Study Guide 13 - Pressure, Denisty and Surface Tension

What is the relation between pressure and height of a collumn of incompressible liquid?

\(\mathrm {P = \rho gh}\)

With \(\rho\) in \(\mathrm{kg/m^3,}\) \(\mathrm{h}\) in \(\mathrm {m}\) and \(\mathrm {g = 9.8 \;m/s^2,}\) what are the units of \(\mathrm {P?}\)

Move to part b.

For mercury \(\mathrm {(density = 13600\; kg/m^3) using\; P = \rho gh}\)
\(\mathrm{0.5454 \times 10^5 = 13600(9.8)h \\ h = 0.41\; m = 410\; mm}\)

The unit of pressure in S.I. is "pascal" (Pa), which is \(\mathrm {N/m^2}\)
Now knowing that

\(\mathrm {1 \;Atm = 1.01\times 10^5 \;Pa}\)
we have:
\(\mathrm {0.540\; Atm = 0.540\times 1.01\times 10^5 = 0.5454\times 10^5 \;Pa}\)

The answers for parts (a), (c) and (d) can now be determined.

For water \(\mathrm {(density\; 1000\; kg/m^3)}\)
\(\mathrm {0.5454 \times 10^5 = 1000(9.8)H \\ h = 5.6\; m}\)

For blood \(\mathrm {(density\; 1050 \;kg/m^3)}\)
\(\mathrm {0.5454 \times 10^5 = 1050(9.8)h \\ h = 5.3 m = 530 \;cm}\)

The relationship between pressure and height of a column of incompressible liquid of density \(\rho \) is:
\(\mathrm {P = \rho gh}\)
where \(\mathrm {P}\) is measured in pascals \(\mathrm {(Pa) \;or \;N/m^2.}\)

For mercury, \(\mathrm{\rho = 13600\;kg/m^3}\)
\(\mathrm {1.00\times 10^3 = 13600(9.8)h \\ h = 7.5\times 10^{-3} m = 7.5\; mm}\)

One atmosphere is \(\mathrm{1.01\times 10^5 \;Pa}\)
Therefore \(\mathrm {1.00\times 10^3 \;Pa = 1.00\times 10^3/1.01\times 10^5 = 9.9\times 10^{-3} \;Atm.}\)

For water, \(\mathrm {\rho = 1000\; kg/m}\)
\(\mathrm {1.00\times 10^3 = 1000(9.8)h \\ h = 0.102\; m}\)

For blood, \(\mathrm {\rho = 1050\; kg/m^3}\)
\(\mathrm{1.00\times 10^3 = 1050(9.8)h \\ h = 0.097 \;m = 9.7 \;cm}\)

What is the relation between pressure and height of a column of incompressible liquid?

\(\mathrm {P = \rho gh}\)

\(\mathrm {P = (1000)(9.8)(0.123 - 0.083) = 392\; Pa}\)

Is the pressure in B; higher than, lower than or the same as that in A?

It is Lower than as can be seen from the manometer \(\mathrm {M_1.}\)

What is the pressure difference P?

\(\mathrm {Pressure \;difference = P = \rho gh \\ P_B = P_A - \rho gh = 1.13 \times 10^5 - (1000)(9.8)(0.10) = 1.12\times 10^5 \;Pa \;(gauge)}\)

The absolute pressure is higher than that in chamber B.

\(\mathrm {P_{ABS} = P_{GA} + 1.01 \times 10^5 \;Pa \\ For\; chamber\; B: P_{ABS} = 1.01 \times 10^5 + 1.12 \times 10^5 = 2.13 \times 10^5 \;Pa}\)

The pressure in C higher than that in chamber B.

\(\mathrm {So\; the\; pressure\; in\; C\; is\; P_B + \rho gh \\ P_C = 1.12\times 10^5 + (820)(9.8)(0.122) = 1.13 \times 10^5\; Pa\; (gauge) \\ P_{C, ABS} = 1.01 \times 10^5 + 1.13 \times10^5 = 2.14 \times 10^5 \;Pa.}\)

The pressure in D the same as that in chamber C.

Notice that the pressure calculated for C and D is the same as that given for A. This means that x = 0. This is the answer to part (b).
 

As indicated above \(\mathrm {x = 0.}\)

The pressure in A is higher than the atmospheric pressure.

The pressure in A is higher by the amount \(\mathrm {P = \rho gh}\)
Since the gauge pressure in A is \(\mathrm {1.13 \times 10^5 \;Pa,}\) then:
\(\mathrm {\rho gh = 1.13\times 10^5}\)

\(\mathrm { H = 1.13 \times 10^5/(13600 \times 9.8) = 0.848\; m = 848 \;mm}\)

The Barometric equation is:

\(\mathrm{ P = P_0e^{-\frac{mgh}{k_bT}}}\)

where \(\mathrm {k_B = 1.38\times 10^{-23}}\) J/K is Boltzmann's constant and m is the mass of the average air molecule in kg.

\(\mathrm {m = 29 \times 10^{-3}(kg/mole)/6.02\times 10^{23} (molecules/mole) = 4.82\times 10^{-26} kg}\)

Substituting

\(\mathrm {0.15 = e ^{-\frac{(4.82 \times 10^{-26})(98)h}{(1.38 \times 10^{23})(300)}}= e^{- 1.141 \times 10^{-4 h}}}\)

\(\mathrm {\ell n (0.15) = -1.897 = -1.141 x 10^{-4} h}\)

\(\mathrm {h = 16.6 \times 10^3 m = 16.6\; km}\)

How many surfaces does this soap bubble have?

It has 2, an inside and an outside air-water interface.

What is the excess pressure inside such a bubble?

\(\mathrm{4\gamma /r}\)

There is \(\mathrm {2\gamma/r}\) for each surface so;
\(\mathrm {\Delta P = 4\gamma /r = 4(7.30\times 10^{2)}/0.50\times 10^{-5} = 0.584\times 10^5 \;Pa}\)

 The total pressure inside the bubble \(\mathrm {= P_{out} + \Delta P}\)
\(\mathrm {= 1.2 \times 1.01\times 10^5 + 0.584\times 10^5 = 1.80\times 10^5 \;Pa \\ = 1.80\times 10^5/1.01 \times 10^5 = 1.78\; Atm}\)

To achieve a pressure of 1.2 Atm. in an open tank the liquid column must provide 0.2 Atm since the atmosphere already provides 1 Atm.
\(\mathrm {\rho gh = (1000)(9.8)h = 0.2 \times 1.01 \times 10^5 \\ h = 2.06\;m}\)

How many surfaces does this soap bubble have?

It has only one air-water interface.
Therefore \(\mathrm {\Delta P = 2\gamma /r = 2(7.30 \times 10^{-2})/0.5 \times 10^{-5} = 0.292 \times 10^5 \;Pa}\)

The external pressure on the bubble is \(\mathrm {1.2\;Atm. = 1.2 \times 1.01\times 10^5 = 1.212 \times 10^5 \;Pa}\)

The total absolute pressure inside the bubble =
\(\mathrm {1.212 \times 10^5 + 0.292 \times 10^5 = 1.50 \times 10^5 \;Pa}\)

\(\mathrm {\Delta P_L/\Delta P_S = (4\gamma /r_L)/(4\gamma /r_S) = r_S/r_L}\)

\(\mathrm {V_S/V_L = (r_S/r_L)^3 = 1/2}\)

\(\mathrm {r_S/r_L = (1/2)^{1/3} = 0.79}\)

Therefore \(\mathrm {\Delta P_L/\Delta P_S = 0.79}\)

The angle is indicated in the figure; it is greater than \(\mathrm {90^\circ}\).

The angle greater than 90 degrees is shown in red
The direction of the surface tension force is shown in green. It will pull the surface down.