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A-R-Q
 Post subject: [CHMA] How deep is the snorkler?
PostPosted: Mon, 5 Dec 2011 00:15:59 UTC 
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Snorkler takes syringe filled with 16mL of air from the surface, where the pressure is 1.0atm, to an unknown depth. The volume of air in syringe at this depth is 7.5mL. What is the pressure at this depth? If the pressure increases by 1 atm for every 10m of depth, how deep is the snorkler?

Attempt:
Using
P1V1 = P2V2
I found out that P_2 is equal to about 2.13 atm.

However, I don't how to retrieve the depth..
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Shadow
 Post subject: Re: [CHMA] How deep is the snorkler?
PostPosted: Mon, 5 Dec 2011 01:59:20 UTC 
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A-R-Q wrote:
Snorkler takes syringe filled with 16mL of air from the surface, where the pressure is 1.0atm, to an unknown depth. The volume of air in syringe at this depth is 7.5mL. What is the pressure at this depth? If the pressure increases by 1 atm for every 10m of depth, how deep is the snorkler?

Attempt:
Using
P1V1 = P2V2
I found out that P_2 is equal to about 2.13 atm.

However, I don't how to retrieve the depth..


Use the formula for the volume of a syringe, which is in the shape of a cylinder.

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outermeasure
 Post subject: Re: [CHMA] How deep is the snorkler?
PostPosted: Mon, 5 Dec 2011 05:57:41 UTC 
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A-R-Q wrote:
Snorkler takes syringe filled with 16mL of air from the surface, where the pressure is 1.0atm, to an unknown depth. The volume of air in syringe at this depth is 7.5mL. What is the pressure at this depth? If the pressure increases by 1 atm for every 10m of depth, how deep is the snorkler?

Attempt:
Using
P1V1 = P2V2
I found out that P_2 is equal to about 2.13 atm.

However, I don't how to retrieve the depth..


That's what the part I highlighted in red is for.

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\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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