Using data (Pressure vs. density) to find molecular mass .

In summary, the gas had an unknown molecular mass and the student attempted to solve the problem using Ideal Gas Law. However, the data did not appear to be in a linear fashion and the slope obtained from the linear fit was not accurate. The student then used Kaleidograph to plot the data and found that the slope obtained was 1.8243 which gave the gas a molecular mass of 44.9095.
  • #1
terp.asessed
127
3

Homework Statement


Use the data:
P/bar
a: 0.1000
b: 0.5000
c: 1.000
d: 1.01325
e: 2.000
vs.
density/(g/L)
a: 0.1771
b: 0.8909
c: 1.796
d: 1.820
e: 3.652...for an unknown gas at 300K to determine the molecular mass of the gas...

Homework Equations


PV = nRT (b/c of very low pressure)

The Attempt at a Solution


..so, first I plotted the data on a graph, and it seemed like a linear line...considering Pressure is very low, I assumed ideal gas law, as in PV = nRT where n = mass/(molecular mass):

PV = (mass/mm)RT --> mm = (mass/V)(RT/P) = (density)*RT/P...so, I chose one middle data point (c), where:

density = 1.796 g/L
P = 1bar= 0.9869 atm
in addition to given values:
R = 0.082058 L*atm/(mol*K)
T= 300K

and I got mm = 44.80g/mol, which is not the same as 44.10 on the back of the textbook...could someone hint where I made the mistake?
 
Physics news on Phys.org
  • #2
terp.asessed said:

Homework Statement


Use the data:
P/bar
a: 0.1000
b: 0.5000
c: 1.000
d: 1.01325
e: 2.000
vs.
density/(g/L)
a: 0.1771
b: 0.8909
c: 1.796
d: 1.820
e: 3.652


...for an unknown gas at 300K to determine the molecular mass of the gas...

Homework Equations


PV = nRT (b/c of very low pressure)

The Attempt at a Solution


..so, first I plotted the data on a graph, and it seemed like a linear line...considering Pressure is very low, I assumed ideal gas law, as in PV = nRT where n = mass/(molecular mass):

PV = (mass/mm)RT --> mm = (mass/V)(RT/P) = (density)*RT/P...so, I chose one middle data point (c), where:

density = 1.796 g/L
P = 1bar= 0.9869 atm
in addition to given values:
R = 0.082058 L*atm/(mol*K)
T= 300K

and I got mm = 44.80g/mol, which is not the same as 44.10 on the back of the textbook...could someone hint where I made the mistake?

What you did looks very good. BUT... You made an important assumption when you constructed the plot. For a real gas, where is this approximation likely to be best? Can you extrapolate the data back to this point?
 
  • #3
Wait, I thought this problem was related to ideal gas? I am sorry, but how am I to apply real gas in the question? Does it mean I have to find a 'general' equation as in y = mx +b?
 
  • #4
terp.asessed said:
Wait, I thought this problem was related to ideal gas? I am sorry, but how am I to apply real gas in the question? Does it mean I have to find a 'general' equation as in y = mx +b?

I think that I assumed that you plotted something that you did not.

What happens when you plot MM (just like you calculated) versus Pressure for all of the data? Do that first, and share the graph with us, and we will think about what to do next.
 
  • #5
This is supposed to be real experimental data, so there is supposed to be some uncertainty involved. So no single data point should deliver the exact value of the molar mass. But, if you plot the density as a function of the pressure, the slope of the line you get should be your best estimate of mm/RT. The best straight line you plot should not pass through each and every data point. From the slope of this line, what value do you get for the molar mass?

Incidentally, include the additional data point 0,0.

Chet
 
Last edited:
  • #6
Hello I did what you all have suggested, and got a slope of value 1.8145 = mm/RT from the graph of density as a function of pressure, but only got mm of 44.668, NOT 44.10...

1.8145 = mm/(RT) = mm/[(0.082058)(300K)]
mm = 44.668

So, I tried for another slope W/OUT an additional data point (0,0) and got slope of 1.8243, which gave mm = 44.9095...

Where am I making mistake?
 
  • #7
terp.asessed said:
Hello I did what you all have suggested, and got a slope of value 1.8145 = mm/RT from the graph of density as a function of pressure, but only got mm of 44.668, NOT 44.10...

1.8145 = mm/(RT) = mm/[(0.082058)(300K)]
mm = 44.668

So, I tried for another slope W/OUT an additional data point (0,0) and got slope of 1.8243, which gave mm = 44.9095...

Where am I making mistake?
I had Kaleidograph do a linear fit to determine the slope with (0,0) included, and I got 1.8243. It looks like you did everything correctly.

But, when I used this value in the formula, I got mm=45.5, rather than 44.9. I think you may have neglected the difference between bars and atm.

I wouldn't worry about this too much. Your methodology is correct. That's all that really matters. We have no idea how they got their result in the back of your textbook. I have confidence in what you did.

Chet
 
  • Like
Likes terp.asessed
  • #8
Thank you very much! Knowing that not only me but also you got the value makes me feel better :)
 

1. How do you determine molecular mass using pressure and density data?

The molecular mass can be calculated by using the ideal gas law equation, PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is the temperature. By rearranging the equation to solve for n, and then dividing the mass of the gas by the number of moles, the molecular mass can be obtained.

2. What type of data is needed to find molecular mass using pressure and density?

The two main types of data needed are pressure and density. These can be measured using specialized equipment such as a manometer and a densitometer. Additionally, the temperature of the gas must also be known.

3. Can any gas be used to calculate molecular mass using this method?

Yes, this method can be used for any gas that behaves according to the ideal gas law. This includes most gases at standard temperatures and pressures, but may not be applicable to high pressure or low temperature gases.

4. Are there any limitations to using pressure and density data to find molecular mass?

There can be limitations to this method if the gas does not behave according to the ideal gas law, or if the pressure and density measurements are not accurate. Additionally, this method may not be suitable for mixtures of gases.

5. How can the accuracy of the calculated molecular mass be ensured?

The accuracy of the calculated molecular mass can be ensured by using precise equipment to measure pressure and density, and by taking multiple measurements to account for any errors. It is also important to use appropriate units and to ensure that the gas is behaving according to the ideal gas law.

Similar threads

  • Biology and Chemistry Homework Help
Replies
7
Views
2K
  • Classical Physics
Replies
8
Views
2K
  • Biology and Chemistry Homework Help
Replies
3
Views
2K
  • Biology and Chemistry Homework Help
Replies
2
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
14
Views
1K
  • Biology and Chemistry Homework Help
Replies
11
Views
3K
Replies
19
Views
1K
  • Biology and Chemistry Homework Help
Replies
6
Views
8K
  • Biology and Chemistry Homework Help
Replies
1
Views
2K
  • Biology and Chemistry Homework Help
Replies
4
Views
4K
Back
Top