Question

A tank with an attached manometer contains water at 20°C. The atmospheric pressure is 100 kPa. There is a stopcock located 1 m from the surface of the water in the manometer. The stopcock is closed, trapping the air in the manometer, and water is added to the tank to the level of the stopcock. Find the increase in elevation of the water in the manometer assuming the air in the manometer is compressed isothermally.

Answer 1

Answer:

Change in the level of liquid is 0.084 m

Explanation:

As we know that amount of gas as well as the temperature of the gas is constant here

So initial pressure of gas is same as atmospheric pressure

initial volume of gas is

$V = 1 S$

S = area of the tube

Final volume of the gas

$V_f = (1 - \Delta L)S$

Now by pressure equilibrium condition we have

$P_o + \rho g (1) = P + \rho g (\Delta L)$

so we have

$P = P_o + \rho g(1 - \Delta L)$

now we can use the condition of isothermal process

$P_i V_i = P_f V_f$

$P_o( 1 S) = (P_o + \rho g(1 - \Delta L))(1 - \Delta L) S$

$10^5 = (10^5 + 10^3(10)(1 - \Delta L))(1 - \Delta L)$

Solving above equation we have

$\Delta L = 0.084 m$

#Learn

Topic : manometer

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