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Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: G O D I V A 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 quasistatic and isothermal.
This is what I did but not sure if it is right:
V_{1} = 0.3 cm^{2} * 15 cm = 4.5 cm^{3}
V_{2} = 0.3 cm^{2} * 90 cm = 27 cm^{3}
I did W = RTln(V_{2}/V_{1}) = 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.
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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.

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?

Since the expansion is isothermal, we can apply Boyle's Law:
P_{1}V_{1} = P_{2}V_{2}

Since the expansion is isothermal, we can apply Boyle's Law:
P_{1}V_{1} = P_{2}V_{2}
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 = E_{K} = 0.5mv^{2}, I just dont know how to find the work.