How Does Adding Mercury Affect Pressure in a Manometer?

[Cesium]

Homework Statement

A mercury-filled manometer with two unequal-length arms of the same cross-sectional area is sealed off with the same pressure p in the two arms. With the temperature constant, an additional 10.0cm³ of mercury is admitted through the stopcock at the bottom. The level on the left increases 6.00cm and that on the right increases 4.00cm. Find the pressure p.

Then there is a picture of a two-armed manometer (sorry don't have a scanner but it's simple enough). At the initial pressure p, the left arm has 50cm in length filled with gas and the right arm has 30cm in length filled with gas.

Homework Equations

Boyle's Law: P₁V₁ = P₂V₂

The Attempt at a Solution

P₁ is what we are solving for. So first find the cross sectional area, A. 10.0cm³ of Hg was added and that went entirely into the two arms. So:

6A + 4A = 10
A = 1cm²

Then apply Boyle’s law to each arm:

Let P_l be the final pressure of the left arm and P_r be the final pressure of the right arm.

50P₁ = (50-6)P_l
30P₁ = (30-4)P_r

Now here comes the equation I’m not sure is entirely justified. 10.0cm³ of Hg was added which is the same as adding 10/A cm Hg or 10 cm Hg or 100 torr of pressure to the system, right? So we can write:

P₁ + 10 = P_l + P_r

Combining the 3 equations, I can easily solve for the final answer, but I wanted to make sure my logic is correct. Thanks.

 

[anathema]

Your logic is correct. Adding 10.0cm3 of Hg is equivalent to adding 100 torr of pressure to the system. This can be seen by using the ideal gas law, PV = nRT, where n is the number of moles of gas and R is the gas constant. Since the volume and temperature are constant, we can write:

P1V1 = P2V2
P1 = P2 - 100 torr

Since the pressure in both arms increases by the same amount, we can write:

Pl = P1 + 100 torr
Pr = P1 + 100 torr

Substituting these values into the equations you have already set up, we get:

50P1 = (50-6)(P1+100)
30P1 = (30-4)(P1+100)

Solving for P1, we get:

P1 = 80 torr

So the pressure in the system is 80 torr.

 

[kerimek]

I can say that your logic is correct. You correctly applied Boyle's law to find the cross-sectional area and the final pressures in each arm. Your reasoning for adding 10 cm Hg or 100 torr of pressure to the system is also correct. Overall, your approach to solving this problem is sound and your final answer should be accurate. Good job!