Calculating percentage change in volume at constant pressure.

[waqaszeb]

Homework Statement

The coefficient of the thermal expansion of ethanol is given by:

\alpha (per degree Celsius) = 1.0414 x 10^(-3) + 1.5672 x 10^(-6) t + 5.148 X 10^(-8) t^2

Calculate the percentage change in volume when ethanol is heated from 0 to 100 (celsius) at constant pressure.

Homework Equations

alpha = (1/V)(partial(V)/(partial(T))(at constant P)

The Attempt at a Solution

I simply rearranged the above equation and integrated with limits T1 and T2:

delta(V) = (alpha/2)(100)^2 = 5.123..something but the answer is 13.8 %

Can someone also comment on the units of alpha? Its confusing because the units t and t^2 are included. From a mathematical point of view, we can't just add them..( or that's what i think at least).

 

[rude man]

Homework Statement

The coefficient of the thermal expansion of ethanol is given by:

\alpha (per degree Celsius) = 1.0414 x 10^(-3) + 1.5672 x 10^(-6) t + 5.148 X 10^(-8) t^2

Calculate the percentage change in volume when ethanol is heated from 0 to 100 (celsius) at constant pressure.

Homework Equations

alpha = (1/V)(partial(V)/(partial(T))(at constant P)

The Attempt at a Solution

I simply rearranged the above equation and integrated with limits T1 and T2:

delta(V) = (alpha/2)(100)^2 = 5.123..something but the answer is 13.8 %

Can someone also comment on the units of alpha? Its confusing because the units t and t^2 are included. From a mathematical point of view, we can't just add them..( or that's what i think at least).

You're wise to question units (dimensions). Too few people do that, including a former PhD boss of mine! Yet it's a very powerful way to check your computations as you go along.

The answer of course is that the t coefficient has units of K^-1and the t² coefficient has units of K^-2. All terms must be dimensionless as is α.

Just out of curiosity, why did you use t instead of T in your expression for α?

You need to integrate to get ΔV/V over 0 to 100 deg C.

 

[waqaszeb]

You're wise to question units (dimensions). Too few people do that, including a former PhD boss of mine! Yet it's a very powerful way to check your computations as you go along.

The answer of course is that the t coefficient has units of K^-1and the t² coefficient has units of K^-2. All terms must be dimensionless as is α.

Just out of curiosity, why did you use t instead of T in your expression for α?

You need to integrate to get ΔV/V over 0 to 100 deg C.

Hey, thanks for you help! I got ΔV/V = (alpha)(T dt), which I integrated from 0 to 100. The answer I get now is 20.86..I'm not sure what I'm doing wrong..

As for the units of alpha..I don't know why its t and not T. This is a question from Chemical Thermodynamics by Peter A. Rock.

 

[rude man]

Hey, thanks for you help! I got ΔV/V = (alpha)(T dt), which I integrated from 0 to 100. The answer I get now is 20.86..I'm not sure what I'm doing wrong..

I don't see your integral. What is it?

As for the units of alpha..I don't know why its t and not T. This is a question from Chemical Thermodynamics by Peter A. Rock.

I assume t is in Kelvin, not centigrade. Double-check this!

 

[waqaszeb]

I don't see your integral. What is it?



I assume t is in Kelvin, not centigrade. Double-check this!

I integrated both (ΔV/V)= (alpha) dt [from 0 to 100] = alpha (100) = wrong answer
and (ΔV/V) = (alpha)(T) dt [ from 0 to 100) = alpha x 2 x ( 100^2 - 0^2 ) = 20.86

the answer in the back of my textbook says the answer is 13.8 %

 

[rude man]

V/V. What was your expression for Δ

I integrated both (ΔV/V)= (alpha) dt [from 0 to 100] = alpha (100) = wrong answer
the answer in the back of my textbook says the answer is 13.8 %

OK, I got the same answer (13.8%) if I assume t is in deg C.

You did right by setting ΔV/V = αΔt or, let's be sophisticated physicists, dV/V = αdt. I need to see your work step-by-step to see what the error was you made, because you took the right tack in integrating this equation.

 

[waqaszeb]

Here's what I did:

dv/V = \alpha dt = \alpha T (100-0) [ after integration ] = \alpha (100-0) = \alpha (100) = 10.43 for me..

I added the three quantities of alpha and then multiplied by 100. I also multiplied each individual quantity to 100 and then added them together. the answer is still wrong..

 

[rude man]

Here's what I did:

dv/V = \alpha dt = \alpha T (100-0) [ after integration ] = \alpha (100-0) = \alpha (100) = 10.43 for me..

I added the three quantities of alpha and then multiplied by 100. I also multiplied each individual quantity to 100 and then added them together. the answer is still wrong..

dV/V is a differential fractional increase in volume. You did not compute the total fractional change in volume.

And you did not integrate α(t)dt from 0 to 100C either.

Integrate both sides of this equation correctly & what do you get?

 

[waqaszeb]

Got it. Thank you so much!