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**1. 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 10.0cm

^{3}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.

**2. Homework Equations**

Boyle's Law: P

_{1}V

_{1}= P

_{2}V

_{2}

**3. The Attempt at a Solution**

P

_{1}is what we are solving for. So first find the cross sectional area, A. 10.0cm

^{3}of Hg was added and that went entirely into the two arms. So:

6A + 4A = 10

A = 1cm

^{2}

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.

50P

_{1}= (50-6)P

_{l}

30P

_{1}= (30-4)P

_{r}

Now here comes the equation I’m not sure is entirely justified. 10.0cm

^{3}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:

P

_{1}+ 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.

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