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GAS LAWS

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					      GAS LAWS-FOR CLOSED SYSTEMS
LAW OF                                            LAW OF GAY-
                             INCREASE AVERAGE
CHARLES                       KINETIC ENERGY(T)   LUSSAC




                                    CAUSES
PV1    =      PV2                                 P 1V 1   =    P 2V 2
T1             T2                                 T1            T2
TEMP IS IN KELVIN                                 TEMP IS IN KELVIN
                               INCREASE OF
K = C + 273                     MOMENTUM          K = C + 273
                                    CAUSES


 INCREASED                                                  INCREASED
VOLUME IN AN                                               PRESSURE IN
EXPANDABLE                   INCREASE COLLISION                A RIGID
 (CONSTANT
                    CAUSES                      CAUSES      (CONSTANT
                                 FREQUENCY
 PRESSURE)                                                    VOLUME)
 CONTAINER                                                  CONTAINER
                    GAS LAWS
INCREASE EXTERNAL            LAW OF BOYLE
    PRESSURE

                             P 1V 1   =    P 2V 2
       CAUSES


                                T           T
                             TEMP IS IN KELVIN
  INCREASE OF
    EXTERNAL                 K = C + 273
   COLLISIONS
       CAUSES




                                 DECREASED
                                VOLUME IN AN
   EXPANDABLE       CAUSES       EXPANDABLE
CONTAINER PUSHED                CONTAINER AT
 INTO SMALLER V                   CONSTANT
                                TEMPERATURE
•   THE LAWS OF GAY-LUSSAC (P and V) And the law of Charles (V and T) ARE
    DIRECT PROPORTIONS – AS ONE INCREASES THE OTHER INCREASES
    DIRECTLY.




BOYLES LAW IS AN INVERSE PROPORTION, AS P INCREASES, VOLUME
  DECREASES. PV=k


•   P




               V
   PROBLEM SOLVING, GAS LAWS
• STEP ONE: ANALYZE. Read the problem an set up
  your givens.
• STEP TWO: CHECK UNITS, CONVERT NOW SO YOU
  DO NOT FORGET       !
• STEP THREE: CRITICAL THINKING THE LOGIC! –
  DECIDE WHAT IS CONSTANT AND WHAT IS
  CHANGING, THEN DECIDE WHAT LAW TO APPLY.
• STEP FOUR: MAKE A LOGICAL ESTIMATION BASED
  ON THE APPROPRIATE GAS LAW, THEN USE THE
  EQUATION.
• STEP FIVE, CHECK, SUBSTITUTE IN YOUR ANSWER
  TO SEE IF THE EQUATION CORRECTLY SOLVES FOR
  A GIVEN.
EXAMPLE. CALCULATE THE VOLUME OF A 1.0L BALLOON THAT
  IS PRESSURIZED FROM 1.0 ATM TO 2.0 ATM AT CONSTANT
                        TEMP.
•   STEP ONE: ANALYZE. Read the problem an set up your givens.
    –   P#1 = 1.0ATM   LOGIC – PRESSURE AND VOLUME
    –   P#2 = 2.0ATM   CHANGE,AT CONSTANT TEMP, THEREFORE
    –   V#1 = 1.0L     THIS IS BOYLE’S LAW – WHICH IS AN
    –   V#2 = X        INVERSE PROPORTION.

    – AT CONSTANT TEMPERATURE


•   STEP TWO:CHECK UNITS – NET NEEDED HERE!
•   STEP THREE: CRITICAL THINKING THE LOGIC! – DECIDE WHAT IS
    CONSTANT AND WHAT IS CHANGING, THEN DECIDE WHAT LAW TO
    APPLY.
    –   P#1 = 1.0ATM       PRESSURE IS INCREASING (DOUBLES),
    –   P#2 = 2.0ATM       THEREFORE VOLUME WILL DECREASE


    – V#1 = 1.0L
                       VOLUME WILL DECREAS (HALVED),
    – V#2 = X
                       BECAUSE PRESSURE DOUBLED.
•   STEP FOUR: MAKE A LOGICAL ESTIMATION BASED ON THE
    APPROPRIATE GAS LAW, THEN USE THE EQUATION.
     – -BECAUSE PRESSURE DOUBLES, THE VOLUME WILL DECREASE
       (IVERSELY PROPORTIONAL CHANGE). As the initial volume is 1.0 L the
       final volume should be 0.50 L
•   P 1V 1   =   P 2V 2   à (1.0 atm)(1.0 L) = (2.0 atm)( X) à X= 0.50L
•    T1          T2


•   CHECK (FOR MATH ERRORS, THIS WILL NOT FIND LIGICAL ERRORS
    AT ALL TIMES.)
     – P1V1 = P2V2 à (1.0 ATM)(1.0L) = (2.0 ATM)(0.50L)

				
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posted:7/14/2013
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