How to Calculate change in volume produced by a piston?

[spacealgae]

Homework Statement

Hello all! I need a little bit of help with my physics homework! It asks me to calculate the change in volume produced by the piston, and it gives me the equation:
∆V= Ax∆d

Homework Equations

The Attempt at a Solution

Now I know that A= 1x 10-^16m^2
And ∆d=1700nm
My only problem is that I'm confused on how to calculate the answer with the two given variables! Help!
∆V=1x10-^16m^2x1700nm=??

 

[Bystander]

What do the prefixes mean?

 

[spacealgae]

Oh sorry, it means to the 'power of', so to break it down, I believe the variable is:
1 times 10 to the power of -16 m to the power of 2

 

[SteamKing]

Oh sorry, it means to the 'power of', so to break it down, I believe the variable is:
1 times 10 to the power of -16 m to the power of 2

Yeah, but what does 'nm' mean?

 

[spacealgae]

Newton-Meter

 

[spacealgae]

1. 
   A unit of the meter-kilogram-second system equal to the energy expended, or work done, by a force of one Newton acting through a distance of one meter and equal to one joule.

 

[Mark44]

Yeah, but what does 'nm' mean?

Newton-Meter

I don't think so, in the context of this problem. Here it means "nanometer" I believe, or ##10^{-9}## meter.

One problem that you're having is that your units don't match. The cross-sectional area of your piston is in units of square meters (##m^2##), but the piston's stroke is given in units of nanometers (nm). In your calculation you need to be working in the same units, in powers of meters or of nanometers.

 

[Mark44]

1. 
   A unit of the meter-kilogram-second system equal to the energy expended, or work done, by a force of one Newton acting through a distance of one meter and equal to one joule.

Not relevant. Volume is in units of ##(length)^3## -- no force involved.

 

[spacealgae]

That's weird, because in my notes, it talks about Newton Meter, not Nanometer. Ok, let's say it is in Nanometer though, how would I go about the calculation then?

 

[SteamKing]

That's weird, because in my notes, it talks about Newton Meter, not Nanometer. Ok, let's say it is in Nanometer though, how would I go about the calculation then?

How many nanometers are in 1 meter?

 

[spacealgae]

1000000000?

 

[SteamKing]

1000000000?

It's OK to use scientific notation, so people aren't counting zillions of zeroes.

So, if you have Δd = 1700 nm, how many meters is that?

 

[spacealgae]

Ok, in scientific notion, that would be 1.7e-6?

 

[SteamKing]

Ok, in scientific notion, that would be 1.7e-6?

Yes.

 

[Mark44]

That's weird, because in my notes, it talks about Newton Meter, not Nanometer. Ok, let's say it is in Nanometer though, how would I go about the calculation then?

Units for Newton-meters are usually abbreviated as n-m. The abbreviation for nanometers is just nm, no hyphen.

In any case, since you're calculating volume, all of the units have to be length units -- no forces

Ok, in scientific notion, that would be 1.7e-6?

Why are you so tentative? Don't you trust your work? Also, be sure to include units.

∆V= Ax∆d

With expressions such as this one, don't write x to indicate multiplication -- it is too easily confused as being a variable.

Have you finished this problem yet. All you need to do is to multiply the area (A) by the linear distance the piston moves (∆d). The main difficulty in this problem is converting to the same units.

 

[BvU]

That is a really small piston you have there ! What is it made of ?
The stroke of ## 1.7 \ 10^{-6}## m is also really really small !

 

[spacealgae]

Ok, so I would just times 1.7x10^3*1x10^16m^2?

 

[spacealgae]

which would equal 1.87x10^-13m^2?

 

[spacealgae]

• Which leads me to my final question, which is to take the volume I just calculated, and calculate the work done by the gas molecules, and the equation for that is: W=Px*∆V
• and I know that the volume is equal to 1.87x10^-13m^2, and the pressure is 0.25 A t m, so how would I multiply those two together then?

 

[spacealgae]

Wait my mistake, it would be 1.7x10^3*1x10^-16=1.7x10^-13

 

[BvU]

Homework Statement

.. It asks me to calculate the change in volume produced by the piston, and it gives me the equation:
∆V= Ax∆d

Instead of spraying this thread with shots, why not provide the complete problem statement with all variables and given/known data ?
I'm afraid that, because of your subsequent posts, I start to suspect the values for A and ##\ \Delta d \ ##you mention .

 

[spacealgae]

sorry, I just actually figured it out (I think) and got excited. But here we go from the top:
How do I calculate the change in the volume produced by the piston?
Given formula:
∆V=∆x∆d
What I know: distance =1700nm
Area of piston=1x10^-16m^2
What I think I figured out, through the help of this forum, (thank you everyone by the way) was to covert 1700nm into scientific notation, which is 1.7x10^3.
I multiplied 1.7x10^3 with the area of the piston, 1x10^-16m^2, and got the answer 1.7x10^13m^2.
Now, my next question is how to calculate the work done by the gas molecules.
Given formula: W=Px∆V
What I know: Volume=1.7x10^13m^2
A t m (atmospheric pressure)=0.25 A t m

 

[SteamKing]

sorry, I just actually figured it out (I think) and got excited. But here we go from the top:
How do I calculate the change in the volume produced by the piston?
Given formula:
∆V=∆x∆d
What I know: distance =1700nm
Area of piston=1x10^-16m^2
What I think I figured out, through the help of this forum, (thank you everyone by the way) was to covert 1700nm into scientific notation, which is 1.7x10^3.

You've missed an important distinction here. 1 nm = 10 ^-9 m. It's not enough to write 1700 in scientific notation if you lose the nm in the process.

I multiplied 1.7x10^3 with the area of the piston, 1x10^-16m^2, and got the answer 1.7x10^13m^2.

Aside from the error in the distance traveled, 10 ^-16 ⋅ 10³ ≠ 10¹³

Don't you know how to multiply two numbers in scientific notation?

Now, my next question is how to calculate the work done by the gas molecules.
Given formula: W=Px∆V
What I know: Volume=1.7x10^13m^2
A t m (atmospheric pressure)=0.25 A t m

Let's get the basic geometry of a cylinder and fundamental unit operations clear before we tackle work done.

It's the grade school stuff that needs attention here.

 

[spacealgae]

You've missed an important distinction here. 1 nm = 10 ^-9 m. It's not enough to write 1700 in scientific notation if you lose the nm in the process.

Aside from the error in the distance traveled, 10 ^-16 ⋅ 10³ ≠ 10¹³

Don't you know how to multiply two numbers in scientific notation?

Let's get the basic geometry of a cylinder and fundamental unit operations clear before we tackle work done.

It's the grade school stuff that needs attention here.

Yes, yes, I realized I made a mistake and forgot to convert nm to meters. Physics isn't my strong suit, be patient with me. So, once I convert 1700nm to meters, that would be 1.7x10^-6. So I just need to multiply that to 1x10^-16m^2 to get 1.7x10^-22. I'm confused as to if I would add the m^2 at the end of the answer though.

 

[SteamKing]

Yes, yes, I realized I made a mistake and forgot to convert nm to meters. Physics isn't my strong suit, be patient with me. So, once I convert 1700nm to meters, that would be 1.7x10^-6. So I just need to multiply that to 1x10^-16m^2 to get 1.7x10^-22. I'm confused as to if I would add the m^2 at the end of the answer though.

Since you're multiplying meters by square meters, what units would the answer be in? Remember, you are calculating the change in volume.

 

[spacealgae]

cubic meters?

 

[SteamKing]

cubic meters?

Yes.

 

[spacealgae]

Ok, so the final answer would be 1.7x10^-22m^3?

 

[SteamKing]

Ok, so the final answer would be 1.7x10^-22m^3?

Yes, for the change in volume.

 

[spacealgae]

Awesome! Thank you very much! So now that's all figured out how would I multiply that by the atmospheric pressure, 0.25 A t m? Would I convert the A t m to scientific notation and then multiply?

 

[SteamKing]

Awesome! Thank you very much! So now that's all figured out how would I multiply that by the atmospheric pressure, 0.25 A t m? Would I convert the A t m to scientific notation and then multiply?

It's a little more involved than that.

First, what are the units of work?

BTW, the unit 'atmosphere' is abbreviated 'atm.', not 'a t m'.

 

[spacealgae]

I believe that units of work are in Joules. And sorry, I know it's 'atm', not 'a t m', but that's just how my homework had it, so I thought I just would just keep it like that. An error in my homework grammar, I guess.

 

[SteamKing]

I believe that units of work are in Joules.

Yes. But joules are known as derived units in SI, which means they are made up of other fundamental units. What are the fundamental units which make up a joule?

 

[Mark44]

So, once I convert 1700nm to meters, that would be 1.7x10^-6.

Again, you're omitting the units in your answer. 1.7 x 10^-6 in what units?

When you're working problems in physics, it's very important to keep track of the units.

 

[BvU]

Area of piston=1x10^-16m^2

Can someone convince me that an exercise with such a "piston" is for real ?