Quick physics question.

Energize

Energize

Energize

Caporegime
Joined
12 Mar 2004
Posts
29,962
Location
England
Can anyone tell me how to work out the pressure here (Pa)?

"A balloon at atmoshperic pressure is taken to a depth of 50m in water that has a temp of 285k and density of 1100kg/m3".

Do I just do 9.8*.5*1100 to work out the pressure difference?
 
Energize said:
Can anyone tell me how to work out the pressure here (Pa)?

"A balloon at atmoshperic pressure is taken to a depth of 50m in water that has a temp of 285k and density of 1100kg/m3".

Do I just do 9.8*.5*1100 to work out the pressure difference?
Click to expand...

Pressure at depth of 50m is atmospheric pressure + pressure due to water above you.

we don't have any dimensions for the balloons so there must be a clever way here.

sid
 
Is that the full question mate? have you worded it right? do you want the pressure inside the balloon at a depth of 50m yeah?

sid
 
This is the full question.

"A balloon contains .05 m3 of hydrogen at an atmospheric pressure of 1.01x10^5 Pa and a temperature of 300K. The balloon is taken to a depth of 50m in sea water that has a temperature of 285K and a density of 1100Kg/m3. Determine the volume of the balloon at this depth."

With the pressure I can use the ideal gas equation to work out the volume. I assume thats what the book wants me to do anyway.
 
Yes use ideal gas law, you can work out the pressure in the balloon from hydrostatic pressure matching that of the balloon and thats given by roe.g.h as you have pointed out.
 
The thing is though, when I add the atmospheric pressure, and the pressure from the calculation I 1.06x10^5 Pa which gives me .45m3 as the answer, but the book says it's 7.6x10^-3 m3 , though the book has been wrong before.
 
Sorry I meant .045m3.

I see where I went wrong now, I was doing .5x9.8x1100 instead of 50x9.8x1100. :rolleyes: Still I get 7.49609375x10^-3 at least 0.05 away from their answer.
 
Last edited:
Jonnycoupe said:
Is the sea water given as 1100 kg/m^3?

G at 9.8 seems a bit lame rather than 9.81.

Have you added Pa?
Click to expand...

Stop being pedantic, if your going to assume 9.81 why not 9.807?

Its a rounding number i rarely use anything but 9.8.

also G is not the same a acceleration caused by mavity (g)

G is the gravitational constant.

KaHn
 
KaHn said:
Stop being pedantic, if your going to assume 9.81 why not 9.807?

Its a rounding number i rarely use anything but 9.8.

also G is not the same a acceleration caused by mavity (g)

G is the gravitational constant.

KaHn
Click to expand...

I was just trying to help clarify some rounding if hes out by a bit. Wind your neck back in.

This is physics, the whole point of it is pedantics. :p

I use 9.81 everyday I need to and I did throughout my A level course and degree.

I not too sure what your waffling on about with regards to the definition of G either.

0.007496 is pretty close to 0.0076, I would just assume that you've done it correctly and put it down to rounding errors.
Click to expand...

EXACTLY. Kahn has just been a pleb; your rather flippant stance is both suprising and disappointing.
 
Last edited:
Jonnycoupe said:
EXACTLY. Kahn has just been a pleb; your rather flippant stance is both suprising and disappointing.
Click to expand...

Excuse me? Your attempt to be pedantic was shot down, if I was Energize I would take their answer and take the percentage error and explain the error rather than aiming to get their answer.

And the fact you are using G as 9.81 (9.807 ms^-2) is incorrect aswell G is the Gravitational Constant (G) not the acceleration caused by mavity (g)

So if you want to play pedantics, USE THE CORRECT TERMANOLOGY OR YOU WILL LOSE MARKS IN EXAMS!

http://en.wikipedia.org/wiki/Gravitational_constant

KaHn
 
OP,

Ignoring the willywaving going on....

Take the sum:

1100 x 9.8 x 50
change it to
1100 x 9.81 x 50

Now is that any closer to the answer stated in the textbook...?

Thats all I bloody replied for, worth bothering try to help ....on OCUK? No chance.
 
Look what you are missing from my post is that the numbers, aslong as they are close will not be marked down upon (unless you are doing 2 + 2 etc), if you did a degree you should know that the terminology and units add a lot to the exam questions.

If you were doing F=ma you cant just say 4x20= 80 as you will be marked down for not including units.

Same as if you said G = 9.807 ms^-2, it is incorrect and you will be marked down on it.

I was trying to point out that the rounding errors do not matter but correct terminology does.

If they say prove x = 4.80293645 Pa

And you get x = 4.804552 Pa

Just do your value as a percentage as theirs and comment on that you think it is jut rounding errors.

Aslong as you state what you are doing you will show an understanding of the question and will pick up marks if you get the numbers wrong.

KaHn
 
You must log in or register to reply here.
Back
Top Bottom