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Boyle's Law Data Anaylsis
Boyle's law is one of the gas laws and basis of derivation for the Ideal gas law, which describes relationship between the product pressure and volume within a closed system as constant when temperature remains at a fixed measure; both entities remain inversely proportional. The law was named for chemist and physicist, Robert Boyle who published the original law in 1662. The law itself can be defined as:
"For a fixed amount of gas kept at a fixed temperature, P and V are inversely proportional."
Wikipedia.com
The mathematical equation for Boyle's law is:
PV = K
where:
P is the pressure of the gas,V is the volume of the gas,
K is a constant value representative of the pressure and volume of the system.
The data for this study is Robert Boyle's original data from 1662 and is taken form the Classic Chemistry Papers at //web.lemoyne.edu/~GIUNTA/index.html
Excel was used to graph Boyle's data as a scatter plot of volume vs. pressure. Excel also calculated the trend line, power regression was choosen for the best fit curve. The exponent on this equation is very close to negative one, this illustrates mathematically an inverse relationship. Anything raised to the power of negative one equals its reciprocal (x-1 = 1/x).
Graphing volume vs. the inverse of pressure (1/P) should result in a linear trend line. This illustrates that as the volume approaches zero Boyle's law would predict that the inverse of pressure would approach zero.
As is illustrated by the graphs as Pressure increases the volume decreases; the pressure and volume of a gas sample at a given temperature are inversely proportional to each other. The law assumes that the temperature remains constant; so with V1 = original volume, V2 = new volume, P1 = original pressure and P2 = new pressure Boyle's law can also be expressed as:
V1 = P1 V1 P2 | V1 P1 = V2 P2 |
According to Boyle’s law, the shape of the graph between pressure and reciprocal of volume is _______.
According to Boyle’s law, the shape of the graph between pressure and reciprocal of volume is Straight line.
Concept: Fundamental Laws of Gases - Pressure and Volume Relationship or Bolye's Law
Is there an error in this question or solution?
31st Oct 2019 @ 2 min read
The graph of Boyle's law is known as pressure-volume graph or PV curve. It is as follows:
As observed from the graph above, pressure increases with a decrease in volume, and vice versa. Thus, pressure is inversely proportional to volume. Other parameters (temperature and amount of gas) are constant in the graph above.
Mathematical explanation
Volume is on the x-axis and pressure, on the y-axis. The equation of the curve is PV = k, which is the equation of Boyle's law. The curve is hyperbolic in nature having two asymptotes: P = 0 (horizontal) and V = 0 (vertical).
Note: An asymptote is a line or curve such that the distance between it and a given curve tends to zero as x and/or y coordinates tends to infinity.
As volume tends to positive infinity, pressure tends to zero, and we get the horizontal asymptote, P = 0.
When volume approaches zero, pressure approaches infinity, and it results in the vertical asymptote, V = 0.
Graphs at different temperatures
Graphs of Boyle's law can be plotted at different temperatures. Each curve in the graphs below is at a constant temperature and such curves are called isotherms.
The above graph is a pressure-volume graph plotted at three different temperatures (T1, T2, and T3). As observed from the graph, with an increase in temperature, curves shift upwards. This is because of increase in the value of k.
The graph of pressure vs inverse volume is a straight line passing through the origin and having the positive slope, k.
The graph above is a straight line parallel to the x-axis. This proves the product of pressure and volume at a constant temperature and amount of gas is constant. The lines in the graph are independent of volume (or pressure).
Logarithmic graphs
The equation of Boyle's law is PV = k. Taking the logarithm to both sides.
The plots are a straight line with the y-intercept of log k.
Associated articles
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"Graphs of Boyle's Law" ChemistryGod, 31st Oct 2019, //chemistrygod.com/boyle-law-graphThanks for your response!
Abdullah
19th Apr 2022
👍👍
John Dalton
28th Jan 2022
Awesome!
Robert Boyle
20th Jan 2022
man what u guys are doing is phenomenal .
and P = pressure of the gas So at constant temperature, if the volume of a gas is doubled, its pressure is halved. PV = constant PV = k PiVi = PfVf Pi is the initial (original) pressure (a) Pi and Pf must be in the same units of measurement (eg, both in atmospheres or both in kPa) (b) A Real Gas is one which approaches Boyle's Law behaviour as the temperature is raised or the pressure lowered. Please do not block ads on this website. Consider an experiment in which a known amount of hydrogen gas in a syringe has a volume of 23 mL at atmospheric pressure (760 mm Hg or 1 atm or 101.3 kPa). You then apply an external pressure of 912 mm Hg (1.2 atmospheres or 121.6 kPa) by pressing down on the plunger in the syringe. The volume of hydrogen gas is then recorded as 19.2 mL. You continue to apply external pressure by pushing the plunger down further, recording the volume of hydrogen gas as shown in the table below: Decreasing the applied pressure increases the volume of the gas. At constant temperature, the volume of a given quantity of gas is inversely proportional to its pressure : where V = volume of the gas
At constant temperature for a given quantity of gas, the product of its volume and its pressure is a constant :
OR
where k is a constant
Vi is its initial (original) volume
Pf is its final pressure
Vf is its final volume
(b) Vi and Vf must be in the same units of measurement (eg, both in litres or both in millilitres) (a) A hypothetical gas which obeys Boyle's Law at all temperatures and pressures is called an Ideal Gas.
No ads = no money for us = no free stuff for you!Graphical Representations of Boyle's Law
Pressure
(mm Hg)* Volume
(mL) Trend 760 23 Increasing the pressure applied to the plunger causes a reduction in the gas volume. 912 19.2 1064 16.4 1216 14.4 1368 12.8 1520 11.5 * A pressure of 760 mm Hg is equal to 1 atmosphere (atm) or 101.3 kilopascals (kPa)
If we plot these points on a graph, the graph looks like the one below:
volume (mL) | Gas Volume versus Pressure Pressure (mm Hg) |
Note that this is not a linear relationship, the line in the graph is curved, it is not a straight line.
But look what happens if we multiply volume and pressure (P × V):
760 | 23 | 1.75 × 104 | P × V is a constant! For this amount of gas at this temperature: P × V = 1.75 × 104 |
912 | 19.2 | 1.75 × 104 | |
1064 | 16.4 | 1.75 × 104 | |
1216 | 14.4 | 1.75 × 104 | |
1368 | 12.8 | 1.75 × 104 | |
1520 | 11.5 | 1.75 × 104 |
For a given amount of gas at constant temperature we now we can write the equation:
P × V = constant
If we divide both sides of the equation by P, we get:
Recall that the equation for a straight line that runs through the point (0,0) is
y = mx
where m is the slope (or gradient) of the line
Then a graph of V against 1/P, should be a straight line with a slope (or gradient) equal to the value of the constant.
The table below shows what happens if we calculate 1/P for each volume, V, in the experiment above and then graph the results:
11.5 | 1520 | 6.6 × 10-4 | As gas volume (V) increases, the value of 1/P increases. As gas volume (V) decreases, the value of 1/P decreases. |
12.8 | 1368 | 7.3 × 10-4 | |
14.4 | 1216 | 8.2 × 10-4 | |
16.4 | 1064 | 9.4 × 10-4 | |
19.2 | 912 | 1.1 × 10-3 | |
23 | 760 | 1.3 × 10-3 |
By plotting these points on a graph, we can see that the relationship is linear:
volume (mL) | Gas Volume versus 1/Pressure 1/Pressure (1/mm Hg) |
We now have a simple method for determining the value of the constant:
Recall that we can calculate the slope (gradient, m) of a straight line using two points on the line
Choosing the points (0.00094,16.4) and (0.0013,23)
m = (23 - 16.4)
(0.0013 - 0.00094)= (6.6)
(0.00036)= 1.8 × 104
and the equation for this straight line is
This equation then allows us to calculate the volume of the gas at any pressure, as long as we use the same amount of gas and keep the temperature the same.
Let us say we have a specific amount of gas and keep the temperature constant, then initially at pressure Pi the gas has a volume of Vi and we know that:
PiVi = constant
If we maintain the same temperature and the same amount of gas, but change the pressure to Pf, then the new gas volume will be Vf, and
PfVf = the same constant
So, as we long as we use the same amount of gas at the same temperature:
PiVi = constant = PfVfthat is:
PiVi = PfVf
This means that if we know the initial conditions (Pi and Vi), and, we know the final pressure (Pf), we can calculate the final volume (Vf):
or we can calculate the final pressure (Pf) if we know the final volume (Vf):
Similarly, if we know the final conditions (Pf and Vf), and, we know the initial pressure (Pi), we can calculate the initial volume (Vi):
or we can calculate the initial pressure (Pi) if we know the initial volume (Vi):
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Question : A certain mass of gas occupies a volume of 2.5 L at 90 kPa pressure.
What pressure would the gas exert if it were placed in a 10.0 L container at the same temperature?
Solution:
(Based on the StoPGoPS approach to problem solving.)
- What is the question asking you to do?
Calculate final gas pressure
Pf = ? kPa - What data (information) have you been given in the question?
Extract the data from the question:
Conditions of the experiment: constant amount of gas at constant tempertaure.
Vi = initial gas volume = 2.5 L
Pi = initial gas pressure = 90 kPa
Vf = final gas volume = 10.0 L - What is the relationship between what you know and what you need to find out?
Because the amount of gas and temperature are constant, we can use Boyle's Law:
PiVi = PfVf
Rearrange this equation by dividing both sides by Vf:
PfVf
Vf= PiVi
VfPf = PiVi
Vf - Substitute in the values and solve for Pf
Pf = PiVi
Vf= 90 × 2.5
10.0= 22.5 kPa - Is your answer plausible?
Consider that the volume has increased from 2.5 to 10, an increase of 10/2.5 = 4.
If the volume increases 4 times, then the pressure must decrease and the new pressure will be 1/4 of the initial pressure, that is, 1/4 × 90 = 22.5 kPa
Since this is the same value as we calculated above, we are reasonably confident that our answer is correct. - State your solution to the problem "final gas pressure":
Pf = 22.5 kPa
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Question : 4.5 L of gas at 125 kPa is expanded at constant temperature until the pressure is 75 kPa.
What is the final volume of the gas?
Solution:
(Based on the StoPGoPS approach to problem solving.)
- What is the question asking you to do?
Calculate final volume
Vf = ? L - What data (information) have you been given in the question?
Extract the data from the question:
Conditions: constant amount of gas and temperature.
Pi = 125 kPa
Vi = 4.5 L
Pf = 75 kPa - What is the relationship between what you know and what you need to find out?
Because the amount of gas and temperature are constant, we can use Boyle's Law:
PiVi = PfVf
Rearrange this equation by dividing both sides by Pf:
PfVf
Pf= PiVi
PfVf = PiVi
Pf - Substitute in the values and solve for Vf
Vf = PiVi
Pf= 125 × 4.5
75= 7.5 L - Is your answer plausible?
Pressure has decreased by a factor of 75/125 = 0.6 Therefore volume must have increased by a factor of 1/0.6 = 1.667 Final volume must be 1.667 × 4.5 = 7.5 L
Since this value is the same as the one we calculated above, we are reasonably confident that our answer is plausible.
- State your solution to the problem "final gas volume ":
Vf = 7.5 L
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