Question

Can someone work out a part of a problem for me?

ln(1.389)=4726.97 (1/T1 -2.846x10^-3)

Answer 1

@agent0smith
@cowgirl1
@ganeshie8
@Rushwr
@whpalmer4

Answer 2

I can give the full problem if needed but I can solve the problem until to this point.

Answer 3

Are you needing the full problem?

Answer 4

At what temperature does ethanol boil on a day in the mountains when the barometric pressure is 547 mmHg? The heat of vaporization of ethanol is 39.3 kJ/mol and its normal boiling point is 78.3°C.(R = 8.314J/ K • mol)

Answer 5

@Rushwr In case you needed it.

Answer 6

@Photon336

Answer 7

I can get about half way there then I get stuck.

Answer 8

Boiling happens when the vapor pressure is the same as the external pressure.

Answer 9

Right.

Answer 10

I get that. It's the math I'm having issues with. Bleh.

Answer 11

normal boiling point is at 760 mmHG or 1 atm at 78.3°C

Answer 12

I'll show how far along I can get.

Answer 13

ln(760/547mmHg) =(3.93x10^3 J/mol)/(8.314 J/K*mol) (/T1 - 1/351.3)
ln(1.389) = 472.97 (T1 -2.846 x 10^-3)

Answer 14

At this point I get stuck

Answer 15

oh I see

\[\ln(\frac{ p }{ p_{0} }) = \frac{ \delta H }{ R }(\frac{ 1 }{ T_{1} }-\frac{ 1 }{ T_{2} })\]

Answer 16

Yeah. Sorry. I'm not good at putting it on here.

Answer 17

You need to solve for T_{2}

Answer 18

i would re-write the equation and solve for T_{2} and then plug in your numbers. YOU should always rewrite the equation first then plug in your values.



whenever we have a logarithm where two numbers are divided we can re-write it as shown below. so

Ln(x/y) = ln(x)-ln(y)

\[Ln(p_2)-Ln(p_{1}) = (\frac{ \delta H }{ R })(\frac{ 1 }{ T_{1} }-\frac{ 1 }{T_{2} })\]

our next step is to isolate T_2 The temperature we're looking for. we can divide both sides by delta H/R

This is what we get.

\[\frac{ R(\ln(p_{2})-\ln(p_{1}) }{ \Delta H } = \frac{ 1 }{ T_{1} }-\frac{ 1 }{ T_{2} }\]

now we need to find out what 1/T_{2} is that's the temperature right? so we isolate that by subtracting 1/T1 from both sides.

\[\frac{ 1 }{ T_{1} }-\frac{ R(\ln(p_{2})-\ln(p_{1}) }{ \Delta H } = \frac{ 1 }{ T_{2} }\]

then you would get 1/T_{2} BUT note that we can take the reciprocal of that and get T_{2}

\[\frac{ 1 }{ T_{2} } = (\frac{ 1 }{ T_{2} })^{=1} = T_{2}\]

Answer 19

Can you show how to do it without rewriting the formula? We were not taught to do it that way so it's kinda confusing.

Answer 20

\[\ln (760/547) = 39300/8.314 (1/T1 - 1/351.3)\]

Answer 21

\[\frac{ 1 }{ T_{1} } - \frac{ R(\ln(p2)−\ln(p1)}{ ΔH } = \frac{ 1 }{ T_{2} }\]

Answer 22

\[\ln (1.389) = 4726.97 (\frac{ 1 }{ T1 }-2.846x10^{-3})\]

Answer 23

@staldk3 my point was that so you can just plug in the numbers where the variables are in the end so that you don't get confused. once you've solved for your variable it becomes eaiser

Answer 24

The part above is where I'm lost. I can get to that part at least.

Answer 25

So put everything together first:

P_0 = 760 atm
T_1 = 351K
p_2 = 547 atm
T_2 = ?


We don't know what T_2 is so we need to solve our equation to find it.
That's what I did above.

I used to do it the way you did but you can easily get confused. If you solve for the variable you need first you can just plug in all your numbers at the end.

which I did here



\[\frac{ 1 }{ 351K } - \frac{ 8.314*\ln(547)-\ln(760) }{ 39300 } = (0.0017)^{-1} = 594K\]

Answer 26

But I solved the equation for 1/T2 step by step. the most important thing is to take a look at how I did each step.

Answer 27

Why is it 547 - 760?

Answer 28

And how did it go from (.0017)^-1 to 594K?

Answer 29

remember it's Ln(p2)-Ln(p1)

pressure 2 is 560 that's the pressure that we're at right?
and pressure 1 is 760 because that's the original pressure

Answer 30

Okay. Right

Answer 31

@staldk3 when we solve this equation we get 1/T2 not T2 that's why we need to take take the recrirpocal.

\[(1/T_{2}) = (1/T_{2})^{-1} = T_{2}\]

Answer 32

I see.

Answer 33

So, remember go over my derivation step by step on your own

Answer 34

and then after plug in the numbers.

Answer 35

Alright. I will

Answer 36

you'll find that once you know what you need to solve for FIRST. and you solve the equation for that variable you can easily keep track of what's going on.

Answer 37

Ya. I was doing fine with the other problems like this but for some reason this one tripped me up

Answer 38

it's the math

Answer 39

Yeah.

Answer 40

It's always the math

Answer 41

like the more complicated the formula, you have to do the algebra and manipulations first. if the formula was easier you could just plug in your numbers.

Answer 42

Yeah. I prefer plug and chug.

Answer 43

try going over this again and if you have questions tag me.

Answer 44

Alright. Thanks for the help

Answer 45

Let me reinforce what @Photon336 said: much, much better to rearrange the formula before you plug in your numbers. You are really asking for trouble if you start plugging in numbers first.

Answer 46

i have to admit that is true-I sucked at doing formulas last semester because of that !

Answer 47

yeah @cowgirl1 I used to plug in the numbers first. Until I took gen physics I in college.

Answer 48

yeah I did a one on one with my teacher and she was like no wonder you keep screwing them up...lol...this semester has been much easier