Air Pressure Problems: Calculating Volume & Mass

[Robert Alan]

Homework Statement

A tube with air in it is marked off by 1cm marks up to 10cm. 0.3kg mass is pushing a plunger down to compress the air. The area of the tube is 10cm^2.

a. What is the pressure inside the sealed tube? 0.295Pa
b. What is the volume of air in the tube? 20cm^3
c. If the mass is increased to 0.6kg, what is the new volume?
d. How much mass is required to squeeze the plunger down to 2cm?

Homework Equations

F/A = F2/A2

The Attempt at a Solution

I tried to find an equation that had both volume and mass in it and was stumped. The new force of the mass is 0.6kg(9.8)N. After that I have no clue. Please help!

 

[turin]

Do you know about the ideal gas law? Does the temperature change in this problem?

Don't forget that there is already an abmient pressure (both inside and outside the tube).

 

[Robert Alan]

No this is just a cylinder with air in it and something pushing down with a force. I think the problem has to be solved for the initial pressure P1 and using P1V1=P2V2 solve for the second pressure then extract the force out and solve for mass. But I'm getting some goofy answers and I'm sure I've made errors in the math involving converting cubic centimeters to cubic meters. I'm still working on it lol.

 

[queenofbabes]

Since the question does not mention temperature I suppose you will not need the ideal gas equation. Show us your working in full?

 

[Robert Alan]

Okay give me a second to organize it.

 

[Robert Alan]

m1=0.3kg
Area=10cm^2 or 0.001m^2
P=F/A = 2.94 / 0.001 = 2940Pa
F=0.3 * 9.8 = 2.94N
Volume = Bh = 10x8 (currently at the 8cm mark) = 80cm^3

Now the mass increases to 0.6kg

F= 5.88N

That's as far as I got. I'm trying to figure out howto get this in P1V1=P2V2.

 

[turin]

Since the question does not mention temperature I suppose you will not need the ideal gas equation.

Some version of it is required. Some relation between pressure and volume. Sometimes one must use physical intuition (or a larger list of formulas that are implicitly derived from a single formula) when some relevant piece of information (e.g. the value of the temperature) is not explicitly given. IMHO, physical intuition is better than mindlessly applying random formulas when solving physics problems.

 

[turin]

P=F/A = 2.94 / 0.001 = 2940Pa

It looks like you're calculating gauge pressue (i.e. neglecting atmospheric pressure). The pressure that appears in [Boyle's?] law is absolute pressue, not gauge pressure.

(currently at the 8cm mark)

How did you obtain this little tidbit of information?

 

[Robert Alan]

It looks like you're calculating gauge pressue (i.e. neglecting atmospheric pressure). The pressure that appears in [Boyle's?] law is absolute pressue, not gauge pressure.

How did you obtain this little tidbit of information?

The 8cm mark was given to us in the problem. After the 0.6kg mass it went down to the 6cm mark. So based on that I assumed 0.9kg would put it at 4cm and 1.2kg at 2cm. It turned out to be the correct answer. But as more pressure builds up it would take more mass to push the plunger down, at least in my mind it would. *Shrug*

 

[turin]

The 8cm mark was given to us in the problem.

OK, future reference, don't omit information from the problem statement. For instance, without that info (or, e.g., the volume in the tube without the mass), you cannot answer part b.

... as more pressure builds up it would take more mass to push the plunger down, ...

I'm not sure what you mean, but the relationship is certainly not linear.