U-shaped manometer question(a tricky one)

[kushite]

Homework Statement

la u-shaped monometer has a cross section of 1cm^3, one side is closed, and the other other is open to the atmosphere, the barometer pressure is 1x10^5 pa. the manometer is filled with mercury, at a constant tempreture someone want to increase the pressure of the closed amount air to 3x10^5pa. what is the volume of the mercury to be added to the other open leg of the manometer? 9cm is given as the height of the air column at the closed leg.

Homework Equations

My problem is the answer doesn't match whatever i get, even if my reasoning is good

The Attempt at a Solution

because the pressure is to be increased at the air column, that column does not move, it remains at its height, so 1.2(air density)x9.8x0.09=atmospheric pressure+pressure due to the new height of the added mercury(h x 9.8x13600)

The should be the 'h'x the crossection. but the answer in my book is 162cm^3, i can't fo the life of understand why. PLEASE HELP

 

[anathema]

Hello,

Thank you for sharing your problem with us. It seems like you are on the right track with your solution, but there may be a few small errors in your calculations. Let's break it down step by step:

1. The manometer has a cross section of 1cm^3, which means that the area of the cross section is 1cm^2.
2. The closed side of the manometer has a height of 9cm, which means that the volume of air in that side is 9cm^3.
3. The pressure in the manometer is initially 1x10^5 pa, which is the same as the atmospheric pressure on the open side.
4. To increase the pressure on the closed side to 3x10^5 pa, we need to add more mercury to the open side.
5. We can use the equation P = ρgh to calculate the pressure due to the added mercury. In this case, the density (ρ) of mercury is 13600 kg/m^3, the height (h) is unknown, and the gravitational acceleration (g) is 9.8 m/s^2.
6. Plugging in these values, we get 3x10^5 pa = (13600 kg/m^3)(9.8 m/s^2)(h). Solving for h, we get h = 2.19x10^-4 m.
7. Now, we need to convert this to centimeters to match the units of the other side of the equation. 2.19x10^-4 m = 0.0219 cm.
8. Finally, we can calculate the volume of mercury needed by multiplying the area of the cross section (1cm^2) by the height (0.0219 cm). This gives us a volume of 0.0219 cm^3.
9. However, this is the volume of mercury for one side of the manometer. Since we need to add mercury to both sides, we need to double this value. This gives us a final answer of 0.0438 cm^3 or 4.38x10^-5 m^3.

I hope this helps clarify the solution and why it may be different from the answer given in your book. It's always good to double check your calculations and units to make sure everything is correct. Keep up the good work!