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A cylindrical tank filled to a height of 25 feet with tribromoethylene has been pressurized to 3 atmospheres (Psurface= 3 atmospheres). The total pressure at the bottom of the tank is 5 atmospheres. Determine the density of tribromoethylene in units of kilogrms per cubic meter.

User Jose Paez
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1 Answer

2 votes

Answer:


\rho=2710.957\ kg.m^(-3)

Step-by-step explanation:

Given:

height of the given liquid in the tank,
h=25\ ft=7.62\ m

pressure at the surface of the liquid,
P_(surf)=3\ atm

pressure at the bottom of the liquid,
P_(botm)=5\ atm

So the pressure due to height of the liquid column:


\Delta P=P_(botm)-P_(surf)


\Delta P=5-3


\Delta P= 2\ atm=202650\ Pa

Now as we know that the pressure due to the height of liquid column is given as:


\Delta P=\rho.g.h

where:


\rho= density of the liquid


g= acceleration due to gravity


202650=\rho* 9.81* 7.62


\rho=2710.957\ kg.m^(-3)

User Bennidhamma
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