Question

5. An open end manometer constructed from a U-shaped tube was operated using an oil which has a density of 1.088 g ml-1. In a particular measurement, the level in the end connected to the gas manifold, on which the experiment was being conducted, measured 76.2 cm above the U-neck, while the level in the open end (to the atmosphere) was 23.8 cm above the U-neck. The outside air pressure in the laboratory was measured as 754 torr. What is the pressure in the gas manifold? For reference, the density of mercury is 13.533 g ml-1 at room temperature. a. 42.0 torr b. 230 torr c. 272 torr d. 516 torr e 712 torr

with steps please

Answer 1

Q5. In the manometer tube, pressure at U-neck must be same for both tubes.

For left tube, Total pressure = P_g + P_o-g

where P_g = pressure of gas

P_o-g = pressure due to oil at gas side

For right tube, Total pressure = P_a + P_o-a

where P_a = atmospheric pressure = 754 torr

P_o-a = pressure due to oil at atmosphere side

Since total pressure must be same,

P_g + P_o-g = P_a + P_o-a ...(1)

P_o-g = (height of oil) * (density of oil) * (acceleration due to gravity)

where height of oil = 76.2 cm = 0.762 m

density of oil = 1.088 g/mL = 1088 g/m³

acceleration due to gravity = 10 m/s²

Substituting the values,

P_o-g = (0.762 m) * (1088 g/m³) * (10 m/s²)

P_o-g = 8290.56 Pa

P_o-g = 8290.56 Pa * (1 torr / 135.33 Pa)

P_o-g = 61.3 torr

P_a-g = (height of oil) * (density of oil) * (acceleration due to gravity)

where height of oil = 23.8 cm = 0.238 m

density of oil = 1.088 g/mL = 1088 g/m³

acceleration due to gravity = 10 m/s²

Substituting the values,

P_a-g = (0.238 m) * (1088 g/m³) * (10 m/s²)

P_a-g = 2589.44 Pa

P_a-g = 2589.44 Pa * (1 torr / 135.33 Pa)

P_a-g = 19.1 torr

Substituting the values in equation (1)

P_g + 61.3 torr = 754 torr + 19.1 torr

P_g = 754 torr + 19.1 torr - 61.3 torr

P_g = 712 torr

Pressure of gas is 712 torr