Question

1. An open tube manometer is connected to a gas tank (see Fig. (1). The mercury is 39 cm higher on the right side than on the left when a barometer nearby reads 75 cm Hg. What is the absolute pressure of the gas? Express the answer in cm Hg, atm., Pa and lb/in?. PP 1-y- D-P To Tank Fig.ci)

Answer 1

The question is based on the principle of hydrostatic pressure calculation. Hydrostatic pressure is the pressure that is exerted by a fluid at equilibrium at a given point within the fluid, due to the force of gravity. Hydrostatic pressure increases in proportion to depth measured from the surface because of the increasing weight of fluid exerting downward force from above. Mathematically, it is given as ,

Po SL S lenity o१ ।११ d

P =P+h * P* g

here P1 = pressure at lower point.[Pa]

P0 = pressure at higher point.[Pa]

h= depth of the liquid.[m]

$\rho_l$ = density of the liquid. [kg/m3]

g = 9.8 [m/s2]

Now in the problem we have to calculate the pressure of gas in the tank. It can calculated by using the above law.

The diagram is shown below;

Pressure t atm Po 7 1. Pressue Presstere

Now Applying the Hydrostatic law between Point 1 and 2

At point 1 , pressure is Pg [cm of Hg]   which is the pressure of the gas.

At point 2 , pressure is P2 [cm of Hg] which is an assumed point at the bottom of u tube manometer.

y1 is the depth of the liquid column between point 1 and point 2.

so Now using the hydrostatic law, we get

,

P =P+h * P* g

$P_1 = P_g + y_1*\rho_l * g$ . ..................... 1

NOTE : p1 is greater than Pg because point 2 is lower position than point 1. so from hydrostatic law it should be greater.

Now, similarly, we have to apply the hydrostatic law between point 2 and point 3.

At point 3 , pressure is P0 [cm of Hg]   which is the pressure of the atmosphere.

At point 2 , pressure is P2 [cm of Hg] which is an assumed point at the bottom of u tube manometer.

y2 is the depth of the liquid column between point 3 and point 2.

so Now using the hydrostatic law, we get

,

P =P+h * P* g

$P_1 = P_0 + y_2*\rho_l * g$........................ 2

So as we can see that left hand side of equation 1 and 2 is same so by equation the right hand side of equation 1 and 2 we get,

,

Pi Poy2Pg Pg+ y1 * P*g 9

so we get,

$P_0 + y_2*\rho_l * g = P_g + y_1*\rho_l * g$,

by rearranging it we get,

,

Pg= Po + ¥2 * pi* g-y1* P* g

   ................... 3

P Po(y2 y1)* pg

NOTE : We have been given that difference between right hand side of column height and left hand side column height = 39 cm

so, the pressure will be

$(y_2 - y_1)*\rho_l * g$   = 39 [cm of Hg] ,

and The barometer reading is given to us as   75 [cm of Hg].

NOTE : barometer is a pressure measuring device which measure the atmospheric pressure in absolute terms. mercury barometer is commonly used.

So The atmospheric pressure   = P0 = 75 [cm of Hg].

so we have all the values we can directly calculate the Pg by replacing the values in equation obtained.

,

P Po(y2 y1)* pg

$P_g$ [cm of Hg]   =   75 [cm of Hg]   + 39 [cm of Hg]  

                          =   114 [cm of Hg]   .............. {ans.}

                            =    114 [cm of Hg] * 0.01316 [atm] / 1 [cm of Hg]    =   1.50 [atm]   ............ {ans.}

                           =   1.50 [atm] *   101325 [Pa] /1 [atm]   = 151987.5 [Pa] ............{ans.}

                            =    1.5 [atm] * 14.6959 [lb/in2] / 1 [atm]   = 22.043 [lb/in2] ...............{ans.}

so the pressure of the gas in the tank is 1.5 atm or 114 cm of Hg.