Thermodynamics in action

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.

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dextercioby

Homework Helper
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.

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.

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cdhotfire

O and thank you for replying. dextercioby

Homework Helper
cdhotfire said:
Hmm, does not seem I learned those law, 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.
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.

cdhotfire

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

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