hw01 - thermo 1-solutions

kumar (kk24268) – HW01 - Thermo 1 – mccord – (51580) 1

This print-out should have 10 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.

001 10.0 pointsWhich of the following is true of a generalthermodynamic state function?

1. The change of the value of a state func-tion is independent of the path of a process.correct

2. The change in the value of the state func-tion is always positive for endothermic pro-cesses.

3. The change in the value of a state func-tion is always negative for a spontaneous re-action.

4. The value of the state function remainsconstant.

5. The value of a state function does NOTchange with a change in temperature of aprocess.

Explanation:A change in a state function describes a

difference between the two states. It is inde-pendent of the process or pathway by whichthe change occurs.

002 10.0 pointsAn isolated system can only exchange energywith the surroundings.

1. True

2. False correct

Explanation:An isolated system cannot exchange energy

nor matter with the surroundings.

003 10.0 pointsAn endothermic reaction corresponds to onein which

1. change in energy is positive; heat isevolved.

2. change in energy is negative; heat isevolved.

3. change in energy is positive; heat is ab-sorbed. correct

4. change in energy is negative; heat is ab-sorbed.

Explanation:Endothermic reactions require energy to

flow from the surroundings into the system.In other words, heat is absorbed by the sys-tem. ∆H, or the change in energy, is positivefor endothermic reactions since the amount ofenergy in the system increased.

004 10.0 pointsConsider the reaction

2H2(g) + O2(g) → 2H2O(g)

at constant pressure. Which response is truefor the reaction?

1. Work may be done on or by the systemas the reaction occurs, depending upon thetemperature.

2. Work is done by the system as it occurs.

3. No work is done as the reaction occurs.

4. Work is done on the system as it occurs.correct

Explanation:ni = 3 mol gas nf = 2 mol gas

For P = const,

w = −P ∆V = −(∆n)RT ,

∆n = nf − ni = −1 .

Therefore w will be positive. When w ispositive, work is done on the system, andprogresses as the reaction progresses.

005 10.0 points


A chemical reaction takes place in a containerof cross-sectional area 100 cm2. As a result ofthe reaction, a piston is pushed out through 23cm against an external pressure of 573 torr.What is the value for w for this reaction?(Sign does matter.)

Correct answer: −176 J.

Explanation:

006 (part 1 of 2) 10.0 pointsA gas sample in a piston assembly expands,doing 540 kJ of work on its surroundings atthe same time that 384 kJ heat is added to thegas. What is the change in internal energy ofthe gas during this process?

Correct answer: −156 kJ.

Explanation:The change in internal energy ∆U is given

simply by summing the two energy terms in-volved in this process. We must be careful,however, that the signs on the energy changesare appropriate; in this case, internal energywill be added to the gas sample by heating(so q is positive), but the gas does work on itssurroundings as it expands (so w is negative):w = −540 kJ q = 384 kJ

∆U = q + w

= 384 kJ + (−540 kJ)

= −156 kJ .

007 (part 2 of 2) 10.0 pointsWill the pressure of the gas be higher or lowerwhen these changes are completed?

1. lower correct

2. higher

Explanation:If the heat added had exactly matched the

amount of energy lost due to the work of thegas (i.e., if q had been equal but oppositein sign to w), then ∆U would have been 0and the temperature of the gas would nothave changed. Because less heat was added,however, and the gas was allowed to expand

further, the temperature of the gas wouldhave had to decrease, and consequently thepressure of the gas would be lower at the end.

008 10.0 pointsA 100 W electric heater (1W = 1 J/s) oper-ates for 8 min to heat the gas in a cylinder. Atthe same time, the gas expands from 1 L to10 L against a constant atmospheric pressureof 4.162 atm. What is the change in internalenergy of the gas?

Correct answer: 44.2046 kJ.

Explanation:Pext = 4.162 atm Vini = 1 L1 L · atm = 101.325 J Vfinal = 10 LIf the heater operates as rated, then the to-

tal amount of heat transferred to the cylinderwill be

q = (100 J/s) (8 min) (60 s/min)

= 48000 J = 48 kJ

Work will be given by

w = −Pext∆V

in this case because it is an expansion againsta constant opposing pressure:

w = −(4.162 atm) (10 L− 1 L)

= −37.458 L · atm

Convert to kilojoules (kJ)

w = (−37.458 L · atm)(101.325J/L · atm)

= −3795.43 J = −3.79543 kJ

The internal energy change is

∆U = q + w

= 48 kJ + (−3.79543 kJ)

= 44.2046 kJ

009 10.0 pointsA piece of metal of mass 20 g at 109 ◦C isplaced in a calorimeter containing 41.7 g ofwater at 22◦C. The final temperature of themixture is 55◦C. What is the specific heat


capacity of the metal? Assume that there isno energy lost to the surroundings.

Correct answer: 5.33111J

g ·◦ C.

Explanation:mmetal = 20 g mH2O = 41.7 gTH2O = 22◦C Tfinal = 55◦C

Tmetal = 109◦C CH2O = 4.184J

g ·◦ C

qlost metal = −qgained water

mmCm∆Tm = −mH2OCH2O∆TH2O

Cmetal = −

mH2OCH2O (Tfinal − TH2O)

mmetal (Tfinal − Tmetal)

= −

(41.7 g)(

4.184 Jg·◦C

)

(20 g) (55◦C− 109◦C)

× (55◦C− 22◦C)

= 5.33111J

g ·◦ C.

010 10.0 pointsWhen 0.473 g of compound X is burned com-pletely in a bomb calorimeter containing 3000g of water, a temperature rise of 0.222◦C isobserved. What is ∆Urxn for the combustionof compound X? The hardware component ofthe calorimeter has a heat capacity of 3.38kJ/◦C. The specific heat of water is 4.184J/g ·◦C, and the MW of X is 56.0 g/mol.

Correct answer: −418.746 kJ/mol.

Explanation:mX = 0.473 g MWX = 56.0 g/molmwater = 3000 g ∆T = 0.222◦CCs = 4.184 J/g ·◦C HC = 3.38 kJ/◦C

qcal = qwater + qhardware

qcal = Cs ·mwater ·∆T + Chardware ·∆T

Then divide the heat by the number ofmoles, which is mass/MW. Also rememberthat because the temperature increased in thecalorimeter, the reaction must be exothermicand the value of q (and therefore ∆E) is neg-ative.

qrxn = −qcal

hw01 - thermo 1-solutions

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