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PV=nRT (This is so hard... for me) Chemistry

FrankP

By FrankP
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FrankP

FrankP

Posted
Posted

As usual, I am having a real tough time with this...

 

 

I have attached my work I have done and the questions.

 

PV=nRT where R is a constant @ ((.08026 L) x (atm)/ (K) x (mol) )

 

 

  • So if I could air out my thought process here to so you guys could help me determine what I am doing wrong in my thinking. I am looking at the column of air in which the bubble passes as a prism so I can gather the volume. I know that the density of both H20 and Hg are given and are 1.025g/cm3 & 13.6g/cm3 respectively.
  • So now that I have density I can gather some information in terms of D=(Mass/Volume)

Therefore:

  • H20 ((1.025g)/1cm3) or ((1.025g)/1mL)
  • ((13.6g)/1cm3) or ((13.6g)/1mL)

Using that to further explain the column of space that the bubble travels you want to find total moles of H20 present in that space. So finding the overall volume of space is necessary.

 

I am not sure about Volume this is the 2 things I have tried:

  1. I looked to find V in terms of 100 ft3 because it is unreasonable to assume that the space which the bubble passes through is 2D (at least that's how my brain thinks). 100 ft3 = 2.832 x 106 cm3 and using that information you can gather total moles of H20 contained in that area.
  2. Use the density as a conversion factor as stated above. Where: D(H20) => ((1.025g)/1cm3) or ((1.025g)/1mL). If you take that factor and multiply it to get moles per liter in doing so take the conversion factor and plug it in like this (image attached #2) it's a conversion ratio between grams of H2O in the conversion => Liters => Divide by molecular weight of water. This should get you # of moles per liter. Then you need to find out how many moles are present without the liters. So you end up with .0569mol H2O

 

 

OK... so now I am done with my ability to rationalize the answer. The reason I find myself so confused is that we are not given a size of the bubble or temperature. I want to be able to say that temperature should be consistent even though I know there would most likely be negligible temperature difference in terms of this. I know the temperature has to be greater than freezing. However, I'm not sure if there is a specific temp that is assumed for problems like this in the value of water. If that is the case I am making this harder...

 

The second point of confusion is the relevance to Hg Mercury as ((13.6g)/1cm3) I know that barometers are usually measured in units of mercury as a standard and so I know that I need to use that to eventually help me solve the problem but how?

 

 

Thanks as usual

 

 

 

post-54062-0-49757300-1479324857_thumb.jpg

Curiatron

Curiatron

Posted
Posted

Interesting problem. Here is my contribution. Hopefully others can critique me too:

 

1. Problem statement says to assume that none of the air dissolves into the water. Therefore, does the quantity change? Do you then really need to know how many moles there are?

 

2. Re. your question on knowledge of temperature - look at the left and right hand sides of the ideal gas law equation (PV=nRT). Does a change in pressure and/or volume (left hand side) require a corresponding change to the quantity and/or temperature of the gas (right hand side)? Is there a way that you can keep the right hand side the same while changing the P and V values on the left hand side? For example, if we said the right hand side (nRT) is 6, are there different combinations of P and V that we can find to keep the right hand side (i.e. n,R,T) equal to 6?

 

 

Sensei

Sensei

Posted
Posted (edited)

The reason I find myself so confused is that we are not given a size of the bubble or temperature.

You don't have to have size of bubble.

Question is about how many times volume of gas at sea level will be bigger, than below sea level, when water press it.

f.e. if bubble has V1=1m^3 at -100m, and V2=2m^3 at 0m

or it has V1=1cm^3 at -100m, and V2=2cm^3 at 0m,

it's exactly the same ratio. V2/V1= 2/1 = 2 times bigger (just illustration).

 

Answer yourself:

What is pressure at some depth below sea level?

How to calculate it?

What is pressure at sea level?

How these two things are connected?

Edited by Sensei
FrankP

FrankP

Posted
  • Author
Posted

Interesting problem. Here is my contribution. Hopefully others can critique me too:

 

1. Problem statement says to assume that none of the air dissolves into the water. Therefore, does the quantity change? Do you then really need to know how many moles there are?

 

2. Re. your question on knowledge of temperature - look at the left and right hand sides of the ideal gas law equation (PV=nRT). Does a change in pressure and/or volume (left hand side) require a corresponding change to the quantity and/or temperature of the gas (right hand side)? Is there a way that you can keep the right hand side the same while changing the P and V values on the left hand side? For example, if we said the right hand side (nRT) is 6, are there different combinations of P and V that we can find to keep the right hand side (i.e. n,R,T) equal to 6?

 

 

This is an interesting spin on how to think about this problem are you suggesting you set one of the variables such as T = 1 since it is not given in order to help solve for P?

You don't have to have size of bubble.

Question is about how many times volume of gas at sea level will be bigger, than below sea level, when water press it.

f.e. if bubble has V1=1m^3 at -100m, and V2=2m^3 at 0m

or it has V1=1cm^3 at -100m, and V2=2cm^3 at 0m,

it's exactly the same ratio. V2/V1= 2/1 = 2 times bigger (just illustration).

 

Answer yourself:

What is pressure at some depth below sea level?

How to calculate it?

What is pressure at sea level?

How these two things are connected?

I understand what you are saying about comparing arbitrary V1 to V2 I get the concept but to me that is not accurate. I am prob wrong but when you don't know the displacement of the bubble you can not determine accurately how much larger it is. It is at best an educated guess. But that's not my main source of confusion.

 

To answer your questions

  1. What is the pressure below sea level I don't know. I believe you are asking me what I asked back again.
  2. I don't know what it is BECAUSE I don't know how to calculate it
  3. I know the pressure at sea level (1 atm)
  4. Pressure at 100' depth and pressure at sea level are connected because the amount of pressure being exerted on the outside is less as depth decreases, therefore the volume of the bubble will increase. As the bubble approaches the surface.

Let me re-explain my confusion because I kind of feel like this happens a lot here, I ask a question explicitly state where my confusion is. Instead of being shown how to approach the problem I am asked the same question that I asked, with different words. I don't understand (PV=nRT) enough to deduce this kind of information I just learned it on Wednesday for the first time (Incase there is time zone differences I posted this question Wednesday, there is no way this is considered beginner problems).

 

I am not given temperature and I am not given pressure at the 100' depth. Therefore there are 2 unknowns in the equation. It mathematically is impossible to solve with 2 unknowns (unless an arbitrary number is substituted for the one variable). In my example which is attached, I tried to substitute (T=1) basically making the equation PV=nR [or PV=nR(1) ] to try and solve for pressure at depth. This did not spit an answer back that made any sense, however looking back maybe that's all I am being asked to do. (But I highly doubt it with my teacher)

studiot

studiot

Posted
Posted (edited)

First where does it say in your question to use PV = nRT?

 

Have you heard of Boyle's Law?

 

https://www.google.co.uk/search?hl=en-GB&source=hp&biw=&bih=&q=boyles+law&gbv=2&oq=boyles+law&gs_l=heirloom-hp.3..0i10l10.1109.3797.0.4047.10.10.0.0.0.0.297.1439.2j7j1.10.0....0...1ac.1.34.heirloom-hp..0.10.1439.pgJ9APDpU9M

 

 

I am not sure you need the densities at all, unless you want to calculate the increase in pressure with depth from first principles.

 

https://www.google.co.uk/search?hl=en-GB&source=hp&biw=&bih=&q=pressure+increases+with+depth&gbv=2&oq=pressure+increase&gs_l=heirloom-hp.1.2.0l10.1485.4485.0.8422.17.13.0.4.4.0.219.1391.2j7j1.10.0....0...1ac.1.34.heirloom-hp..3.14.1516.aGzLh9kyPPY

 

Ask for more help if you need it.

 

:)

Edited by studiot
FrankP

FrankP

Posted
  • Author
Posted (edited)

Well we just touched on Boyles law in passing today we hadn't actually done any problems with it so I guess that might help me let me check it out and get back to you...

 

Boyles law assumes that 2 of the variables remain constants right? Assuming an inverse relationship to volume and pressure..

 

Oh man... lol lemme see if this works

 

___________________________________________________________________

Ok so I get what you said here P1V1=K1=P2V2 or more ideally --> (P1V1=P2V2)

 

I get this equation but my question is I don't know the pressure or the volume

 

but now im stuck just with a different equation I am able to get moles of H2 O and moles of Mercury using the density and that is the only mathematical calculations I can perform without using google to tell me the pressure at a depth of 100 feet. Which I do not want to do because I know that I won't be able to use google on my exam.

Edited by FrankP
studiot

studiot

Posted
Posted

So you are looking for first principles.

 

Have you seen the equation for the pressure in a column of fluid?

 

P = density x g x height of column

FrankP

FrankP

Posted
  • Author
Posted (edited)
post-54062-0-34606300-1479326991_thumb.png

No I have never seen that so its

 

P= (1.025/1cm^3) x (grams) x (100')

 

Where Pressure and Grams are unknown

 

the only other way I can think to get grams is this:

 

(Density CF) * (1ml -> L CF) * (MW H2O CF) = (Moles H2O / L)

 

(Moles H2O / L) *(1liter/ML) * (1ml/1cm3) = (.0569 mol H2O/ Cm3)

 

(.0569 mol H2O/ Cm3) * 2.832x106 = 161140.8g H2O Total in the entire 100 ft3 column.

 

 

To be honest the math I just did above to me sounds horribly wrong it sounds like im just trying to do something to get a value for grams lol that is what is getting me pissed. Im literally playing a guessing game now...

 

the file I attached below has already been attached but I took the number of 56.913 mol/liter (that is a type mole/g sorry about that) that I got and took it one step further to get (g H2O / Cm3 so I could use that conversion factor to multiply by the Cubic volume of the entire 100 foot column.


How about this I tried to use. Your idea to find the total pressure at the bottom I'm not sure if this is right

post-54062-0-86087500-1479412473_thumb.jpg

Edited by FrankP
studiot

studiot

Posted
Posted (edited)

g is the acceleration due to gravity, not grams.

 

Look here and see if you can find and explanation of my formula that you like.

 

[math]P = \rho gh[/math]
Where P is the pressure, rho is the density (it is a greek letter rho), g is the acceleration due to gravity and h is the depth.
What does Boyle's law tell you about the ratio of the gas volume at 100 feet depth to the gas volume at the surface compared to the ratio of the pressures at these points?
Edited by studiot
Sensei

Sensei

Posted
Posted (edited)

g in studiot equation is acceleration 9.81 m/s^2. Not grams.

 

You lack everything what is basic.. What children used to have in seventh grade primary school here..

Maybe you should start from buying primary school physics books and read them all again.. ?

Edited by Sensei
FrankP

FrankP

Posted
  • Author
Posted (edited)

g in studiot equation is acceleration 9.81 m/s^2. Not grams.

 

You lack everything what is basic.. What children used to have in seventh grade primary school here..

Maybe you should start from buying primary school physics books and read them all again.. ?

 

Thanks for that tidbit of your personal opinion there. Contributions to the conversation in a positive direction seem to be your specialty. If you do not plan to assist in the conversation then I will sincerely ask for your resignation in this board. I lack general knowledge because it has been 10 years since I have taken a class which required this knowledge. Also FYI this is a chemistry class.

 

Being that this event is occurring under water I understand that g=gravity and I know 9.8m/s^2 is the value for gravity It would not be my first guess. Had this been a balloon accelerating from a kids hand filled with He gas I would have considered Gravity in the equation so pardon my inability to draw conclusions being that we have not even mentioned gravity one time in my chemistry class 'it must have slipped my mind'...

 

g is the acceleration due to gravity, not grams.

 

Look here and see if you can find and explanation of my formula that you like.

 

[math]P = \rho gh[/math]
Where P is the pressure, rho is the density (it is a greek letter rho), g is the acceleration due to gravity and h is the depth.
What does Boyle's law tell you about the ratio of the gas volume at 100 feet depth to the gas volume at the surface compared to the ratio of the pressures at these points?

 

 

 

ok so I tried this but the problem is I don't understand how I am supposed to integrate gravity. We have never done anything like this.. I have looked on every website and everywhere in my textbook and I have not seen anything that would consider this question chemistry.... So I will try and solve it using the formula you provided me..

 

Boyles law says volume and pressure are inversely related. How to adopt that to 100' depth of seawater I am not sure.

 

I don't understand how to mix the two together, though. How can I multiply (1.025g/cm3)*(9.8m/s^2)*(100ft) all those units are inconsistent. If I convert 9.8m/s^2 to Centimeters its still going to be squared not cubed. and if I convert 100ft to centimeters in height I will have the following units ( Grams, cm3, cm2, cm)

 

i understand when you multiply you can add exponents but does that really apply to this?

Edited by FrankP
Sensei

Sensei

Posted
Posted (edited)

 

Thanks for that tidbit of your personal opinion there. Contributions to the conversation in a positive direction seem to be your specialty. If you do not plan to assist in the conversation then I will sincerely ask for your resignation in this board. I lack general knowledge because it has been 10 years since I have taken a class which required this knowledge. Also FYI this is a chemistry class.

 

That's why you must to relearn everything from scratch. Your teachers right now assume you already know all this. Therefor you have problems, as you don't know what they think, you do. And it'll be worser and worser, the more advanced subjects will be on homework..

 

Ideal gas law, pressure caused by fluid with given height, are physics subjects.

Edited by Sensei
FrankP

FrankP

Posted
  • Author
Posted

 

That's why you must to relearn everything from scratch. Your teachers right now assume you already know all this. Therefor you have problems, as you don't know what they think, you do. And it'll be worser and worser, the more advanced subjects will be on homework..

 

I understand what you are saying. I'm not dumb, I get it, I have 3 weeks left in the semester. The point of me picking up physics books and re-learning everything would make no sense. As it is I am still in the top 5% of this class in terms of grades. This class is an introduction to chemistry. For undergraduate students at a non-technical institution where less than 1% of all majors granted on a yearly basis are in the field of science. For this teacher to assume that any of the students she is teaching know what you are insisting that the average person knows is a fallacy on your end.

 

Please either contribute to the topic or back out of the thread.

studiot

studiot

Posted
Posted

 

 

I don't understand how to mix the two together, though. How can I multiply (1.025g/cm3)*(9.8m/s^2)*(100ft) all those units are inconsistent. If I convert 9.8m/s^2 to Centimeters its still going to be squared not cubed. and if I convert 100ft to centimeters in height I will have the following units ( Grams, cm3, cm2, cm)

 

You can't.

 

You have to convert to a consistent set of units.

You will also have to address the fact that pressure was given to you in yet other units and convert that as well.

 

Perhaps this question is really an exercise in unit conversion?

 

I strongly recommend using the MKS units which are universally accepted in Science (even in the USA).

 

How about attempting my question concerning Boyle's Law?

Ask if the sentence is too long.

 

We can return to the pressure part when we have decided what we want to calculate.

FrankP

FrankP

Posted
  • Author
Posted

 

You can't.

 

You have to convert to a consistent set of units.

You will also have to address the fact that pressure was given to you in yet other units and convert that as well.

 

Perhaps this question is really an exercise in unit conversion?

 

I strongly recommend using the MKS units which are universally accepted in Science (even in the USA).

 

How about attempting my question concerning Boyle's Law?

Ask if the sentence is too long.

 

We can return to the pressure part when we have decided what we want to calculate.

 

Ok sorry for this back and forth at this point I'm probably frustrating everyone.

 

I did try to answer your Boyle's law question as it pertains to pressure and volume.

 

"Boyles law says volume and pressure are inversely related. How to adopt that to 100' depth of seawater I am not sure. " (that is whatI said)

 

 

 

Sidebar: So this question is physics not chemistry?

studiot

studiot

Posted
Posted (edited)

 

Ok sorry for this back and forth at this point I'm probably frustrating everyone.

 

I did try to answer your Boyle's law question as it pertains to pressure and volume.

 

"Boyles law says volume and pressure are inversely related. How to adopt that to 100' depth of seawater I am not sure. " (that is whatI said)

 

 

 

Sidebar: So this question is physics not chemistry?

 

What does Boyle's law tell you about the ratio of the gas volume at 100 feet depth to the gas volume at the surface compared to the ratio of the pressures at these points?

 

Here is a statement of Boyle's Law, in relation to this problem

 

[math]{P_{100ft}}{V_{100ft}} = {P_{surface}}{V_{surface}}[/math]
Even if you did not take the next step, you need to be able to take the data and present it in this form.
Doing this should get you some marks.
Saying I can't do anything gets you none.
So can you rearrange this to give the ratios as asked?
The gas laws are taught and used in Chemistry, Physics and General Science and many other branches of technical activity.
Edited by studiot
Capiert

Capiert

Posted
Posted (edited)

As usual, I am having a real tough time with this...

I have attached my work I have done and the questions.

 

PV=nRT where R is a constant @ ((.08026 L) x (atm)/ (K) x (mol) )

Hi Frank

 

I like Studiot's last formula

P1*V1=P2*V2.

Please notice, P*V is energy (units). *=multiply.

Your constant R is in metric units, so you can not use (cubic) feet. (=Please convert to (cubic) meters.)

K is degrees Kelvin.

1 mole is 22.4 L at STP=standard temperature & pressure (for "any" gas!)

273.15 K (=0 degree C), 1 atm (760 mm Hg=Mercury, =101.3 kPa).

1 L=1 Liter is 1/1000th of a cubic meter.

(Unfortunately?) moles are often given in gram, not kg=kilogram,

because it's more obvious (easier) for gas.

I suppose we can't follow the problem (task)

because we can't read the thumbnail pictures,

the quality is too little.

I can not picture what you have to do exactly,

with your description about a prism,

column of air (no container?),

bubbles passing etc.

Densities are ok. They (also) can be converted to (metric) kg/cubic_meter (units).

But I don't know who does that, because moles are often given in gram from the chemical periodic table (chart, poster) AtWt=atomic weight (in gram (mass);

although weight (Wt=m*g) in Physics is a force (in Newton units, NOT gram).

But science is a bit screwy sometimes. The word "Weight" is ambiguous:

according to the physicists, everybody (including the chemists) are using it for the wrong word, "mass".

The chemists were extra careful & said "weight mass", together,

or "mass weight"

so everybody including the physics, would know what the chemists are talking about.

 

Basically (it is easy=simple) you (are to) equate moles n=P*V/(R*T)

n1=n2

because the mass before (expansion)

is the same as after. (COM=Conservation of mass).

Each symbol receives a number (1 or 2), left side 1, right side 2.

If the temperature stays the same, it(s' values) cancels out,

when brought to 1 side,

leaving you with Studiot's equation:

The ability to find either the new volume or pressure

knowing the older ones, & 1 of the newer.

R=R1=R2 (is a constant).

 

The trick for success (in physics & chemisty)

is to get the units right

for each symbol;

& make sure the formulas

stay an equation (=balanced, equal equality, NOT more, nor less than the other side of the equation. No compromises=exceptions!)

 

I hope that helps.

 

 

I am looking at the (=a?) column of air in which the bubble passes (=rise?) as a prism (=?) so I can gather the volume.

If you would redescribe that (better), maybe we could help.

Maybe all you have to do is type the task exactly as it is written, or else refoto in better quality.

Otherwise we're just guessing, what you have to do, & thus don't know where your problems are.

 

I know that the density of both H20 and Hg are given and are 1.025g/cm3 & 13.6g/cm3 respectively. [*]So now that I have density I can gather some information in terms of D=(Mass/Volume)

Therefore:

  • H20 ((1.025g)/1cm3) or ((1.025g)/1mL)
  • ((13.6g)/1cm3) or ((13.6g)/1mL)
Using that to further explain the column of space (=volume?) that the bubble travels

you want to find total moles of H20 present in that space (=volume).

I do not know yet why you need conversion to moles, if you have volumes.

But (you're probably looking for them) don't worry simply give us a clear copy of your task,

& describe those few sentences better, please.

Maybe we are missing just a few words.

If you state depth, below sea_level please use a minus sign, & in meters (units).

 

So finding the overall (the) volume of (the) space is necessary. I am not sure about Volume, this is the 2 things I have tried:

  • I looked to find V in terms of 100 ft3 because it is unreasonable to assume that the space which the bubble passes through is 2D (at least that's how my brain thinks). 100 ft3 = 2.832 x 106 cm3 and using that information you can gather total moles of H20 contained in that area.
  • Use the density as a conversion factor as stated above. Where: D(H20) => ((1.025g)/1cm3) or ((1.025g)/1mL). If you take that factor and multiply it to get moles per liter in doing so take the conversion factor and plug it in like this (image attached #2) it's a conversion ratio between grams of H2O in the conversion => Liters => Divide by molecular weight of water. This should get you # of moles per liter. Then you need to find out how many moles are present without the liters. So you end up with .0569mol H2O
OK... so now I am done with my ability to rationalize the answer. The reason I find myself so confused is that we are not given a size of the bubble or temperature. I want to be able to say that temperature should be consistent even though I know there would most likely be negligible temperature difference in terms of this. I know the temperature has to be greater than freezing. However, I'm not sure if there is a specific temp that is assumed for problems like this in the value of water. If that is the case I am making this harder... The second point of confusion is the relevance to Hg Mercury as ((13.6g)/1cm3) I know that barometers are usually measured in units of mercury as a standard and so I know that I need to use that to eventually help me solve the problem but how? Thanks as usual
Edited by Capiert
Capiert

Capiert

Posted
Posted (edited)

But I suspect you're (having problems)

wondering about (& trying to figure out)

pressure

P=F/A

force F

per area A,

& why the air is pushing more than a ton (worth)

(per square meter);

or a kilogram (worth)

(per square centimeter)

on our skin;

& that we get an extra atmosphere (worth) of pressure,

each -3 meters (= -10 feet) deep in the water.

That all has to do with the mass

of air above our heads (till outer space),

multiplied by g

is the weight (force),

but it's per (an) area (you choose).

The sea water also has mass (per volume),

but if you use a column (for the depth, =(negative) height)

then the bottom's end

is your (desired) area.

Because

volume=length*area.

Maybe that's the tip you need.

 

Mercury rises 760 mm

in a tube (tall enough,

with "any" diameter, meaning for the "area")

as long as a vacuum above (it=mercury) exists

& the air's pressure can push down

on it's (=mercury's) surface (in the dish, or cup)

to make the mercury go up into the tube.

In other words, the (air's) pressure

is pushing up that weight

Wt=m*g

(of the mercury('s mass m)).

How many grams, depends on the tube's (volume, or)

area

(& the height=length_tall of mercury).

 

Pascals are kilograms per square_meter, pressure.

Newtons are kilogram*meter per second_squared, force, e.g. weight.

Edited by Capiert
1
studiot

studiot

Posted
Posted (edited)

Capiert, if your way of thinking and presentation can help Frank to understanding, well done.

So +1 for encouragement - these are the sort of posts we like to see.

 

However I do not necessarily agree with quite everything you said and personally found it a bit rambling, though mostly along the right lines.

Perhaps a little bit at a time might be better in future?

 

;)

Edited by studiot
Capiert

Capiert

Posted
Posted

So with Studiot's equation

all you have to find

is the pressure

at -100 feet

in sea water.

The "pressure ratio P1/P2",

is the volume factor

(used on V1)

you are looking for.

That means pressures

large/small.

Small=1 atm. (That's a simple denominator of 1, intended to be easy.)

 

Using mercury's density,

we know it rises 0.760 m (height, under vacuum).

Multipling that (height) with a "square meter"

gives us the volume of mercury, lifted.

Thus knowing how many cc=cubic centimeters, (from a volume conversion),

you can find the mercury's mass (in gram, & convert to kg).

With that (mass) you can find its weight (force),

which is in Pascal (units kg/qm), which I gave you from STP.

Now all you need is the depth in meters;

& calculate sea water's mass,

for a similar area (column's volume).

Then calculate its weight.

Since each weight, used the same area (of 1qm=quadratic meter=square_meter),

you can get the pressures.

But remember, the pressure at the sea water's depth

is the extra weight (force, of sea water),

added onto the air's.

Right under the water's surface (a few centimeters),

the pressure is slightly (a tiny bit) more than 1 atm,

(increasing as you go deeper,

till "about" -3 m, where the pressure is "about" 2 atm, etc.).

Except for those large numbers, (kilogram & cc),

you should have enough info to manage (if you are careful).

 

Good luck.

Capiert, if your way of thinking and presentation can help Frank to understanding, well done.

So +1 for encouragement - these are the sort of posts we like to see.

Thanks, just trying to give him a bit (of extra) background,

so he could do it himself.

We'll have to (wait &) see, if it helped. A shot in the dark.

 

However I do not necessarily agree with quite everything you said

Maybe I said it wrong.

Perhaps you would like to comment.

Please do.

 

and personally found it a bit rambling,

Yes, unfortunately my poor style (& haste?=trying too quick).

though mostly along the right lines.

Nice to hear.

Perhaps a little bit at a time might be better in future?

 

;)

Yes.

P.S. I do believe we all need encouragement,

but sometimes we forget that.

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