Question

If the barometeric pressure was 0.97 atm when the mercury level in an open-end manometer connected to a flask of gas was 191 mmHg higher on the side open to the atmosphere than on the flask side, what was the pressure of the gas in the flask in torr? Enter your answer with two decimal places and no units. (Hint: sketch the apparatus.)

Answer 1

Answer:

The pressure of the gas in the flask is 72.79kPa

Explanation:

To solve the problem it is necessary to take into account the concepts related to pressure due to the weight of a fluid.

The equation that represents this pressure is given by,

$P=h\rho g$

Thus the change in pressure due to the weight of a fluid is given by

$\Delta P = h_1\rho g - h_2 \rho g$

Where,

P = Pressure

h = height/depth

g = Gravity acceleration

$\rho$= Density of the fluid

From the values previous given, we have

Density of Mercury: $13.6g/cm^3$

Barometric Pressure: 0.97atm = 737.19 mmHg

Reading of Open Manometer: 191mmHg

Then we have:

$\Delta P = h_1\rho g - h_2 \rho g$

$\Delta P = (h_1- h_2)\rho g$

$\Delta P = (737.19-191) (13.6)(9.8)$

$\Delta P = 72796.20Pa$

$\Delta P = 72.79kPa$

Therefore the pressure of the gas in the flask is 72.79kPa