Boyle’s Law
As
the pressure (P) on a gas increases, what happens to the volume (V)?
Answer:
P↑ , V↓
Boyle
said, “The volume of a given amount of
gas, at a constant temperature, varies inversely with the applied pressure.”
∴ P ∝
1/V
(∝
means “proportional to”; to remove a proportionality symbol and introduce an
equal sign instead, we incorporate a constant (k)):
P
= k (1/V) or k = PV
For
the gas at initial conditions, P1V1 = k
For
the gas at final conditions, P2V2 = k
And,
since a constant is, well, constant, we can equate the k, which allows us to equate
the other halves of the equations:
P1V1
= P2V2
ex. If 15.0 L of gas at 80.0 kPa is compressed
until its pressure is 320 kPa, what is the new volume of the gas?
V1 = 15.0 L P1V1 = P2V2
P1 = 80.0 kPa 80.0 kPa(15.0 L) = 320 kPa(V2)
P2 = 320 kPa V2 = 3.75 L
V2 = ? ∴ the new volume is 3.8 L.
👉 Check out this video for a cool demonstration of Boyle's Law.
=======================================
Pressure
can be measured using several units (Pa, kPa, atm, Torr, mmHg, psi, bar) and
you should be able to convert between all of them, using the relationships on
p. 509.
Let’s practice. Convert (a) 212 kPa into atm, (b) 11.57 psi into mm Hg
Let’s practice. Convert (a) 212 kPa into atm, (b) 11.57 psi into mm Hg
(a)
Using the relationship between kPa and atm (101.325 kPa = 1.00 atm), set up a
ratio:
212
kPa x
----------------- = ------------
101.325 kPa 1.00 atm
----------------- = ------------
101.325 kPa 1.00 atm
x = 2.09 atm
(b)
Using the relationship between psi and mmHg (14.7 psi = 760 mmHg), set up a
ratio:
11.57
psi x
------------- = ---------------
14.7 psi 760.0 mmHg
------------- = ---------------
14.7 psi 760.0 mmHg
x = 598 mmHg
=======================================
Charles’ Law
As
the temperature (T) of a gas increases, what happens to the volume (V)?
Answer:
T↑ , V↑
Charles
said, “The volume of a fixed mass of gas
is proportional to its temperature when the pressure is kept constant.”
∴ V ∝
T → V = kT or k = V/T
For
the gas at initial conditions, V1/T1 = k
For
the gas at final conditions, V2/T2 = k
Which
combines to give:
V1/T1
= V2/T2
Charles
did V-T experiments on various gases and found that regardless of the gas used,
the x-intercept on the V vs. T graph was always -273°C
(see figure 11.15, p. 519). A fellow
scientist, Lord Kelvin used this information to create a new temperature scale,
where TK = TC + 273.15.
For instance, 25.00°C would be equivalent to 298.15 K. For
all gas calculations, we will always use temperatures in K.
ex.
A 2.50 L party balloon is removed from a room at 22.3ºC and taken outside on a cold winter day.
If the outside temperature is -3.6ºC, what is the volume of the balloon
now?
V1 = 2.50 L V1/T1 = V2/T2
T1 = 22.3 + 273.15 = 295.5 K 2.50 L/295.5 K = V2/269.6 K
V2 = ? V2 = 2.28 L
T2 = -3.6 + 273.15 = 269.6 K ∴ the volume is 2.28 L.
V1 = 2.50 L V1/T1 = V2/T2
T1 = 22.3 + 273.15 = 295.5 K 2.50 L/295.5 K = V2/269.6 K
V2 = ? V2 = 2.28 L
T2 = -3.6 + 273.15 = 269.6 K ∴ the volume is 2.28 L.
👉 Check out this video for a cool demonstration of Charles' Law.
Gay-Lussac’s Law
As
the temperature (T) of a gas increases, what happens to the pressure (P)?
Answer:
T↑ , P ↑
Gay-
Lussac said, “The pressure of a fixed
amount of gas, at a constant volume, is directly proportional to its Kelvin
temperature.”
∴P ∝
T → P = kT or k = P/T
For
the gas at initial conditions, P1/T1 = k
For
the gas at final conditions, P2/T2 = k
P1/T1
= P2/T2
ex.
A gas, at 225 kPa, has a temperature of 26ºC.
If the pressure is decreased to 101 kPa, what is the new temperature of
the gas?
P1 = 225 kPa P1/T1 = P2/T2
T1 = 26 + 273.15 = 299 K 225 kPa/299 K = 101 kPa/T2
P2 = 101 kPa T2 = 134.22 K
T2 = ? ∴ the temperature is 134 K.
P1 = 225 kPa P1/T1 = P2/T2
T1 = 26 + 273.15 = 299 K 225 kPa/299 K = 101 kPa/T2
P2 = 101 kPa T2 = 134.22 K
T2 = ? ∴ the temperature is 134 K.
👉 Check out this video for a cool demonstration of Gay-Lussac's Law.
Homework:
Practice p. 514 # 1, 4, 9
Learning Check p. 518 # 16
Practice p. 522 # 11, 14
Practice p. 525 # 21, 25
Student Questions:
1. I can't get 1 or 4 from page 514. Not sure what I'm doing wrong. Also, the answer to page 518 #16 is not at the back of the book so I can't check if I did it right.
No problem. I did all the answers, because I'm just nice like that. :)
