A gas vessel is attached to an open- end manometer filled with a nonvolatile liquid of density 0.993 g/mL as shown below.

The difference in heights of the liquid in the two sides of the manometer is 32.3 cm when the atmospheric pressure is 765 mmHg. Given that the density of mercury is 13.6 g/mL, the pressure of the enclosed gas is atm.
A) 0.976
B) 1.08
C) 1.04
D) 1.01
E) 0.993

Question

A gas vessel is attached to an open- end manometer filled with a nonvolatile liquid of density 0.993 g/mL as shown below.

The difference in heights of the liquid in the two sides of the manometer is 32.3 cm when the atmospheric pressure is 765 mmHg. Given that the density of mercury is 13.6 g/mL, the pressure of the enclosed gas is atm.
A) 0.976 
B) 1.08 
C) 1.04 
D) 1.01 
E) 0.993

Quizplus · Accepted Answer

First, we need to determine the pressure difference between the gas inside the vessel and the atmospheric pressure. To do this, we need to calculate the difference in pressure between the two sides of the manometer.

The pressure difference due to the liquid in the manometer is given by:

ΔP = ρgh

where ρ is the density of the liquid, g is the acceleration due to gravity, and h is the difference in height of the liquid in the two sides of the manometer.

In this case, we are given that h = 32.3 cm and ρ = 0.993 g/mL. The acceleration due to gravity is approximately 9.81 m/s2. Using these values, we can calculate the pressure difference:

ΔP = (0.993 g/mL)(9.81 m/s2)(0.323 m) = 3.18 kPa

Next, we need to convert the atmospheric pressure from mmHg to kPa so that we can add it to the pressure difference. We have:

765 mmHg = 101.3 kPa (since 1 mmHg = 0.133 kPa)

Therefore, the pressure in the vessel is:

Pvessel = Patm + ΔP = 101.3 kPa + 3.18 kPa = 104.48 kPa

Finally, we need to convert this pressure to atm:

Pvessel = 104.48 kPa × (1 atm/101.325 kPa) ≈ 1.03 atm

The closest answer choice is C) 1.04, which is the correct answer.

A Gas Vessel Is Attached to an Open- End Manometer