# Homework Help: Thermodynamics in action

1. Jan 13, 2005

### cdhotfire

Okay, i found this problem to be pretty tough, maybe someone can lend me a hand.

Theres a container with a piston, and theres a gas on the bottom of the piston.

State 1
$A=1.2 x 10^{-2}m^{2}$
$T_1=0^{\circ{}}C$
$V_1=1.5 x 10^{-3}m^{3}$
$P_1=1.02 x 10^{5}Pa$

State 2
Theres a 150kg weight ontop to the piston, wish causes the piston to go down.
$T_2=0^{\circ{}}C$

State 3
The cylinder is brought in contact with boiling water raising temperature to $100^{\circ{}}C$
The weight goes up, the gas pushes up the piston, and the piston pushes up the weight.
$T_3=100^{\circ{}}C$

State 4
The block is removed, and the gas expands to fill the container, which is a cylinder.
$T_4=100^{\circ{}}C$

All I need is how to get this started I think I can get it going from there.

The truth is I dont even know how to start this, looked at it for like 30 minutes, maybe I missed something.

But any help would be appreciated.

Thxs.

Last edited: Jan 13, 2005
2. Jan 13, 2005

### dextercioby

You need to understand the fact that any thermodynamical transformations that take place in the gas (probably assumed ideal) are assumed quasistatical and reversible.So u can apply the law of Mendeleev and Clapeyron.

You didn't say what the problem was asking you... :tongue2:

Daniel.

3. Jan 13, 2005

### cdhotfire

Hmm, does not seem I learned those laws, onless my book and teacher are calling them by different names. O, and I asked to get this started, for example finding pressure on state 2. From what I see, is that the 150 kg weight does work on the gas? I might be wrong. But if so it does 1470 J, so I got Work. Work is $\Delta W= P \Delta V$. I can see that pressure goes up, and volume goes down. But I have no clue as in how to find either of those.

Last edited: Jan 13, 2005
4. Jan 13, 2005

### cdhotfire

O and thank you for replying.

5. Jan 13, 2005

### dextercioby

That "A" is it the surface of the piston??If so,u can compute the pressure the piston is exeritng on the gas...Temperature remains constant,so you'd have to apply
$$P_{1}V_{1}=P_{2}V_{2}$$

Daniel.

6. Jan 13, 2005

### cdhotfire

Hmm, I didnt think of that. I think I can take it from here. Thank you very much.