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Gas and pressure question

Hi,

Could someone help me with this question?
A diver at the bottom of a lake of depth d fills a syringe with an ideal gas and seals the nozzle. The piston remains free to move. The volume of the gas in the syringe at the bottom of the lake is 90cm3 . As the diver returns to the surface, the temperature of the gas does not change. At the surface of the lake the gas in the syringe is at atmospheric pressure and the volume of the gas is 720 cm3 . What is the volume of the gas in the syringe at a depth d/2 ?

A 160 cm3 B 180 cm3 C 206 cm3 D 225 cm3 E 288 cm3 F 315 cm3 G 360 cm3 H 405 cm3
(edited 1 month ago)
Original post by TwisterBlade596
A diver at the bottom of a lake of depth d fills a syringe with an ideal gas and seals the nozzle. The piston remains free to move. The volume of the gas in the syringe at the bottom of the lake is 90cm3 . As the diver returns to the surface, the temperature of the gas does not change. At the surface of the lake the gas in the syringe is at atmospheric pressure and the volume of the gas is 720 cm3 . What is the volume of the gas in the syringe at a depth d/2 ?
A 160 cm3 B 180 cm3 C 206 cm3 D 225 cm3 E 288 cm3 F 315 cm3 G 360 cm3 H 405 cm3


I found the constant (720) due to pV = constant, and then found the pressure at the bottom of the lake. I then realised that pressure is proportional to the height, so halving the height would halve the pressure (800,000 / 2 = 400,000 Pa). I then subbed 400,000 Pa into the pV = constant equation and got an answer of 180 cm3. The actual answer is 160 cm3.
Reply 3
Original post by TwisterBlade596
I found the constant (720) due to pV = constant, and then found the pressure at the bottom of the lake. I then realised that pressure is proportional to the height, so halving the height would halve the pressure (800,000 / 2 = 400,000 Pa). I then subbed 400,000 Pa into the pV = constant equation and got an answer of 180 cm3. The actual answer is 160 cm3.

So atmospheric pressure is 100,000 pascals and therefore the change is pressure going to the bottom of the lake is 720/90 * 10^5 - 100,000 = 700,000. So pressure at d/2 = 100,000 + 700,000/2 = 450,000 and the volume is
720,000 * 10^2 / 450,000 = 160
(edited 1 month ago)
Original post by mqb2766
So atmospheric pressure is 100,000 pascals and therefore the change is pressure going to the bottom of the lake is 720/90 * 10^5 - 100,000 = 700,000. So pressure at d/2 = 100,000 + 700,000/2 = 450,000 and the volume is
720,000 * 10^5 / 450,000 = 160

Oh okay thank you! I didn’t consider the change in pressure.
Reply 5
Original post by TwisterBlade596
Oh okay thank you! I didn’t consider the change in pressure.

I guess that was the gotcha behind the question, otherwise it was pretty much gcse.

I didnt put it in, but a sketch would probably help and a simple thought experiment about what would happen (your approach) if it was -d (so in the air).
(edited 1 month ago)

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