One mole of an ideal gas is contained in a circular cylinder

[CapnShanty]

From 2010 thread:

One mole of an ideal gas (p=RT/v) is contained in a circular cylinder that is equipped with a piston at one end. The cylinder has radius 3 dm. while the gas is held at the fixed temp T=293.2K the piston is slowly pulled out of the cylinder at the rate of 2 dm per sec. If h is the distance from the inner surface of the piston to the base of the cylinder, then the volume of the cylinder is hA, where A is the area of the circular base. Initially, h = 10dm.

1. suppose the piston is moved outward at the rate of 2dm per sec. write the volume of the cylinder as a funcion of t=time.

2. Remembering that the temp of the ideal gas is held fixed while the piston is withdrawn, wirte the pressure, p, as a funciton of the volume v, and then as a function of the time, t.

3. how long does it take for the pressure to decrease to 50 per of its original value?

4. how far has the piston moved when this pressure is reached? label the units in your answer.

I have the same problem (oddly enough) and can't do it; I have for the volume of the cylinder as a function of time h(t)=10+2t, but I don't know how to write the pressure as a function of the volume or the time.
Any help would be appreciated.

 

[Deleted member 4993]

I have the same problem (oddly enough) and can't do it; I have for the volume of the cylinder as a function of time h(t)=10+2t, but I don't know how to write the pressure as a function of the volume or the time.
Any help would be appreciated.

Boyle's Gas law says \(\displaystyle PV=nRT\), as you probably know.

The initial volume of the cylinder is \(\displaystyle 90\pi \;\ dm^{3}=\frac{9\pi}{100} \;\ m^{3}\) when the piston is at h=10 dm.

\(\displaystyle R=8.31 \;\ \frac{J}{mol\cdot K}\) when the volume is in cubic meters and the pressure is in Pascals.

Since the volume is given in dm^3, we can convert to m^3. There are 1000 dm^3 in 1 m^3

Thus, for n=1 mole, \(\displaystyle P=\frac{nRT}{V}\Rightarrow \frac{(8.31)(293.2)}{\frac{9\pi}{100}}=8617.3277 \;\ Pa\)

As the piston is pulled OUT starting from 10 dm, the volume can be expressed as \(\displaystyle V=9\pi(10+2t) \;\ dm^{3}\)

or \(\displaystyle \frac{9\pi (t+5)}{500} \;\ m^{3}\)

There's a start. Can you finish now?. Make sure my conversion is correct.

Did you read Galactus's response above?

He tells you (gives you) the expression for pressure as a function of time!

Please share your work with us ...

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...Before-Posting

 

[CapnShanty]

Sorry

Did you read Galactus's response above?

He tells you (gives you) the expression for pressure as a function of time!

Please share your work with us ...

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...Before-Posting

That's my mistake. I'm a college sophomore taking calc 2 and they're giving us this algebra review exercise and I haven't done this for about five years, so I'm at a loss. I tried to apply Galactus' (8.31)(293.2)/(9pi/100) deal, by putting the volume equation (9pi(10+2t)) in on the bottom of that instead of 9pi/100, but I got nowhere. I have pressure as a function of volume, but I don't know how to get pressure as a function of time, which is what I need, and it obviously isn't what I guessed it would be, so I don't know how to find how long it would take for the pressure to decrease by 50%, or how far the piston has moved by then.
Sorry again and thanks for any assistance.