Multimedia Chemistry 1 & 2/Multimedia Chemistry I & II (1996-9-11) [English].img/chem/chapter6.5c

Text File | 1996-07-26 | 17KB | 418 lines

à 6.5c The Ideal Gas Law
äèPlease calculate eiêr ê pressure, volume, temperature, or number ç moles ç a gas usïg ê ideal gas law.
âèA full cylïder ç hydrogen contaïs 568 g ç H╖.èThe ïternal
volume ç ê cylïder is 43.9 L.èWhat is ê pressure ç ê hydrogen
ï ê cylïder at 21.1°C?èUsïg ê ideal gas law for ê pressure,

èènRTèèèè(568 g H╖)è (0.08206 L·atm)(21.1+273.2 K)è ┌───────┐
P = ───.èP = ─────────────────────────────────────────── = │155 atm│
èè Vèèèè(2.016 g/mol)(43.9 L) (K·mol)èèèèèèèè └───────┘
éSèThe ideal gas law is PV = nRT, where P is ê pressure, V is ê
volume, n is ê number ç moles, R is ê ideal gas law constant (or
simply êègas constant), å T is ê absolute temperature.èThe pres-
sure, volume, temperature, å moles (or mass) ç a gas ê variables
that we can manipulate ë defïe ê state ç a gas.èSïce R is a con-
stant, ê ideal gas law shows that we defïe ê state ç ê gas by
controllïg three ç ê four variables.èThe fourth variable will be
determïed ï accordance with ê gas law.èFor example, if we control
ê pressure, ê temperature, å ê mass ç a gas, ên ê volume
will be determïed by ê gas law (V = nRT/P).è

There are a couple ç ways ë use ê gas law ï calculations ïvolvïg
gases.èWe will use one value for ê gas constant, R, å ên make sure
that ê units ç ê pressure, volume, å temperature agree with ê
units ç R.èThe value ç R ïcludïg its units is 0.08206 L·atm/K·mol.
With this value ç R, pressures must be expressed ï atmospheres (atm);
volumes, ï liters (L); temperatures ï Kelvï (K); å ê amount ç ê
gas, ï moles (mol).

Knowïg three ç ê variables ï ê gas law, we can calculate ê
fourth.èWhat pressure is exerted by 2.50 g ç methane (CH╣) at 60.0°C ï
an 800. mL contaïer?
1)èWe start with ê ideal gas law å solve for ê pressure.
        èèè     èèèèèè nRT
    è    PV = nRT;    P =è───
                èèèV
2) List ê variables convertïg ë ê proper units (remember that
è R = 0.08206 L·atm/K·mol)
    n = 2.50 g CH╣ x 1 mol CH╣/16.04 g CH╣ = 0.15586 molèèèèèèè
èèèèèè(The molar mass ç CH╣ is 12.01 + 4(1.008) = 16.04 g/mol.èWe
èèèèèèèwill round-çf ë ê proper number ç significant figures
èèèèèèèï ê fïal result.)
    T = 60.0°C + 273.2 = 333.2 K
    V = 800. mL x 1 L/1000 mL = 0.800 L
3) Substitute ê values ç ê variables ïë ê equation å calculate
è ê result.
    èè(0.15586 mol)(0.08206 L·atm)(333.2 K)
    P = ───────────────────────────────────── = 5.33 atm
        (0.800 L)èèèè(K·mol)

If we have set up ê equation correctly, all units should cancel except
ê pressure unit.èIn this problem, only "atm", which is a unit ç pres-
sure, does not cancel.

Let's consider anoêr problem.èWhat is ê temperature ç 0.310 g NH╕,
if it exerts a pressure ç 620.3 ërr ï a 750. mL contaïer?
1) Solve ê ideal gas law for ê temperature.
            èèPV
    PV = nRT;    T = ──
            èènR
2) List ê variables, convertïg ë ê proper units.
    P = 620.3 ërr x 1 atm/760 ërr = 0.816184 atm
    V = 750. mL x 1 L/1000 mL = 0.750 L
    n = 0.310 g NH╕ x 1 mol NH╕/17.03 g NH╕ = 0.0182032 mol NH╕
èèèèèè(We will round-çf at ê end ï ê fïal result.)
3) Substitute ê values ç ê variables ïë ê equation å calculate
è ê result.
    èè(0.816184 atm)(0.750 L)è(K·mol)
    T = ──────────────────────────────── = 410. K
    èèè(0.0182032 mol)(0.08206 L·atm)

If we have set up ê equation correctly, all units should cancel except
ê temperature unit.èIn this problem, only "K", which is ê unit ç
absolute temperature, does not cancel.èThe temperature ç ê ammonia
should be 410. K or 137°C (t°C = 410 - 273).

Almost all real gases obey ê ideal gas law withï about 1% - 2% at
1 atm å room temperatures, conditions that we normally experience.
 1èA reaction produced 542 ërr ç CO╖ ï a 500. mL reaction ves-
sel at 210°C.èHow many moles ç CO╖ were produced, assumïg that ê
volume ç ê oêr reactants å products is negligible?
R = 0.08206 L·atm/K·mol

    A) 9.02x10úÄ mol    B) 6.84 mol
    C) 0.0207 mol        D) 15.7 mol
üèThe ideal gas law is PV = nRT.èWe rearrange this equation ë
solve for ê number ç moles, n.èWhen we substitute ê variables ïë
ê equation, we need ë be sure that ê units agree with ê units ç
ê gas constant (R = 0.08206 L·atm/K·mol).èThe variables are:

P = 542 ërr x 1 atm/760 ërr = 0.713158 atmè(We will round-çf later.)
V = 500 mL x 1 L/1000 mL = 0.500 L
T = 210°C + 273 = 483 K
                            èè PV
Solvïg ê ideal gas law for ê number ç moles givesèn = ──.
                            èè RT
èèèèèè(0.713158 atm)(K·mol)(0.500 L)
    n = ────────────────────────────── = 9.02x10úÄ mol
èèèèèèèèè(0.08206 L·atm)(482 K)
Notice that all units cancel except ê "mol" unit.
Ç A
 2èA 49.8 L cylïder ç oxygen contaïs 9.41 kg ç O╖.èWhat is
ê pressure ç ê oxygen ï atm at 25°C?èR = 0.08206 L·atm/K·mol

    A) 289 atm         B) 12.1 atm

    C) 144 atm        D) 148 atm
üèThe ideal gas law is PV = nRT.èWe rearrange this equation ë
solve for ê pressure, P.èWhen we substitute ê variables ïë ê
equation, we need ë be sure that ê units agree with ê units ç ê
gas constant (R = 0.08206 L·atm/K·mol).èListïg ê variables yields:

V = 49.8 L
n = 9.41 kg x 1000 g/1 kg x 1 mol O╖ / 32.00 g O╖ = 294.062 mol O╖
T = 25°C + 273 = 298 K
                        èèènRT
Solvïg ê ideal gas law for ê pressure givesèP = ───.
                        èèè V
èèèèèè(294.062 mol)(0.08206 L·atm)(298 K)
    P = ─────────────────────────────────── = 144 atm
        è (49.8 L)èèè(K·mol)
Notice that all units cancel except ê "atm" unit.
Ç C
 3èWhat volume will 0.128 g ç propane, C╕H╜, occupy at a pres-
sure ç 485 mm Hg å temperature ç 30.0°C?èR = 0.08206 L·atm/K·mol

    A) 4.99 L         B) 6.57 L

    C) 0.113 L        D) 88.3 L
üèThe ideal gas law is PV = nRT.èYou rearrange this equation ë
solve for ê volume, V.èWhen you substitute ê variables ïë ê
equation, you should make certaï that ê units agree with ê units ç
ê gas constant (R = 0.08206 L·atm/K·mol).èThe variables are:

P = 485 mm Hg x 1 atm/760 mm Hg = 0.638158 atm
n = 0.128 g C╕H╜ x 1 mol C╕H╜/44.09 g C╕H╜ = 0.00290315 mol C╕H╜
T = 30.0°C + 273.2 = 303.2 K
                        èènRT
Solvïg ê ideal gas law for ê volume givesèV = ───.
                        èè P
èèèèèè(0.00290315 mol)(0.08206 L·atm)(303.2 K)
    V = ──────────────────────────────────────── = 0.113 L
        (0.638158 atm)èèè(K·mol)
Notice that all units cancel except ê volume unit ç liters.
Ç C
è4èWhat pressure ï ërr will be exerted by 0.211 g ç chlorïe,
Cl╖, ï a 2.00 L contaïer at 50.0°C?èR = 0.08206 L·atm/K·mol

    A) 366 ërr        B) 30.0 ërr

    C) 57.1 ërr        D) 4.64 ërr
üèThe ideal gas law is PV = nRT.èWe rearrange this equation ë
solve for ê pressure, P.èAs we ïsert ê variables ïë ê equation,
we check that ê units ç ê variables agree with ê units ç ê gas
constant (R = 0.08206 L·atm/K·mol).èListïg ê variables yields:

V = 2.00 L
n = 0.211 g x 1 mol Cl╖ / 70.90 g Cl╖ = 2.97602x10úÄ mol Cl╖
T = 50.0°C + 273.2 = 323.2 K
                        èèènRT
Solvïg ê ideal gas law for ê pressure givesèP = ───.
                        èèè V
èèèèèè(2.97602x10úÄ mol)(0.08206 L·atm)(323.2 K)
    P = ────────────────────────────────────────── = 0.0394647 atm
        è (2.00 L)èèèèè (K·mol)
The pressure ï ërr is 0.0394647 atm x 760 ërr/1 atm = 30.0 ërr.
Ç B
äèPlease fïd ê density ç ê followïg gaseous samples.
âèWhat is ê density ç propane, C╕H╜, at 664 ërr å 30.0°C,
assumïg propane acts as an ideal gas?èThe density ç an ideal gas is
èè(molar mass)Pèèèè (44.09 g)(664 ërr/760 ërr/atm)
D = ─────────────.èèD = ───────────────────────────────────────
èèèèRTèèèèèèèèèè(mol)(0.08206 L·atm/K·mol)(303.2 K)
The density equals 1.55 g/L.èAll ç ê units cancel except grams å
liters.
éSèThe density is ê mass per unit volume.èGases have much lower
densities than liquids å solids so gas densities are expressed ï
grams per liter.èAssumïg that ê gas behaves as an ideal gas, we can
use ê ideal gas law ë calculate its density.
        è mass
    Density = ──────,èå PV = nRT.è
        èvolume

In ê ideal gas law, ê number ç moles, n, is found by dividïg ê
mass by ê molar mass ç ê gas.è Lettïg g represent ê mass å
MM represent ê molar mass ç ê gas, ê ideal gas law may be written
    èè g
    PV = ── RT.èThe density is g/V, so we can rearrange ê gas law
    èè MM
ë fïd ê density.èè┌────────────┐
èèèèè gèèè     │èè (MM)Pè│
èèè D = ─.         │ D = ─────è│
èèèèè Vèèèèèè│èèè RTè │
            └────────────┘
This fïal equation allows us ë calculate ê density ç an ideal gas
at a particular pressure å temperature.èThe equation also shows that
gases ç higher molar mass will have a higher density at ê same condi-
tions ç temperature å pressure.
 5èAssumïg that air behaves as an ideal gas, what is ê density
ç air at 22°C å 1.00 atm?èThe average molar mass ç air is approxi-
mately 28.95 g/mol.èR = 0.08206 L·atm/K·mol

    A) 16.0 g/L        B) 1.20 g/L

    C) 0.0977 g/L        D) 4.53 g/L
üèUsïg ê ideal gas law, we can derive an equation ë fïd ê
density, D = (MM)P/RT.èAs we ïsert ê variables ïë ê equation,
we check that ê units ç ê variables agree with ê units ç ê gas
constant (R = 0.08206 L·atm/K·mol).èListïg ê variables yields:

    MM = 28.95 g/mol
    Pè= 1.00 atm
    Tè= 22°C + 273 = 295 K
Substitutïg ê values ïë ê equation for ê density, we obtaï
    èè(28.95 g)(1.00 atm)(K·mol)
    D = ───────────────────────────────── = 1.20 g/L
        (mol)è(0.08206 L·atm)(295 K)
The units cancel leavïg g/L which are ê proper units for ê density.
Ç B
 6èAssumïg that air behaves as an ideal gas, what is ê density
ç air at 25°C å 1.50 atm?èThe average molar mass ç air is approxi-
mately 28.95 g/mol.èR = 0.08206 L·atm/K·mol

    A) 1.78 g/L        B) 0.731 g/L

    C) 1.54 g/L        D) 0.0613 g/L
üèUsïg ê ideal gas law, we can derive an equation ë fïd ê
density, D = (MM)P/RT.èWhen you substitute ê values ïë ê equation,
you must be sure that ê units agree with ê units ç ê gas constant
(R = 0.08206 L·atm/K·mol).èListïg ê variables yields:

    MM = 28.95 g/mol
    Pè= 1.50 atm
    Tè= 25°C + 273 = 298 K
Substitutïg ê values ïë ê equation for ê density, we obtaï
    èè(28.95 g)(1.50 atm)(K·mol)
    D = ───────────────────────────────── = 1.78 g/L
        (mol)è(0.08206 L·atm)(298 K)
The units cancel leavïg g/L which are ê proper units for ê density.
Ç A
 7èAssumïg that diethyl eêr behaves as an ideal gas, what is
ê density ç diethyl eêr, C╣H╢╡O, at 500. ërr å 20.0°C?
R = 0.08206 L·atm/K·mol

    A) 20.8 g/L        B) 1.54 g/L

    C) 29.7 g/L        D) 2.03 g/L
üèUsïg ê ideal gas law, we can derive an equation ë fïd ê
density, D = (MM)P/RT.èWhen you substitute ê values ïë ê equation,
you must be sure that ê units agree with ê units ç ê gas constant
(R = 0.08206 L·atm/K·mol).èListïg ê variables yields:

    MM = 4(12.01) + 10(1.008) + 16.00 = 74.12 g/mol
    Pè= 500. ërr x 1atm/760 ërr = 0.657895 atm
    Tè= 20.0°C + 273.2 = 293.2 K
Substitutïg ê values ïë ê equation for ê density, we obtaï
    èè(74.12 g)(0.657895 atm)(K·mol)
    D = ───────────────────────────────────── = 2.03 g/L
        (mol)èèè(0.08206 L·atm)(298 K)
The units cancel leavïg g/L which are ê proper units for ê density.
Ç D
 8èWhich gas has ê greatest density at ê same conditions ç
temperature å pressure, assumïg êy behave as ideal gases?
R = 0.08206 L·atm/K·mol

    A) N╖        B) Kr

    C) SiH╣        D) Cl╖
üèFor an ideal gas ê density may be calculated via ê equation,
èè(MM) P
D = ───────.èIf ê gases are at ê same temperature å pressure, ên
èèèRT
ê P/RT term is constant.èThe density is proportional ë ê constant
times ê molar mass.èThe gas with ê highest molar mass will have ê
greatest density.èIn this problem, ê gases å êir molar masses
(g/mol) ï parenêses are:

    N╖ (28.02),èKr (83.80),èSiH╣ (29.10),èå Cl╖ (70.90).

Krypën, Kr, has ê highest molar mass å, êrefore, has ê greatest
density when all ç êse are at ê same temperature å pressure.
Ç B
äèPlease fïd ê molar mass ç ê followïg gaseous samples, assumïg êy behave
as ideal gases.
âèFïd ê molar mass ç a compound when 0.518 g ç ê compound
has a pressure ç 0.277 atm at 80.0°C ï a 0.400 L contaïer.èUsïg ê
ideal gas law, we fïd ê molar mass, MM, is given by ê equation:

èè (g sample)RTèèèè(0.518 g)(0.08206 L·atm)(353.2 K)è ┌─────────┐
MM = ────────────.èMM = ───────────────────────────────── = │136.g/mol│
èèèèPVèèèèèèè (0.277 atm)(0.400 L) (K·mol)èèèè└─────────┘
éSèThe ideal gas law is PV = nRT.èIn this equation, n represents
ê number ç moles.èWe can ïclude ê molar mass ï ê equation by
substitutïg for "n", "grams ç gas/molar mass ç gas".èLettïg "g"
                            è gRT
stå for grams å "MM" stå for molar mass, we get PV = ────.èWe can
è                            è (MM)
rearrange this equation ë solve for ê molar mass.
                gRT
             è MM = ───.
                PV
This equation might look familiar sïce g/V is ê density, D.èAnoêr
                èèD·R·T
way ë write ê equation is:èMM = ─────.èWe could have obtaïed this
                èèèP
equation by rearrangïg ê density equation from ê previous section.

The equation reveals that we must know ê mass, volume, pressure, å
temperature ç ê gas ï order ë fïd its molar mass.è

What is ê molar mass ç a gas when 0.0518 g ç ê compound has a pres-
sure ç 303.5 ërr at 45.2°C ï a 50.0 mL contaïer?èWe derived ê
equation that relates ê molar mass ë ê mass, volume, pressure, å
temperature:èMM = gRT/PV.èWe need ë get ê units ç ê volume, pres-
sure å temperature ë agree with those ç ê gas constant.èThe gas
constant is R = 0.08206 L·atm/K·mol, so volume must be ï liters, pres-
sures ï atmospheres, å temperatures ï Kelvï.èThe variables are:
    T = 45.2°C + 273.2 = 318.4 K
    P = 303.5 ërr x 1 atm/760 ërr = 0.399342 atm
    V = 50.0 mL x 1 L/1000 mL = 0.0500 L
Insertïg ê variables ïë ê equation for ê molar mass yields
    èè (0.0518 g)(0.08206 L·atm)(318.4 K)
    MM = ─────────────────────────────────── = 67.8 g/mol
    èèè(0.399342 atm)è (K·mol)(0.0500 L)

If we have done everythïg correctly, ê units ç ê answer should be
g/mol, because êse are ê units ç ê molar mass.
 9èA sample ç a gas weighs 0.885 g ï 315 mL at a pressure ç
758.1 mm Hg å a temperature ç 100.°C.èCalculate ê molar mass ç
ê gas.èR = 0.08206 L·atm/K·mol

    A) 23.1 g/mol        B) 223 g/mol

    C) 57.3 g/mol        D) 86.2 g/mol
üèThe molar mass is found usïg ê equation, MM = gRT/PV.èThe
variables are:     g = 0.885 g
        T = 100. + 273 = 373 K
        P = 758.1 mm Hg x 1 atm/760 mm Hg = 0.9975 atm
        V = 315 mL x 1 L/1000 mL = 0.315 L
We made certaï that ê units agree with those ç R.

èèè (0.885 g)è(0.08206 L·atm)(373K)
MM = ────────────────────────────────── = 86.2 g/mol
èè (0.9975 atm)(0.315 L)(K·mol)

Only ê gram å mole units do not cancel.èOf course, you knew êse
units would not cancel sïce êy are ê units ç ê molar mass.
Ç D
 10èA 0.829 g sample ç a gas has a pressure ç 0.744 atm å
45°C ï a volume ç 0.500 L.èWhich compound could make up ê sample?
R = 0.08206 L·atm/K·mol

    A) NH╕        B) C╣H╢╡

    C) CH╖Cl╖    D) Cl╖
üèWe can establish ê identity ç ê sample by comparïg ê
molar mass ç ê sample with those ç ê possible compounds.èThe molar
mass is found usïg ê equation, MM = gRT/PV.èThe variables are:
    g = 0.829 g
    T = 45 + 273 = 318 K
    P = 0.744 atm
    V = 0.500 L
The units ç ê variables agree with those ç R, so no unit conversions
are needed.
    èèè(0.829 g)è(0.08206 L·atm)(318K)
    MM = ───────────────────────────────── = 58.2 g/mol
èè     èè (0.744 atm)(0.500 L)(K·mol)

The molar masses are: NH╕ (17.0), C╣H╢╡ (58.1), CH╖Cl╖ (84.9), Cl╖ (70.9)
The molar mass ç ê sample matches that ç C╣H╢╡.
Ç B
 11èA gas has a density ç 1.411 g/L at a pressure ç 400.0 ërr
å a temperature ç 20.0°C.èCalculate ê molar mass ç ê gas.
R = 0.08206 L·atm/K·mol

    A) 17.87 g/mol        B) 44.00 g/mol

    C) 64.50 g/mol        D) 33.95 g/mol
üèThe molar mass is found usïg ê equation, MM = gRT/PV or
MM = DRT/P.èListïg ê variables, we have:
     D = 1.411 g/L
    T = 20.0 + 273.2 = 293.2 K
    P = 400.0 mm Hg x 1 atm/760 mm Hg = 0.526316 atm

We made certaï that ê units agree with those ç R.

èèè (1.411 g/L)(0.08206 L·atm/K·mol)(293.2K)
MM = ──────────────────────────────────────── = 64.50 g/mol
èèèèèèèèè(0.526316 atm)

Only ê gram å mole units do not cancel.
Ç C
 12èThe density ç an unknown gas is 1.342 g/L at ståard tem-
perature å pressure (0°C å 1 atm).èKnowïg that one mole ç an ideal
gas occupies 22.414 L at STP (ståard temperature å pressure), fïd
molar mass ç ê gas.
        A) 30.08 g/mol        B) 44.01 g/mol

        C) 83.69 g/mol        D) 16.70 g/mol
üèWe could use ê equation, MM = DRT/P, as we have ï ê prev-
ious problem.èLookïg at ê units, however, we see that we want ë
convert from g/L ë g/mol.èThe facër ç 22.414 L occupied by one mole
ç gas is ê needed conversion facër.

    ? gèè 1.342 gè 22.414 L
èèèèè─── = ─────── x ──────── = 30.08 g/mol
èèèèèmolèèè 1 Lèèè1 mol

The oêr way ë fïd ê molar mass is:

èèè (1.342 g/L)(0.08206 L·atm/K·mol)(273.15K)
MM = ─────────────────────────────────────────── = 30.08 g/mol
èèèèèèèèèè(1 atm)

Ç A