Finding Relationship Information from Graph (direct proportionality)

[MohammadG]

First of all sorry if this is the wrong section!

For a physics assignment, I am given a graph of which I am to gather data from.

Graph 1:
In an experiment to investigate how the celcius temperature, T, affects the volume, V, of an enclosed sample of air at a constant pressure, a students graphed results appear as shown below.

Questions:

(1) Is V directly proportional to T?
(2) Is ΔV directly proportional to ΔT?
(3) From the graph, determine an expression for V in terms of T.
(4) Hence determine the value of T when V = 0.

My thoughts:
(1) V is not directly proportional to T as the graph does not pass through the origin (0,0). This graph represents linear dependence.

(2) ΔV is directly proportional to ΔT as a straight line is produced. If you double V, T will double. If you half V, T will half...etc.

(3) I'm kind of stuck for this one. I don't know how to determine as expression. I know that when V = 80, T = 0. So if I use the equation of a straight line, I get y = mx + c,

80= mx + 0
m = -80?

which seems wrong to me, I'm probably approaching it in the wrong way. I don't know what else to say. Is V is not directly proportional to T, the statement I need.

(4) For this I would obviously need to do question 3. Then plug in V = 0 and solve for T. I would then use extrapolation to show the value of T when V = 0.

So that's what I need help with! If you could help me with question 3 I would be so greatful! Also could you please check my logic with questions 1 and 2.

Thank you!

 

[cepheid]

Welcome to PF!

y is the *dependent variable*, which is volume in this case.

x is the *independent variable*, which is temperature in this case.

Therefore, V = mT + C, where m is the slope (in cubic metres PER degree Celsius) and c is the y-intercept (i.e. it is the volume occupied when T = 0).

when T = 0, V = 80

80 = m*0 + c

c = 80

V = mT + 80

What is the slope of the line?

 

[rock.freak667]

(3) I'm kind of stuck for this one. I don't know how to determine as expression. I know that when V = 80, T = 0. So if I use the equation of a straight line, I get y = mx + c,

80= mx + 0
m = -80?

Your equation would be V=mT+C and for V=80 when T=0 you would have

80 = m(0)+C

so m is not zero.

 

[MohammadG]

Welcome to PF!

y is the *dependent variable*, which is volume in this case.

x is the *independent variable*, which is temperature in this case.

Therefore, V = mT + C, where m is the slope (in cubic metres PER degree Celsius) and c is the y-intercept (i.e. it is the volume occupied when T = 0).

when T = 0, V = 80

80 = m*0 + c

c = 80

V = mT + 80

What is the slope of the line?

Okay I see how you go that. No slope was given in the data and there is no way of determining what other points lie on the line.

So the only expression I can produce (without having slope) would be:
V = mT + 80, correct?

Your equation would be V=mT+C and for V=80 when T=0 you would have

80 = m(0)+C

so m is not zero.

So for question 4, what would be the value of T when V = 0? Would it be -80?

 

[cepheid]

Okay I see how you go that. No slope was given in the data and there is no way of determining what other points lie on the line.

So the only expression I can produce (without having slope) would be:
V = mT + 80, correct?

Hmm?

You have the line right there in front of you. Why can't you determine the slope visually? Hint: what is the definition of slope?

So for question 4, what would be the value of T when V = 0? Would it be -80?

You can easily answer #4 once you've answered #3, which gives you an expression for volume in terms of temperature. With this expression, you can determine the volume at any temperature and vice versa.

 

[MohammadG]

Hmm?

You have the line right there in front of you. Why can't you determine the slope visually? Hint: what is the definition of slope?

I must be missing something then. Sorry!
m = y2 - y1/x2-x1

or

rise/run

But I have no points besides (0, 80)?

The slope seems to be something like 0.4(cubic metres per degree celsius) visually. What other way is there of finding the slope? Sorry again if I'm missing something obvious.

 

[cepheid]

I must be missing something then. Sorry!
m = y2 - y1/x2-x1

or

rise/run

But I have no points besides (0, 80)?

The slope seems to be something like 0.4(cubic metres per degree celsius) visually. What other way is there of finding the slope? Sorry again if I'm missing something obvious.

Yes, slope is rise over run.

Estimate it from the graph! Draw tick marks at 1/4, 1/2, and 3/4 of the way between 80 cm^3 and 120 cm^3. Maybe even draw tick marks at intervals of 1/8 the distance between the two if that helps you measure it better. This way you can get an approximate visual estimate of what the volume is at T = 100°C

EDIT: Also draw a straight vertical line upward from T = 100°C on the x-axis to the best-fit line. Then draw a horizontal line from that intersection point to the y-axis. Read off the approximate volume there using your extra tick marks.

 

[MohammadG]

Yes, slope is rise over run.

Estimate it from the graph! Draw tick marks at 1/4, 1/2, and 3/4 of the way between 80 cm^3 and 120 cm^3. Maybe even draw tick marks at intervals of 1/8 the distance between the two if that helps you measure it better. This way you can get an approximate visual estimate of what the volume is at T = 100°C

EDIT: Also draw a straight vertical line upward from T = 100°C on the x-axis to the best-fit line. Then draw a horizontal line from that intersection point to the y-axis. Read off the approximate volume there using your extra tick marks.

Cepheid, Thanks you so much for your continued support!

I've managed to make an approximate equation for the graph. My slope ended up being 0.32.

I've also answered question 4 and have gotten an answer of V= -250 when T = 0.
Which is appropriate.

One last question, as the tick mark method isn't entirely accurate we must specify this in the answer, correct? We say the V ≈ -250 when T = 0, right?

Thanks again! :D

 

[cepheid]

Cepheid, Thanks you so much for your continued support!

I've managed to make an approximate equation for the graph. My slope ended up being 0.32.

I've also answered question 4 and have gotten an answer of V= -250 when T = 0.
Which is appropriate.

I'm going to assume that you meant T = -250 when V = 0.

It's *sort of* appropriate, I mean, you'd expect 0 pressure and volume at absolute zero (for an ideal gas), which is 0 K. Your answer is more like 23 K. (Since 0 K = -273°C). But this can be attributed to measurement error.

One last question, as the tick mark method isn't entirely accurate we must specify this in the answer, correct? We say the V ≈ -250 when T = 0, right?

Thanks again! :D

It should be pretty clear from the context (especially to the person who assigned the problem in the first place) that you cannot possibly arrive at an exact value for the slope. However, if you want, you can use the "approximately equal to" sign. The math notation police aren't going to come and arrest you if you don't.

 

[MohammadG]

I'm going to assume that you meant T = -250 when V = 0.

It's *sort of* appropriate, I mean, you'd expect 0 pressure and volume at absolute zero (for an ideal gas), which is 0 K. Your answer is more like 23 K. (Since 0 K = -273°C). But this can be attributed to measurement error.


It should be pretty clear from the context (especially to the person who assigned the problem in the first place) that you cannot possibly arrive at an exact value for the slope. However, if you want, you can use the "approximately equal to" sign. The math notation police aren't going to come and arrest you if you don't.

After interpolating the graph again I got a more acceptable answer of -270. Which is 5 degrees off absolute zero. I've also used the approximately equal sign in my final calculation.

Thank again!