperature ofthe expansion in the engine The temperature Tmd is the temperature of the exhaust Tmd cannot be less than the temperature ofthe surroundings For refrigeration the COP is qLDd Taod 00mg sz qud Thor 700m COP Cycic Refrigeration The refrigeration cycle uses a uid called a refrigerant to move heat from one place to another The key to understanding how it works is recognizing that at the same prssure the refrigerant boils at a much lower temperature than water For exam le the refrierant commonl used in home refrigerators boils between 40 and 50 F as compared to water39s boiling point of 212 F Let39s look at the process to see how boiling and condensing a refrigerant can move hat The process is the same whether it is operating a refrigerator an air conditioner or a hat pump This example illustrates a closedloop system Cycic Refrigeration We39ll begin with the cool liquid refrigerant entering the indoor coil operating as the evaporator during cooling As is name implis refrigerant in the evaporator quotevaporatesquot Upon entering the evaporator the liquid refrigerant39s temperature is between 40 and 50 F and without changing is temperature it absorbs heat as it changes state from a liquid tn a vapor The hat comes from the warm moist room air blown across the evaporator coil As it passes over the cool coil it gives up some of is heat and moisture may condense from it The cooler drier room air is recirculated by a blower into the space to be cooled 9232008 Cyclic Refrigeration The vapor refrigerant now moves into the compressor which is basically a pump that raises the pressure so it will move through the system Once it passes through the compressor the refrigerant is said to be on the quothighquot side of the system Like anything that is put under pressure the increased pressure from the compressor causes the temperature of the refrigerant to rise As it leaves the compressor the refrigerant is a hot vapor roughly 120 to 140 F It now ows into the refrigeranttowater heat exchanger operating as the condenser during cooling Again as the name suggests the refrigerant condenses here As it condenses it gives up heat to the loop which is circulated by a pump Cyclic Refrigeration The loop water is able to pick up heat from the coils because it is still cooler than the 120 degree coils As the refrigerant leaves the condenser it is cooler but still under pressure provided by the compressor It then reaches the expansion valve The expansion valve allows the high pressure refrigerant to quotflashquot through becoming a lower pressure cooled liquid When pressure is reduced as with spraying an aerosol can or a re extinguisher it cools The cycle is complete as the cool liquid refrigerant reenters the evaporator to pick up room heat In winter the reversing valve switches the indoor coil to operate as the condenser and the heat exchanger as the evaporator Cyclic Refrigeration In summary the indoor coil and refrigeranttowater heat exchanger is where the refrigerant changes phase absorbing or releasing heat through boiling and condensing The compressor and expansion valve facilitate the pressure changes increased by the compressor and reduced by the expansion valve Vaporcompression cycle A simple stylized diagram of the refrigeration cycle 1 condensing coil 2 expansion valve 3 evaporator coil 4 compressor Cyclic Refrigeration A re 39igeralian cycle describes the changes that take place in the refrigerant as it alternately absorbs and rejects heat as it circulates through a refrigerator Heat naturally ows from hot to cold Work monsmmasucg Is applied to cool a space by won commasqu mammarqu pumping heat from a lower temperature heat source into a higher temperature heat sink Insulation is used to reduce reduce the work and energy required to achieve and maintain a lower temperature in the cooled space

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

#### "When you're taking detailed notes and trying to help everyone else out in the class, it really helps you learn and understand the material...plus I made $280 on my first study guide!"

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.