PHY202
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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The problem bottle engine with 3 liters operating temp 200-300?
Pretaining to my earlier quetion, I found the class notes on the bottle engine. Can I use P1/T1 = P2/T2 to find the pressure and then use V1/T1=V2/T2 to find volume and then graph the PV Curve?
I've deleted your preceding question, which should be answered in the process of answering this one.
To keep subscripts straight let P0, V0 and T0 be the values of P, V and T in the original state, use P1, V1, T1 for the state when water first reaches the top of the tube, and P2, V2, T2 for the state when the temperature reaches its maximum value. We will refer to these as States 0, 1 and 2.
You might be given either the height to which water is to be raised, or the temperature at which water reaches its max height in the tube.
If you are given the height to which water is to be raised, you would find P1 and use this to find T1. Volume is constant during the first phase of heating, so as you have concluded P/T is constant and P1 / T1 = P0 / T0.
If you are given information about the temperature in state 1, then you would use the same relationship to find the pressure.
Between state 1 and state 2, the pressure remains the same as water flows out at the height achieved in state 1. So V / T remains constant, and as you have expressed V2 / T2 = V1 / T1.
Between State 0 and State 1, volume is constant. Between State 1 and State 2 pressure is constant. So the P vs. V graph is a vertical line between the point (V0, P0) and (V1, P1), and a horizontal line between the point (V1, P1) and (V2, P2).