Transcript Unit: Gas Laws

DO NOW:
Complete on the BACK of the HW WS!
1. A gas has a pressure of 0.12 atm at 21.0 °C.
What is the pressure at standard temperature?
2. If I have 5.6 L of gas in a piston at a pressure of
1.5 atm and compress the gas until its volume is
4.8 L, what will the new pressure inside the
piston be?
3. If I have 45 liters of helium in a balloon at 250 C
and increase the temperature of the balloon to
550 C, what will the new volume of the balloon
be?
Unit: Gas Laws
Combined Gas Law
After today you will be able to…
• Explain the effect on gas
properties using the Combined
Gas Law
• Calculate an unknown pressure,
temperature, or volume by
solving algebraically
The Combined Gas Law
The combined gas law is a
single expression that
combines Boyle’s, Charles’s,
and Gay-Lussac’s Laws.
•This gas law describes the
relationship between
temperature, pressure, and
volume of a gas.
•It allows you to do
calculations where only the
amount of gas is constant.
RB + JC +
JG-L = BFFs!
The Combined Gas Law
Helpful hint: You are able to
get which law you need by
For example, if there is no
IfIfthere
thereisisno
nomention
mention
of
ofpressure
volume in
in
covering
the
variable
that
mention of temperature in the
1
1
2
2
theisproblem,
cover
PV up
up
and
and
you
you
are
are
not
mentioned
in
the
problem!
problem, cover T up and you are
left There
with the
relationship
between
T 4
is
no
need
to
memorize
left with the relationship between
and
andP.V.(aka
(aka
Gay-Lussac’s
Charles’s 2
Law!)
Law!)
1
individual
just memorize
the
P and V. laws,
(aka Boyle’s
Law!)
Combined Gas Law and you can
derive all of the others!
PV PV
=
T
T
The Combined Gas Law
A gas occupies 3.78L at 529mmHg and 17.2°C. At what
pressure would the volume of the gas be 4.54L if the
temperature is increased to 34.8°C?
P1 V1 P2 V2
=
P1= 529mmHg
T
T
1
2
V1= 3.78L
T1= 17.2°C + 273= 290.2K
(P2)(4.54L)
(529mmHg)(3.78L)
P2= ?
=
(307.8K)
(290.2K)
V2= 4.54L
P2 = 467mmHg
T2= 34.8°C + 273= 307.8K
Questions?
Complete WS