Do we have 1/2P(Volume )?

2015-04-01 7:01 pm
From "W = m(V/T).(Va.T) = m(V/T).[(V/2)T] = (1/2)mV^2", do we have 1/2P(Volume )?
W = mas=m(V/T).(Va.T) =m(V/T)/A.(Va.T)A = m(V/T)/A.[(V/2)T]A = (1/2)P(Volume )
P=pressure, A=area
更新1:

I we press an opened plastic bottle, the gas is out. I mean pressure of an object, and volume of the gas in the bottle.

更新2:

Do we have the 1/2 factor?

更新3:

Pressing contains a length. Because we need to calculate the average acceleration and then find the pressure, and need to calculate average change of volume per time.

更新4:

Why the pressure inside will not be afffected by the pressing?

回答 (1)

2015-04-01 11:47 pm
✔ 最佳答案
First, I give below the derviation of KE using a velocity-time graph. Hope it would make you easier to understand:


圖片參考:https://s.yimg.com/rk/HA00458726/o/1096872542.jpg

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Your question in this post.
How could you relate the motion of an object with the pressure and volume of a gas? This is entirely irrational.



2015-04-01 19:36:34 補充:
You need to specify your question clearly.
How do "pressure of an object" and "volume of a gas" relate to each other?

2015-04-01 19:38:46 補充:
Please also note that PV is not equal to energy E.
E = pressure x change of volume. Only there is a change of gas volume leads to work done.

2015-04-01 22:50:59 補充:
Q:Do we have the 1/2 factor?
A: Why do you think there is a (1/2) factor?

2015-04-02 15:21:37 補充:
The behaviour of a gas under compression is different from the siutation on kinetic energy of a moving body that we have been discussing so far.
Perhaps this can be treated as a new problem.

2015-04-02 15:36:41 補充:
In most problems concering an ideal gas, we usually assume the process is quasi-static, i.e. the process proceeds slowly, so that only the end-points are of interest.
In fact, the compression of an ideal gas is decribed by the "ideal gas equation" and the "First Law of Thermodynmaics".


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