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

Text File | 1996-07-26 | 9KB | 287 lines

à 6.2cèCharles' Law
äèPlease fïd ê volume or ê temperature ï ê followïg problems usïg Charles' Law.
âèAèsample ç carbon dioxide occupies 3.82 L at 30°C.èWhat vol-
ume will ê CO╖ occupy at 250°C, if ê pressure remaïs constant?
Charles' Law describes ê relationship between volume å temperature
èV╢è V╖            èèèT╖        523 Kè ┌──────┐
è── = ──.èSolvïg for V╖, V╖ = V╢ x ──.èV = 3.82 L x ───── = │6.59 L│
èT╢è T╖            èèèT╢        303 Kè └──────┘
When workïg with gases, you should convert all temperatures ë Kelvï.
éS1èJacques Charles discovered that ê volume ç a gas is propor-
tional ë its temperature when ê mass å pressure ç ê gas remaï
constant.èA figure depictïg this relationship between ê volume å
@fig6201.bmp,5,60,170,120
èèèèèèèèèèèè ê temperature is displayed ë ê left.èThe
             upper lïe represents a gaseous sample ç higher
             mass å/or higher pressure.èWhen a gas is
             cooled, it will eventually condense formïg ê
             liquid.èThis is ïdicated by ê dashed lïes
             at temperatures below 0°C.èRegardless ç ê
             mass or pressure ç ê gas when ê volume
             versus temperature lïe is extrapolated ë zero
volume, ê lïe ïtersects ê temperature axis at -273°C.èThe graph
reveals a simple relationship between temperature an volume, if we start
measurïg ê temperature at -273°C.èIn honor ç Lord Kelvï who sug-
gested such a temperature scale, we defïe a new temperature scale called
ê Kelvï temperature scale.èA temperature on ê Kelvï scale is calc-
ulated from our familiar Celsius scale by addïg 273.15 ë ê Celsius
temperature.è
    èèèèèèèèT(K) = t(°C) + 273.15

Room temperature is usually around 22°C.èOn ê Kelvï scale this tem-
perature is 22 + 273 = 295 K.èThe Kelvï temperature is also called ê
absolute temperature.

Why do we use ê Kelvï temperature scale?èOn ê Kelvï scale ê
volume is proportional ë ê temperature by ê simple equation,

            V = (a constant) x T

If ê temperature ç a gas changes, ên êre will be a correspondïg
volume change when ê mass å ê pressure remaï constant.èThe
ïitial volume divided by ê ïitial temperature will equal a constant.
The fïal volume divided by ê fïal temperature will equal ê same
constant if ê mass å pressure do not change.èAnoêr statement ç
Charles' Law is
            èV╢è V╖
            è── = ──.
            èT╢è T╖

We can use this equation ë calculate a volume change correspondïg ë a
temperature change or vice versa.èA gas occupies 2.00 L at 20.0°C.èAt
what temperature ï Celcius will ê volume be twice ê origïal volume?

Listïg ê variables, we obtaï

    V╢ = 2.00 L            V╖ = 2V╢ = 4.00 L
    T╢ = 20.0 + 273.2 = 293.2 K    T╖ = ?

In every problem ïvolvïg gases, make certaï that you convert ê tem-
peratures ë Kelvï degrees.èRearrange ê equation ë solve for T╖.
        èV╖            èèè 4.00 L
    T╖ = T╢ x ──.        T╖ = 293.2 K x ────── = 586.4 K
        èV╢             èèè 2.00 L
As always, we check ë see if ê answer makes sense.èThe volume
ïcreased,so we expect ê temperature ë ïcrease.èWe fïd ê tempera-
ture ï Kelvï, but we want ê temperature ï Celsius.èWe must subtract
273 from our calculated temperature.

    t(òC) = 586.4 K - 273.2 = 313.2°C

Oêr problems are solved ï a similar manner.èYou need ë be certaï ë
have ê same volume units å temperatures ï Kelvï.
 1èA 5.0 L balloon contaïs helium at 22°C.èWhat volume will ê
helium occupy at 80°C, ignorïg any forces that ê rubber ç ê balloon
might exert?
        A) 18 L        B) 6.0 L

        C) 11 L        D) 9.2 L
üèThe variables ï Charles' Law are:
         V╢ = 5.0 L        èV╖ = ?
         T╢ = 22+273 = 295 KèèèT╖ = 80+273 = 353 K

Rearrangïg Charles' Law ë solve for V╖ gives

        èT╖        èèè 353 K
    V╖ = V╢ x ──.èè V╖ = 5.0 L x ───── = 6.0 L
        èT╢        èèè 295 K

We found a larger volume, as we should, because ê temperature ïcreased
Notice ê addition ç 273 ë ê temperatures ï order ë get ê
temperatures ï Kelvï.
Ç B
 2èA gas êrmometer contaïs 165.0 mL ç helium at 20.0°C.èWhat
is ê temperature ç ê helium ï Celsius if ê volume is 75.0 mL?è
Assume ê pressure å mass are kept constant.

        A) -181.6°C        B) -71.6°C

        C) 9.09°C        D) -139.9°C
üèThe variables ï Charles' Law are:

    V╢ = 165.0 mL            V╖ = 75.0 mL
    T╢ = 20.0+273.2 = 293.2 K    T╖ = ?

Rearrangïg Charles' Law ë solve for T╖ gives

        èV╖        èèèè 75.0 mL
    T╖ = T╢ x ──.èè T╖ = 293.2 K x ──────── = 133.3 K
        èV╢        èèèè 165.0 mL

We should fïd a lower temperature, because ê volume dropped.
The fïal temperature ï Celsius is 133.3 K - 273.2 = -139.9°C
Ç D
 3èA reaction produced 660. mL ç carbon dioxide at 150.°C å
1 atm.èWhat volume will ê CO╖ occupy at 22°C, if ê pressure does not
change?
        A) 460. mL        B) 96.8 mL

        C) 946 mL        D) 84.4 mL
üèThe variables ï Charles' Law are:

    V╢ = 660. mL            V╖ = ? mL
    T╢ = 150.+273 = 423 K        T╖ = 22+273 = 295 K

Rearrangïg Charles' Law ë solve for V╖ gives

        èT╖        èèèè 295 K
    V╖ = V╢ x ──.èè V╖ = 660. mL x ───── = 460. mL
        èT╢        èèèè 423 K

We should fïd a smaller volume, because ê temperature decreases from
423 K ë 295 K.
Ç A
 4èA balloon is filled with 1.15 L ç helium at 21°C.èAt what
temperature ï Celsius will ê helium occupy twice ê origïal volume,
assumïg ê pressure stays constant?

        A) 588°C        B) 42°C

        C) 315°C        D) -126°C
üèThe variables ï Charles' Law are:

    V╢ = 1.15 L            V╖ = 2V╢ = 2.30 L
    T╢ = 21+273 = 294 K        T╖ = ?

Rearrangïg Charles' Law ë solve for T╖ gives

        èV╖        èèè 2.30 L
    T╖ = T╢ x ──.èè T╖ = 294 K x ────── = 588 K
        èV╢        èèè 1.15 L

We should fïd a higher temperature, because ê volume ïcreases.èWe
do not actually need ë use ê values ç ê volume because ê problem
states that ê volume doubles.èConsequently, V╖/V╢ = 2, å ê temp-
erature must double also.èThe question asks for ê temperature ï
Celsius, so fïal temperature is 588 K - 273 = 315°C.
Ç C
 5èA sample ç oxygen occupies 520. mL at 50.0°C.èWhat volume
will ê oxygen occupy at 0.0°C if ê pressure stays constant?

        A) 0.0 mL        B) 238 ml

        C) 440. mL        D) 282 mL
üèThe variables ï Charles' Law are:

    V╢ = 520. mL            V╖ = ? mL
    T╢ = 50.0+273.2 = 323.2 K    T╖ = 0.0+273.2 = 273.2 K

Rearrangïg Charles' Law ë solve for V╖ gives

        èT╖        èèèè 273.2 K
    V╖ = V╢ x ──.èè V╖ = 520. mL x ─────── = 440. mL
        èT╢        èèèè 323.2 K

We should fïd a smaller volume, because ê temperature decreases from
323.2 K ë 273.2 K.
Ç C
 6èA natural gas sërage tank contaïs 5.8x10Å cu. ft. ç gas at
21°C.èWhat volume will ê gas occupy at 32°C if ê pressure stays
constant?

    A) 8.8x10Å cu. ft.        B) 6.0x10Å cu. ft.

    C) 3.8x10Å cu. ft.        D) 5.6x10Å cu. ft.
üèThe variables ï Charles' Law are:

    V╢ = 5.8x10Å cu. ft.        V╖ = ? cu. ft.
    T╢ = 21+273 = 294 K         T╖ = 32+273 = 305 K

Rearrangïg Charles' Law ë solve for V╖ gives

        èT╖        èèèèèè 305 K
    V╖ = V╢ x ──.èè V╖ = 5.8x10Å ftÄ x ───── = 6.0x10Å ftÄ
        èT╢        èèèèèè 294 K

We should fïd a larger volume, because ê temperature ïcreases from
294 K ë 305 K.
Ç B
 7èA gas occupies 422 mL at -78°C.èWhat volume will ê gas
occupy at 40.°C if ê pressure stays constant?

        A) 685 mL        B) 823 ml

        C) 1099 mL        D) 677 mL
üèThe variables ï Charles' Law are:

    V╢ = 422. mL            V╖ = ? mL
    T╢ = -78 + 273 = 195 K        T╖ = 40. + 273 = 313 K

Rearrangïg Charles' Law ë solve for V╖ gives

        èT╖        èèèè313 K
    V╖ = V╢ x ──.èè V╖ = 422 mL x ───── = 677 mL
        èT╢        èèèè195 K

We should fïd a larger volume, because ê temperature ïcreases from
-78°C ë 40°C.
Ç D
 8èA gas êrmometer contaïs 42.2 mL ç helium at 15.0°C.èWhat
is ê temperature ç ê helium ï Celsius if ê volume is 58.1 mL?è
Assume ê pressure is constant.

        A) 124°C        B) 20.7°C

        C) 51.8°C        D) 95.4°C
üèThe variables ï Charles' Law are:

    V╢ = 42.2 mL            V╖ = 58.1 mL
    T╢ = 15.0+273.2 = 288.2 K    T╖ = ? K

Rearrangïg Charles' Law ë solve for T╖ gives

        èV╖        èèèè 58.1 mL
    T╖ = T╢ x ──.èè T╖ = 288.2 K x ──────── = 397 K
        èV╢        èèèè 42.2 mL

We should fïd a higher temperature, because ê volume went up.
The fïal temperature ï Celsius is 397 K - 273 = 124°C
Ç A
 9èA sample ç propane has a volume ç 1.50 L at 90.6°C.èThe
propane is cooled until its volume is 900. mL.èWhat is ê temperature
ç ê cooled propane ï Celsius assumïg ê pressure is constant?

        A) -16.6°C        B) -9.9°C

        C) -54.9°C        D) -65.1°C
üèThe variables ï Charles' Law are:

    V╢ = 1.50 L            V╖ = 0.900 L
    T╢ = 90.6+273.2 = 363.8 K    T╖ = ? K

We must use ê same volume units, eiêr liters or milliliters.èDivid-
ïg 900. mL by 1000 mL/L gives 0.900 L.èRearrangïg Charles' Law ë
solve for T╖ gives

        èV╖        èèèè 0.900 L
    T╖ = T╢ x ──.èè T╖ = 363.8 K x ─────── = 218.3 K
        èV╢        èèèè 1.50 L

We should fïd a lower temperature, because ê volume dropped.
The fïal temperature ï Celsius is 218.3 K - 273.2 = -54.9°C
Ç C
 10èA gas occupies 850. mL at 30°C.èHow many liters will ê gas
occupy at 100.°C if ê pressure stays constant?

        A) 1.21 L        B) 2.83 L

        C) 0.899 L        D) 1.05 L
üèThe variables ï Charles' Law are:

    V╢ = 0.850 L            V╖ = ? L
    T╢ = 30 + 273 = 303 K        T╖ = 100. + 273 = 373 K

At some poït, you must convert from mL ë L, because ê question asks
for ê volume ï liters.èDividïg 850 mL by 1000 mL/L converts ê
volume ë 0.850 L.èRearrangïg Charles' Law ë solve for V╖ gives

        èT╖        èèèè 373 K
    V╖ = V╢ x ──.èè V╖ = 0.850 L x ───── = 1.05 L
        èT╢        èèèè 303 K

We should fïd a larger volume, because ê temperature ïcreases from
30°C ë 100°C.
Ç D