2. Robert Boyle was an Irish – born gentlemen philosopher who did
research and investigation in physics, chemistry, alchemy, and theology.
Boyle studied the behavior of gases. He expanded on the assumptions of
Rene Descartes, who said that pressure is due to the restless agitation of the air
particles. Boyle proved his law both great and small pressures, using the now-
famous J-tube experiment.
3. Using a J-shaped piece of glass tubing that was sealed on one end that Robert
Boyle employed, he was able to establish the relationship between volume and
pressure. Varying amounts of mercury were added to the J-shaped tubed with gas
trapped in the sealed end of the tube and were used to vary the pressure of the
system. (see the figure above). He systematically varied the pressure and measured
the volume of gas. The measurements were performed using a fixed amount of gas
and a constant temperature. He noticed that when temperature is held constant, the
volume (V) of a given amount of gas decreases as the pressure (P)is increased.
4. This experiment shows the relationship between pressure and volume of gas at
constant temperature. On the contrary, if the pressure that is applied is decreased,
the gas volume becomes larger.
His experiment proved that pressure is inversely proportional to the volume
of gas at constant temperature, that is, the volume decreases with increasing
pressure and vice versa. Mathematically, we can express this relationship as :
P ∝
1
𝑉
Where,
P ⟶ pressure
V ⟶ Volume
∝ ⟶ proportionality sign
5. To change the proportionality sign to equal sign, then:
P = k
1
𝑉
Where k is the proportionality constant. Furthermore,
P =
𝑘
𝑉
; k = PV
For a given sample of gas under two different sets of condition, at constant
temperature,
P1V1 = P2V2 or
where,
P1 = initial pressure V1 = initial volume
P2 = final pressure V2 = final pressure
6. Example 1.
A fixed amount of gas occupies a syringe with a volume of 6.0 L.
The pressure at 25°C is 1.00 atm. What will be the new pressure if the
volume become 3.0 L at the same temperature.
Given:
P1 = 1.00 atm V1 = 6.0 L
P2 = ? V2 = 3.0 L
8. Example 2.
A certain amount of gas at 30 °C occupies a container with an adjustable
volume. It currently has a volume of 11.0 L, with a pressure of 1.25 atm. What
would its volume be if the pressure were adjusted to 1.50 atm?
Given:
P1 = 1.25 atm V1 = 11.0 L
P2 = 1.50 atm V2 = ?
Solution:
V2 =
𝑃1
𝑉1
𝑃2
;
V2 =
(1.25 𝑎𝑡𝑚)(11.0 𝐿)
1.50 𝑎𝑡𝑚
=
(13.75 𝑎𝑡𝑚.𝐿)
1.50 𝑎𝑡𝑚
= 9. 17 L
9. DO THIS!
A gas occupies a container with a pressure of 2.0 atm at a
particular temperature. What is the new pressure if the container
expands to twice its original volume?
Given:
P1 = 2.0 atm V1 = x
P2 = ? V2 = 2x
10. Given:
P1 = 2.0 atm V1 = x
P2 = ? V2 = 2x
Solution:
P2 =
(2.0 𝑎𝑡𝑚)(𝑥)
(2 𝑥)
P2 = 1 atm
Interpretation:
Since the pressure and volume are inversely proportional to each other, any
changes to the volume could cause pressure to change as well. When the volume
increases twice as much its original value, the pressure decreases half as much as its
original value.
11. 2. What is the pressure required to compress 3, 500 L of hydrogen gas at 2.0
atm into a 50-L tank if the temperature is kept constant?
12. 2. What is the pressure required to compress 3, 500 L of hydrogen gas at 2.0
atm into a 50-L tank if the temperature is kept constant?
Given:
P1 = 2.0 atm V1 = 3, 500L
P2 = ? V2 = 50 L
P1V1 = P2V2
Solution:
P2 =
(𝑃1)(𝑉1)
(𝑉2)
P2 =
(2.0 𝑎𝑡𝑚)(3,500𝐿)
(50 𝐿)
= 140 atm
13. 3. A gas occupies a volume of 250 ml at 300mmHg of pressure. What will be its
volume if the pressure is doubled? Assume that the temperature is constant.
1 atm = 760 mmHg
P1V1 = P2V2
Given:
P1 = 300mmHg V1 = 250 mL
P2 = doubled (300mmHg x 2)= 600mmHg V2 = ?
Solution:
V2 =
(300𝑚𝑚𝐻𝑔)(250𝑚𝐿)
(600𝑚𝑚𝐻𝑔)
= 125mL
