Transcript Boyle’s Law

Boyle’s Law - Review
P1V1 = P2V2
Boyle’s Law
P1V1 = P2V2
Pressure can be in any units
Volume can be in any units
Charles’ Law
If temperature increases, volume
increases
If temperature decreases, volume
decreases
This is direct variation (compared to
inverse for Boyle’s law)
Charles Law
V1
V2
=
T1
T2
Volume can be in any units
Temperature must be in Kelvin
To convert Celsius to Kelvin, add 273
(K = C + 273.)
Charles law
For example:
The temperature inside my fridge
is 4 C. If I place a balloon in my
fridge that initially has a temperature
of 22 C and a volume of .5 L, what
will be the volume of the balloon
when it is fully cooled by my
refrigerator?
Charles’ Law
First convert temperatures to Kelvin
4 + 273 = 277K
22 + 273 = 295K
Charles’ Law
Make a list of what you know
V1 =
T1 =
V2 =
T2 =
Charles’ Law
V1 = .5 L
T1 =
V2 =
T2 =
Charles’ Law
V1 = .5 L
T1 = 295K
V2 =
T2 =
Charles’ Law
V1 = .5 L
T1 = 295K
V2 = x
T2 =
Charles’ Law
V1 = .5 L
T1 = 295K
V2 = x
T2 = 277K
Charles’ Law
Put the information into the
equation:
V1 V2
=
T1 T2
.5 = x
295 277
Charles’ Law
Cross multiply to solve for x
.5(277) = x(295)
138.5 = 295x
x = .47 Liters
Charles’ Law
For Example:
A man heats a balloon in the oven.
If the balloon initially has a volume
of 0.4 L and a temperature of 20 C,
what will the volume of the balloon
be after he heats it to a temperature
of 250 C?
Charles’ Law
Convert temperatures to Kelvin
20 + 273 = 293 K
250 + 273 = 523 K
Charles’ Law
Make a list of what you know
V1 =
T1 =
V2 =
T2 =
Charles’ law
Make a list of what you know
V1 = .4 L
T1 = 293 K
V2 = x
T2 = 523 K
Charles’ law
Put the information into the
equation:
V1 V2
=
T1 T2
.4 = x
293 523
Charles’ law
Cross multiply to solve for x
.4(523) = x(293)
209.2 = 293x
x = .71 L