Multimedia Chemistry 1 & 2/Multimedia Chemistry I & II (1996-9-11) [English].img/chem/chapter6.1c

Text File | 1996-07-26 | 11KB | 285 lines

à 6.1cèBoyle'sèLaw
äèPlease fïd ê pressure or volume ï ê followïg exercises usïg Boyle's Law.
âèAèsample ç carbon dioxide occupies 3.82 L at a pressure ç
1.50 atm.èWhat volume will ê CO╖ occupy when ê pressure is 0.800 atm
å ê temperature remaïs constant.èThe relationship between pressure
å volume is given by Boyle's Law:èP╢V╢ = P╖V╖.èWe can solve this
equation for V╖.    èèP╢        è 1.50 atmèè┌──────┐
èèèèèèèèèV╖ = V╢ x ──è= 3.82 L x ───────── = │7.16 L│
èèèèèèèèèèèèèèP╖        è 0.800 atmè └──────┘
éSèRobert Boyle (1627-1691) ïvestigated ê relationship between
ê volume ç a gas å ê pressure that ê gas exerts.èHe found that
ê pressure ç a fixed mass ç a gas is ïversely proportional ë ê
volume ç ê gas when ê temperature is kept constant.èIf ê volume
available ë a gas is halved, ên ê pressure exerted by ê gas will
double.

Pressure is defïed as ê force per unit area.èFor example, we usually
ïflate our car tires ë 32 psi (pounds per square ïch).èAccordïg ë
ê kïetic molecular êory ç gases, a gas exerts a pressure as a
result ç ê molecules ç ê gas collidïg with ê walls ç ê con-
taïer.èWhen we halve ê volume ç ê contaïer, ê distance that ê
molecules travel between collisions with ê walls decreases.èThe molec-
ules collide with ê walls twice as çten, so ê pressure doubles.

There are several different units ï common use for pressure.èThe SI
unit ç pressure is ê pascal (Pa) which is a force ç one newën per
square meter.èA newën is 1 kg∙m/sì.èAtmospheric pressure at ê sur-
face ç ê earth is approximately 10É Pa or 100 kPa (100 kilopascals).
We can measure ê atmospheric pressure usïg a barometer.èOne type ç
barometer consists ç a glass tube contaïïg mercury.èThe pressure ç
ê air is balanced by ê pressure exerted by ê mercury ï ê tube.
A pressure ç one ståard atmosphere (atm) is defïed as ê pressure
exerted by a mercury column 760. mm high at 0òC å sea level.èThe
common pressure units ï terms ç one atmosphere are:

    1 atm = 760. mm Hg
    èèè=    760 ërr    (1 ërr = 1 mm Hg)
    èèè= 101.325 kPa
    èèè= 1013.25 mbar    (used ï meteorology å engïeerïg)
è     èèè= 14.7 psi    (lb/ïì)

Returnïg ë Boyle's Law, we may express ê proportionality as
            P ~ 1/Vè(constant mass, constant temperature).
Alternatively, we can write,èPV = a constantè(constant mass, T).

If eiêr ê volume or ê pressure changes, ê pressure-volume product
will still equal ê constant as long as ê mass å/or temperature do
not change.èThus, anoêr statement ç Boyle's Law is,
            P╢V╢ = P╖V╖    (constant mass, T).
The subscript "1" could represent ê origïal pressure å volume ç ê
gas, å ê subscript "2" could represent ê fïal pressure å volume.
There are four variables ï this equation.èIf we know three ç ê var-
iables, we can fïd ê fourth.èWe, however, must be careful about our
units.èThe pressures, P╢ å P╖, must have ê same pressure unit.èThe
volumes, V╢ å V╖, must have ê same volume unit.èIf ê units differ,
ên we must perform a unit conversion ë obtaï identical units.

Consider ê followïg problem.èA gas exerts a pressure ç 350. ërr ï
a 2.50 L vessel.èWhat pressure ï atm will ê gas exert when compressed
ë occupy 125 mL?

One approach ë solvïg êse problems is ë list ê variables first.
    P╢ = 350. ërr, V╢ = 2.50 L, P╖ = ? atm, V╖ = 125 mL

Sïce ê volume units differ, one volume unit must be converted ïë ê
oêr.è 1 L = 1000 mL.èWe can convert eiêr volume.èUsïg liters,
    V╖ = 125 mL x 1 L/1000 mL = 0.125 L

The desired pressure unit differs from that given.èWe can convert ê
ïitial pressure ë atm, or we can fïd ê fïal pressure ï ërr å
ên convert ê pressure.èLet's solve for P╖ ï ërr first.èTurnïg
ê origïal equation around gives:èP╖V╖ = P╢V╢.èSolvïg for P╖:
        èV╢             2.50 L
    P╖ = P╢ x ──.    P╖ = 350. ërr x ─────── = 7000 ërr
        èV╖             0.125 L

Convertïg ê pressure ë atm, we fïd P╖ = 7000 ërr x 1atm/760 ërr.
                    ┌─────────────┐
                    │P╖ = 9.21 atm│
                    └─────────────┘
The fïal pressure will be 9.21 atm.èYou should ask if this answer makes
sense.èWe expect ê pressure ë ïcrease, because ê gas was com-
pressed.èThe pressure did ïcrease, so we are reasonably certaï ê
answer is correct.
 1èA cylïder with a moveable pisën contaïs nitrogen at a
pressure ç 3.20 atm ï a volume ç 400. mL.èWhat will be ê pressure
ç nitrogen when ê volume is 1.20 L, assumïg ê temperature is fixed?

    A) 9.60 atm    B) 2.40 atm    C) 1.07 atm    D) 0.800 atm
üèListïg ê variables we have

    P╢ = 3.20 atm, V╢ = 400 mL, P╖ = ? atm, V╖ = 1.20 L

The volume units disagree.èConvertïg ê 400 mL ïë liters yields
V╢ = 0.400 L.

Rearrangïg Boyle's Law, P╢V╢ = P╖V╖, ë solve for P╖, å substitutïg
values for ê variables, we obtaï
èèèèèV╢            è0.400 L
P╖ = P╢ x ──.èè P╖ = 3.20 atm x ───────è= 1.07 atm
èèèèèV╖è            è1.20 L
The fïal pressure is 1.07 atm.èWe expect a decrease ï pressure because
ê volume ïcreased.
Ç C
 2èA cylïder with a moveable pisën contaïs air at a pressure
ç 15 psi ï a volume ç 26.5 cu. ï.èHow much pressure must be applied
ë ê pisën ï order ë compress ê air ë 4.0 cu. ï.?

    A) 99 psi    B) 114 psi    C) 2.3 psi    D) 17 psi
üèListïg ê variables, we have

    P╢ = 15 psi, V╢ = 26.5 ïÄ, P╖ = ? psi, V╖ = 4.0 ïÄ

The volume units agree.èRearrangïg Boyle's Law, P╢V╢ = P╖V╖, ë solve
for P╖, å substitutïg values for ê variables, we obtaï
èèèèèV╢            26.5 ïÄ
P╖ = P╢ x ──.èè P╖ = 15 psi x ───────è= 99 psi
èèèèèV╖è            4.0 ïÄ
The fïal pressure is 99 psi.èWe expect an ïcrease ï pressure because
ê volume decreased.
Ç A
 3èA gas exerts a pressure ç 644 ërr ï a volume ç 20.0 mL.
The gas is allowed ë expå until ê pressure is 287 ërr.èWhat is ê
fïal volume ç ê gas?èAssume ê temperature remaïs constant.

    A) 8.91 mL    B) 64.9 mL    C) 44.9 mL    D) 28.9 mL
üèListïg ê variables we have

    P╢ = 644 ërr,è V╢ = 20.0 mL,è P╖ = 287 ërr,è V╖ = ? mL

The pressure units agree.èRearrangïg Boyle's Law, P╢V╢ = P╖V╖, ë solve
for V╖, å substitutïg values for ê variables, we obtaï
èèèèèP╢             644 ërr
V╖ = V╢ x ──.èè V╖ = 20.0 mL x ────────è= 44.9 mL
èèèèèP╖è             287 ërr
The fïal volume is 44.9 mL.èWe expect an ïcrease ï volume because ê
pressure decreased.
Ç C
 4èA gas occupies a 300. mL glass bulb at a pressure ç 740 ërr.
This glass bulb is connected ë an empty 500. mL glass bulb via a closed
sëpcock.èWhat will be ê fïal pressure ï ê system when ê sëp-
cock is opened?

    A) 278 ërr    B) 463 ërr    C) 444 ërr    D) 1233 ërr
üèListïg ê variables, we have

P╢ = 740 ërr,è V╢ = 300. mL,è P╖ = ? ërr,è V╖ = 800. mL

We must be careful about ê fïal volume ï this problem.èWhen ê
sëpcock is opened ê gas will occupy both glass bulbs.èThe ëtal vol-
ume ç ê system is 300 mL + 500 mL = 800 mL.
Rearrangïg Boyle's Law, P╢V╢ = P╖V╖, ë solve for P╖, å substitutïg
values for ê variables, we obtaï
èèèèèV╢            è300. mL
P╖ = P╢ x ──.èè P╖ = 740 ërr x ───────è= 278 ërr
èèèèèV╖è            è800. mL
The fïal pressure is 278 ërr.èWe expect ê pressure ë decrease because
ê volume ïcreased.
Ç A
 5èA large compressed gas cylïder contaïs helium at a pressure
ç 2400. psi.èThe volume ç ê cylïder is 49.8 L.èWhat volume would
ê helium occupy at 14.7 psi (1 atm)?

    A) 230. Lèè B) 1.76x10æ Lèè C) 50.1 Lèè D) 8.13x10Ä L
üèListïg ê variables, we have

P╢ = 2400 psi,è V╢ = 49.8 L,è P╖ = 14.7 psi,è V╖ = ? L

The pressure units agree.èRearrangïg Boyle's Law, P╢V╢ = P╖V╖, ë solve
for V╖, å substitutïg values for ê variables, we obtaï
èèèèèP╢             2400 psi
V╖ = V╢ x ──.èè V╖ = 49.8 L x ─────────è= 8.13x10Ä L
èèèèèP╖è             14.7 psi
The fïal volume is 8.13x10Ä L.èWe expect an ïcrease ï ê volume,
because ê pressure decreased.
Ç D
 6èA gas occupies a glass bulb at a pressure ç 694 ërr.èThis
glass bulb is connected ë an empty 435 mL glass bulb via a closed valve.
After ê valve was opened, ê pressure dropped ë 212 ërr.èWhat is
ê volume ç ê origïal glass bulb?è(Assume temperature is constant.)

    A) 133 mL    B) 235 mL    C) 191 mL    D) 102 mL
üèListïg ê variables, we have

P╢ = 694 ërr,è V╢ = ?,è P╖ = 212 ërr,è V╖ = V╢ + 435 mL

We must be careful about ê fïal volume ï this problem.èWhen ê
valve is opened, ê gas will occupy both glass bulbs.èThe ëtal volume
ç ê system is V╢ + 435 mL.èInsertïg ê values ïë Boyle's Law,
gives:        èP╢V╢ = P╖V╖,
    (694 ërr)(V╢) = (212 ërr)(V╢ + 435 mL)
The units ç ê volume will be "mL".
        694∙V╢ = 212∙V╢ + (212)(435)
        482∙V╢ = (212)(435)
        èèèè (212 ërr)(435 mL)è ┌────────┐
èèèèèèèèèèV╢ = ────────────────── = │ 191 mL │
            èè(482 ërr)èèèè└────────┘
Ç C
 7èA cylïder with a moveable pisën contaïs argon at a pressure
ç 672 ërr ï a volume ç 1.50 L.èThe gas is compressed ë 20.0 mL.
What is ê fïal pressure ç ê argon ï atm?

    A) 50.4 atm    B) 0.872 atm    C) 11.8 atmèè D) 66.3 atm
üèListïg ê variables, we have
P╢ = 672 ërr,èè V╢ = 1.50 L,èè P╖ = ? atm,èè V╖ = 20.0 mL
Neiêr ê pressure units nor ê volume units agree.èConvertïg ë
atmospheres å liters, we obtaï:
P╢ = 672 ërr(1 atm/760 ërr) = 0.884 atm
V╢ = 1.50 L
P╖ = ? atm
V╖ = 20.0 mLx(1 L/1000 mL) = 0.0200 L
Solvïg Boyle's Law for P╖ yields
èèèèèV╢            èè1.50 L
P╖ = P╢ x ──.èè P╖ = 0.884 atm x ────────è= 66.3 atm
èèèèèV╖è            è 0.0200 L
The fïal pressure is 66.3 atm.èWe expect an ïcrease ï pressure
because ê gas was compressed ïë a smaller volume.
Ç D
 8èA gas exerts a pressure ç 1.66 atm ï a volume ç 2.00 L.
The gas is allowed ë expå until ê pressure is 81.3 kPa.èWhat is ê
fïal volume ç ê gas, assumïg ê temperature was constant?
(1 atm = 101.3 kPa)
            A) 6.13 L        B) 4.13 L

            C) 2.97 L        D) 3.07 L
üèListïg ê variables, we have

P╢ = 1.66 atm,èè V╢ = 2.00 L,èè P╖ = 81.3 kPa,èè V╖ = ? L

The pressure units do not match.èOne choice is ë convert ê atm ïë
kPa.èè P╢ = (1.66 atm)(101.3 kPa/1 atm) = 168 kPa
Rearrangïg Boyle's Law, P╢V╢ = P╖V╖, ë solve for V╖, å substitutïg
values for ê variables, we obtaï
èèèèèP╢            168 kPa
V╖ = V╢ x ──.èè V╖ = 2.00 L x ────────è= 4.13 L
èèèèèP╖è            81.3 kPa
The fïal volume is 4.13 L.èWe expect an ïcrease ï ê volume, because
ê pressure decreased.
Ç B
 9èA balloon contaïs 477 mL ç hydrogen at an atmospheric
pressure ç 748.7 mm Hg.èWhat volume will ê hydrogen occupy at
762.2 mm Hg, if ê temperature does not change?

    A) 486 mL    B) 469 mL

    C) 1196 mL    D) 473 mL
üèListïg ê variables, we have

P╢ = 748.7 mm Hg,èè V╢ = 477 mL,èè P╖ = 762.7 mm Hg,è V╖ = ? mL

The pressure units agree, so we can rearrange Boyle's Law, P╢V╢ = P╖V╖,
ë solve for V╖.èSubstitutïg values for ê variables, we obtaï

èèèèèP╢            748.7 mm Hg
V╖ = V╢ x ──.èè V╖ = 477 mL x ───────────è= 469 mL
èèèèèP╖è            762.2 mm Hg

The fïal volume is 469 mL.èWe expect ê volume ë decrease because ê
atmospheric pressure ïcreased.
Ç B
 10èA sërage bulb contaïs nitrogen at an unknown pressure ï a
volume ç 5.00 L.èThe nitrogen expås ë occupy 7.92 L at a pressure ç
65.4 kPa.èAssumïg ê temperature did not change, what was ê ïitial
pressure ç ê nitrogen ï kPa?

    A) 103.6 kPa        B) 169 kPa
    C) 141 kPa        D) 106.6 kPa
üèListïg ê variables we have

P╢ = ? kPa,èè V╢ = 5.00 L,èè P╖ = 65.4 kPa,èè V╖ = 7.92 L

Solvïg Boyle's Law for P╢ yields

èèèèèV╖            è 7.92 L
P╢ = P╖ x ──.èè P╢ = 65.4 kPa x ────────è= 103.6 kPa
èèèèèV╢è            è 5.00 L

The ïitial pressure was 103.6 kPa.èWe expect an origïal pressure ë
have been higher than ê fïal pressure, because ê gas expåed.
Ç A