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Topic: Work done problem (isothermal).  (Read 4189 times)

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Offline G O D I V A

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Work done problem (isothermal).
« on: February 16, 2010, 07:43:00 AM »
If the gun is fired in a horizontal direction, 1.5 m from the ground, what is the theoretical limit for the horizontal displacement that the bullet can reach before it falls to the ground? Note: To obtain the maximum amount of work, assume that the gas expansion is quasi-static and isothermal.

This is what I did but not sure if it is right:

V1 = 0.3 cm2 * 15 cm = 4.5 cm3
V2 = 0.3 cm2 * 90 cm = 27 cm3

I did W = -RTln(V2/V1) = 4439 J

Im not sure if that is right because it seems the volume changes from 4.5 to 27 and the pressure changes as well from 3.95 atm to 1 atm.  Is another formula required because where does the constant volume come into play because they gave me that.


Offline Yggdrasil

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Re: Work done problem (isothermal).
« Reply #1 on: February 16, 2010, 12:03:53 PM »
There's something wrong with this problem since the pressure of the gas is not enough to propel the bullet out of the rifle (the gas can expand only until it reaches atmospheric pressure [1 atm]).  When the gas pressure is 1 atm, the volume is 17.8cm 3, corresponding to the bullet traveling only 44.25cm down the tube.

Offline G O D I V A

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Re: Work done problem (isothermal).
« Reply #2 on: February 16, 2010, 09:12:55 PM »
There's something wrong with this problem since the pressure of the gas is not enough to propel the bullet out of the rifle (the gas can expand only until it reaches atmospheric pressure [1 atm]).  When the gas pressure is 1 atm, the volume is 17.8cm 3, corresponding to the bullet traveling only 44.25cm down the tube.

How did you get that?

Offline Yggdrasil

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Re: Work done problem (isothermal).
« Reply #3 on: February 16, 2010, 11:39:46 PM »
Since the expansion is isothermal, we can apply Boyle's Law:

P1V1 = P2V2

Offline G O D I V A

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Re: Work done problem (isothermal).
« Reply #4 on: February 17, 2010, 02:14:59 AM »
Since the expansion is isothermal, we can apply Boyle's Law:

P1V1 = P2V2

Understood, but wouldnt the expansion of the gas to 59.25cm apply a force to the bullet which propels it further and out of the barell?  This would be the work right, and since W = EK = 0.5mv2, I just dont know how to find the work.

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