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Please show work.
Image for 40) As shown in the figure, an air pocket at the top of a vertical tube, closed at the upper end and open at t
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Answer #1

When we are at the surface, we use the ideal gas equation:

p_0V_0 = nRT_0

when we are at 56 m deper:

p_fV_f= nRT_f

so, using these equation and solving for Vf

V_f = \frac{p_iT_f}{p_fT_i}V_i

where:

p_i = p_0 = 1.0\times 10^5 Pa

p_f = p_0 +\rho g h = 1.0\times 10^5 +(1000)(9.81)(56) = 6.4936 \times 10^5 Pa

T_i = 37 +273 = 310 K

T_f = 7+273 =280 K

V_i = 563 cm^3

we calculate Vf

V_i = 78.31 cm^3

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