COLLEGE OF SAINT MARY     FALL 2008    PHYSICAL CHEMISTRY    

INSTRUCTOR: DR. PETER ILICH        YOUR NAME: _____________________________

EXAM # 1 -- SOLUTIONS

Copyright 2008/2010 by Peter Ilich

 

Mr. Bond's last assignment (in Kazakhstan):

 1. (50 pts)    On a cold, t = 3 °F, late winter day in Kazakhstan, Cyril Bond finds that the pressure in the front left tire on his Aston-Martin-Ford-Aspire (special issue) reads only 17 pounds per square inch.  After consulting with Nurudin, a garage attendant, Cyril lets him add another (change the number) 1.00 kg/cm2 of air to the tire (the attendant is using a metric pressure gauge).

QUESTION: How much higher than the nominal 30 lb/inch2 (psi) would this second pressure be at 115 °F when Cyril hits the streets of Calcutta on his next assignment, six weeks later? List carefully all assumptions made. Show clearly the unit conversions.

SOLUTION:

STRATEGY:

First, ask yourself a question: What is this problem about?

The text of the problem contains the words "pressure", "car tire", and then a cold and a hot day ("temperature").  So it is about gas -- pressure, volume and temperature.   What is the "volume" here?  Let us start by -- a very important part of solving problems -- making assumptions:

ASSUMPTIONS:

In this problem the data are given about the pressure and the temperature of the gas.   The gas is air and since the pressures and temperatures are relatively mild we make first assumption: 

Second, while pressure and temperature are given no mention of the volume and amount of the gas is made.   In this problem the whole of the gas is contained within a tire.   Since no change in the size of the tire is indicated we make second assumption:

Third, we assume that the tire will not leak air (or at least will not leak at any appreciable rate); this allows us to make third assumption:

 

CALCULATION:

For the winter day we write:   p1V1 = n1 R T1    and for the summer day we write:   p2V2 = n2 R T2

The winter day (Kazakhstan):   

T1 = (3 - 32)[°F] /1.8 [°C/°F] + 273.15 = 257.04 [K]

p1 = 20 lb/inch^2 = (20 * 0.454 [kg lb-1])/ (2.54)2 [cm2] = 1.196 [kg cm-2]  the pressure upon first reading

Now, after 1 kg/cm2  pressured air is added, the pressure in the tire is:

p1 + 1.0 = 2.196 [kg cm-2]

NOTE:    Pressure, like temperature (or concentration) is an INTENSIVE property and may not -- and must not ! -- be added or subtracted.  This is GENERALLY true; however, in special circumstances, like here for example, when the volume of the gas "before" and "after" does not change, we may add pressures. 

Now -- the summer day in Calcutta, India:   

T2 = (115 - 32)/1.8 + 273.15 = 319.26 K

Assumption # 2:     No measurable volume change occurred between the two events.  

        V1 = V2

Assumption # 3:    No measurable air leaking occurred between the two events.

        n1 = n2

So we compare the two expressions and eliminate n, V, and R:

p1 / p2 = T1 / T2   and from here we get:

 p2  = p1 * (T2 / T1)

p2(summer) = 2.196`[kg cm-2] * 319.26 [K] / 257.04 [K] = 2.728 [kg cm-2]

Converted to lb/inch2 this pressure reads:

p2 = 2.728 [kg cm-2] * (1.000/0.454) [lb kg-1] * 2.54^2 [inch-2 cm+2 ] = 38.8 [lb/inch2],

That is, the pressure in the tire on a hot Calcutta day will be almost nine pounds per square inch more than the required pressure, 30 psi.

 

A Saturday morning communal event:

2.  (50 pts)    After finishing a 10 k race the runner's (m = 52.4 kg) body temperature is elevated from 36.6 °C to 37.1 °C.  At the race finish stands, she drinks 1.0 L or iced tea (t = 4 °C) which, she feels, cools her down. 

QUESTION:  How much, if any, has her body temperature decreased after imbibing the iced tea?  (Assume that the human body has the heat capacity of water.) 

BONUS question (10 pts) How much has her total body entropy changed due to the tea ingestion (excluding all other processes)?

SOLUTION:

(1)    UNDERSTANDING THE PROBLEM:

    Imagine that at first you do not understand this problem.  What can you do?  Look for "physical-chemistry"  words and expressions.  You will find "temperature", "cools down", "heat capacity".  These will point you to the part of the textbook that describes HEAT.   Check this chapter on heat for major statements and formulas.   You will find that heat, q, is proportional to amount (kg, L, number of moles) and is also proportional to DT, the difference between the "temperature before" and the "temperature after".    The physical chemical  -- not the communal! -- "event" described here is HEAT TRANSFER, that is a process which happens when a warmer and colder body come to contact.

STRATEGY:

(a)    The change in the heat content of the runner, index (1), given:

q(1) = m(1) * C * DT(1)                                         (1)

The change in the heat content of the tea, index (2), is, analogously:

q(2) = m(2) * C * DT(2)                                         (2)

ASSUMPTIONS:

We can (approximately) solve this problem if we assume TWO things:

FIRST ASSUMPTION:     The heat removed from the runner -- due to tea ingestion -- is the same as the heat passed form runner to tea; so we write:

m(1) * C * DT(1) = q(2) = m(2) * C * DT(2)          (3)

Given that heat capacities on both sides are approximately equal (water)  we can express one temperature difference with the help of the other:

DT(2) = [m(1)/m(2)] * DT(1)                                  (4)

This is an intractable equation as it contains two unknown variables, DT(1) and DT(2). 

SECOND ASSUMPTION     :    The maximum temperature difference in this process, 37.1 °C - 4 C = 33.1 °C, equals the sum of the runner's temperature change, DT(1) and the tea's temperature change, DT(2).  (As this is temperature DIFFERENCE we can express it either on Kelvin or Celsius scale.)

So we write:

DT(1) + DT(2) = 37.1 - 4 = 33.1 °C                      (5)

And this is our SECOND EQUATION that will allow us to solve this problem with two unknowns.

 

SOLVING THE PROBLEM:

(a)    The runner's temperature change, DT(1), is down, from 37.1 °C toward 4 °C.    The tea's temperature change, DT(2), is up, from 4 °C toward 37.1 °C.  Neither DT(1) nor DT(2) will be equal to the maximum temperature difference in this process, DTtot = 33.1 °C.  DT(1) will be a much smaller part of the overall temperature change since the runner's mass, m(1), is much larger than the tea's mass, m(2).  We take the expression for ΔT(2), equation (4) and insert it in the equation (5); now we solve this equation for DT(1):

[m(1)/m(2)] * DT(1) + DT(1) = 33.1 °C                (6)

We insert the values for the runner's mass (52.4 kg) and the tea mass  (1.0 kg) to get:

DT(1) = 33.1/ [1 + m(1)/m(2)] = 0.6198 °C ≈ 0.62 °C
 

 

(b)    The entropy change is calculated with the assumption that the heat exchange is reversible (an often used but not a realistic assumption -- the heat will practically NEVER flow from a cold to a hot body):

D S = n Cp ln Tf/Ti

D S = 52.4*103 [g] * 4.184 [J K-1 g-1] * ln (273.15 + 37.1 - 0.62) / (273.15 + 37.1)

D S = -438.57 J K-1
 

Note that the entropy change of the runner is negative, as expected for a system which is cooling down.

Office hour:

3.    (50 pts)    When you go to see and ask your physical chemistry teacher about a homework problem you climb 45 6-1/2" high stairs.  Your body heats up a little and in this process you loose through perspiration an equivalent of 17.3 g of water, between the body (36.6 °C) and the environment (25.0 °C) temperatures.  Taken these two processes only what do you think is your enthalpy change [J]?

SOLUTION:

UNDERSTANDING THE PROBLEM:

    What happens in this problem?  You climb up, that is, you carry your whole body from a lower level up to a higher level in a building; this is called "gravity" work.  By doing this you STORE energy in your system -- you are at a higher level now.   But, during this process you increase your metabolic rate (the details of which we ignore now), perspire and give away heat.  So -- the problems is about the ENERGY BUDGET of your body.  The "system" (your body) function describing the energy changes is ENTHALPY so in order to solve the problem we have to calculate the enthalpy changes cited in the problem.  As no pV work is involved the enthalpy change, ΔH, and the internal energy change, ΔU, will in this case be practically the same.

STRATEGY & SOLUTION:

    We will calculate the enthalpy change as the sum of the contributing energy exchange parts: (1) the energy gain due to climbing the stairs and (2) the energy loss due to loss fo heat through perspiration.

(1)     The energy gain -- climbing up the stairs:

w = mass (body) * g (gravity constant) * Δh (height difference)  

w = 90.0 (or - insert your body mass) [kg] * 9.8 [m s-2] * Δh [m]

We have to express the height (or height difference,  Δh) in meters, [m]:

Δh = 45 * 6.5 [inch] * 2.54 [cm * inch-1] = 7.4295*102 [cm] = 7.43 [m]

So the "gravity work" will be:

w = 90.0 [kg] * 9.8 [m s-2] * 7.43 [m] = 6.56*105 [J]

(2)    The energy loss -- perspiration:

q = 17.3 [g] * 4.184 [J K-1 g-1] * ΔT

What is the temperature difference here?  The difference between the human body temperature, T1 = 273.15 + 36.6 = 309.75 K, and the room temperature, T2 = 273.15 + 25.0 = 298.15 K.  So we have for the heat exchange:

q = 17.3 [g] * 4.184 [J K-1 g-1] * (298.15 - 309.75) [K] = - 8.396*102 [J]

NOTE the sign of the exchanged heat, q.

The enthalpy change is given as:

ΔH = w + q = 6.56*105 - 8.396*102 [J] = +6.55*105 [J].

COMMENT:    Note that this is a rather simplified picture of what is actually going on during such process: in order to climb up 45 stairs you would have to burn food and loose a lot more energy; overall, the energy balances would be negative.  (But, hopefully, consultation with your teacher would be help you finish the homework.)

 

Bloody affair:

4.    (50 pts)    Human blood plasma with hemoglobin (15 g HbA/100 mL blood) carries up to 20 mL (@ 760 mm Hg) of O2.  Given that molecular mass of human hemoglobin, HbA, is 68 kDa (1 Da(lton) = 1 amu = 1.661*10-27 kg) calculate the efficiency of HbA oxygen carrier (e.g. compare the number of moles of O2 per mole of hemoglobin).

SOLUTION:

(1)    UNDERSTANDING THE PROBLEM:

    In simplest terms, dioxygen, O2 molecule, bonds to hemoglobin, HbA molecule; if one dioxygen molecule would bind to every single HbA molecule present in red blood cells the efficiency would be 100 %.  In chemistry, we are using MOLES rather molecules but the question remains the same: If Y moles of O2 will "react" with X moles of HbA what is the efficiency of this binding?  So - find X and Y and divide X by Y another; if the ratio is 1 the binding efficiency is 1 or 100 %, else, it will be LESS.

STRATEGY:

    (a) From the mass of HbA and its molecular mass we find the number of moles of HbA in 100 mL of blood; this will be the number X -- a Chemistry-101 kind of problem.

    (b) From the Ideal Gas Law we find the number or moles of O2 in 20 mL of gas, at human body temperature, this will be our number Y.

CALCULATION:

(a)    For the number of moles of hemoglobin in 100 mL blood we have:

n(HbA) = 100.0 [g] / 6.8*104 [g mol-1] = 2.21*10-4 [mol]

(b)    From the Ideal Gas Law we find for the number of moles of dioxygen:

n = pV / RT

Assuming the normal pressure (= 101325 Pa) and the normal human body temperature (= 36.6 + 273.2 = 309.6 K) and the volume expressed in m3 we find n:

n(O2) = 101325 [N m-2] * 2.0*10-5 [m3] / 8.314 [J K-1 mol-1] * 309.6 [K] = 7.9*10-4 [mol]

n(O2) / n(HbA) = 3.6

NOTE:  IN order to calculate the true efficiency one must take into account that every HbA molecule sis a TETRAMER, that is, it contains FOUR binding sites, so the overall efficiency of binding O2 per hemoglobin molecule is 3.56//4 = 0.89 or 89 %.  Pretty high, wouldn't you say?

 

More bonuses:

5.    Answer the following questions:

(5 pts)    When water cools down it freezes; when it's warmed up it melts; both processes occur at 0 °C and both involve heat transfer.  Assuming this is the only component of the enthalpy change of a water sample what signs have the enthalpy changes during these two processes?

ANSWER:    Water is the SYSTEM.  When it freezes if gives away heat, therefore ΔH = -q.  It is the opposite with melting -- we heat water, we add energy to it and ΔH = +q.

(5 pts)    You drive over a rough surface and a tire on your bike deflates.  What sign is the enthalpy change of the air in the tire?   What sign is the entropy change?

ANSWER:    The air is flowing out and mixing with the surrounding air so its entropy increase, ΔS > 0.

(5 pts)    You add 100 mL of 1.0 % NaCl aqueous solution to 200 mL of 2.0 % salty solution.  Is this a change in a cumulative (extrinsic) or specific (intensive, intrinsic) property (or both)?  Specify clearly.

ANSWER:    Adding two volumes, 100 mL and 200 mL to give (approximately) 300 mL is a cumulative or EXTENSIVE  change.  The final concentration, however, will not be a simple sum of the two concentrations, so this will be an INTENSIVE change.  (NOTE: Intensive, not intense.)

(5 pts)    In the so-called 64-bit architecture computers the number of memory bytes is (263-1) /8.  How many gigabytes is this?

N = # bytes = (263-1) /8 ≈ (9.2233720*1018 - 1) / 8 = 1.115292150*1018 = 1.153*109 Gb.

This is a lot of "addressable space", about hundred million times larger than the largest computer data storage device we are using today.  For this reason the higher-end computers today are so-called 64-bit architecture devices as they will not any time soon run out of the theoretical (virtual) number of storage bins.   Most of our desktops and laptops are 32-bit devices which means that the highest number of bytes ("addresses") they can access is (232-1) /8 = 536 870 912, a large number indeed but already "small" in many applications, e.g. in chemical, physical, meteorological and materials science research.

NOTE: Solve any three problems.

 

COLLEGE OF SAINT MARY     FALL 2008    PHYSICAL CHEMISTRY EXAM # 1

YOUR NAME: _____________________________