Question #32172

A thin walled cylinder of mass m height h and cross section area S is filled with gas and floats on the surface of water. As a result to leakage from lower part of cylinder the depth of its submergence has increased by a. Determine the initial pressure x in cylinder if initial atm pressure is p atm.

Expert's answer

A thin walled cylinder of mass m height h and cross section area S is filled with gas and floats on the surface of water. As a result to leakage from lower part of cylinder the depth of its submergence has increased by a. Determine the initial pressure x in cylinder if initial atm pressure is p atm.

Solution:

Connection between the pressures p1 and p2 (p1=pinitialp_1 = p_{\text{initial}} - pressure before entering water, p2p_2 - pressure after entering water).



Mendeleev-Clapeyron equation for gas in the first and in the second case:


p1V=vRT;V=hSp_1 V = vRT; V = h * Sp2V=vRT;V=(ha)Sp_2 V = vRT; V = (h - a) * S(1)=(2):p1h=p2(ha)(1) = (2): p_1 h = p_2 (h - a)p2=p1(hha)p_2 = p_1 \left(\frac{h}{h - a}\right)


Newton's second law for the cylinder:


Fatm+mgFgas=0F_{atm} + mg - F_{gas} = 0Fatm=pSF_{atm} = p * SFgas=p2SF_{gas} = p_2 S


(2') and (5) and (4) in (3): pS+mgp1(hha)S=0pS + mg - p_1 \left(\frac{h}{h - a}\right) S = 0

p1(hha)=mgS+pp_1 \left(\frac{h}{h - a}\right) = \frac{mg}{S} + pp1=mgS+phha=(mgS+p)(1ah)p_1 = \frac{\frac{mg}{S} + p}{\frac{h}{h - a}} = \left(\frac{mg}{S} + p\right) \left(1 - \frac{a}{h}\right)


Answer: initial pressure p1=x=(mgS+p)(1ah)p_1 = x = \left(\frac{mg}{S} + p\right) \left(1 - \frac{a}{h}\right).

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