calculus help please..

[jso1226]

This is a claculus lab question...

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.
> i put v=hA= 2t x pi 3^2=18t pi... i dont think its right at all...


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 Remembering that the temp of the ideal gas is held fixed while the piston is withdrawn, wirte the pressure, p, as a funciton of the volume v, and then as a function of the time, t.

> what does it mean by "as a function of volume and time"?

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 no idea how to figure this out... please help me!!

 

[Deleted member 4993]

This is a claculus lab question...

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.

First write "h' as a function of time:

h = 10 + 2*t

Now use V = A * h
> i put v=hA= 2t x pi 3^2=18t pi... i dont think its right at all...


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 Remembering that the temp of the ideal gas is held fixed while the piston is withdrawn, wirte the pressure, p, as a funciton of the volume v, and then as a function of the time, t.

> what does it mean by "as a function of volume and then time"? Do you remember ideal gas law? It is given at the begining!

P[sub:1aop6bjg]1[/sub:1aop6bjg]*V[sub:1aop6bjg]1[/sub:1aop6bjg] = P[sub:1aop6bjg]0[/sub:1aop6bjg]*V[sub:1aop6bjg]0[/sub:1aop6bjg]

P[sub:1aop6bjg]1[/sub:1aop6bjg] = P[sub:1aop6bjg]0[/sub:1aop6bjg]*V[sub:1aop6bjg]0[/sub:1aop6bjg]/V[sub:1aop6bjg]1[/sub:1aop6bjg] = P[sub:1aop6bjg]0[/sub:1aop6bjg]*V[sub:1aop6bjg]0[/sub:1aop6bjg]/[A*(10+2t)]

there...



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

Once you get (2) - you should get the others

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


i have no idea how to figure this out... please help me!!

 

[galactus]

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.