To solve the problem, we will follow these steps:
### Step 1: Understand the Initial Conditions
We have an open glass tube immersed in mercury, with a length of 8 cm extending above the mercury level. The tube is then closed and sealed.
### Step 2: Raise the Tube
The tube is raised vertically by an additional 46 cm. Therefore, the total length of the tube that is now above the mercury level is:
\[
\text{Total height above mercury} = 8 \, \text{cm} + 46 \, \text{cm} = 54 \, \text{cm}
\]
### Step 3: Define Variables
Let \( x \) be the height of the mercury column that rises in the tube when it is raised. The new height of the air column above the mercury will then be:
\[
\text{Height of air column} = 54 \, \text{cm} - x
\]
### Step 4: Apply Pressure Conditions
The pressure at the top of the air column inside the tube is equal to the atmospheric pressure minus the height of the mercury column:
\[
P = P_0 - x
\]
Where \( P_0 = 76 \, \text{cm of Hg} \).
### Step 5: Use the Ideal Gas Law
Using the ideal gas law, we can write:
\[
P_1 V_1 = P_2 V_2
\]
Where:
- \( P_1 = 76 \, \text{cm} \) (initial pressure)
- \( V_1 = 8 \, \text{cm} \) (initial volume)
- \( P_2 = 76 - x \, \text{cm} \) (final pressure)
- \( V_2 = 54 - x \, \text{cm} \) (final volume)
### Step 6: Set Up the Equation
Substituting the known values into the equation gives:
\[
76 \times 8 = (76 - x)(54 - x)
\]
### Step 7: Expand and Rearrange
Expanding the right side:
\[
608 = (76 \times 54) - (76x + 54x - x^2)
\]
Calculating \( 76 \times 54 \):
\[
76 \times 54 = 4104
\]
So, we have:
\[
608 = 4104 - (130x - x^2)
\]
Rearranging gives:
\[
x^2 - 130x + (4104 - 608) = 0
\]
This simplifies to:
\[
x^2 - 130x + 3496 = 0
\]
### Step 8: Solve the Quadratic Equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -130, c = 3496 \):
\[
x = \frac{130 \pm \sqrt{(-130)^2 - 4 \cdot 1 \cdot 3496}}{2 \cdot 1}
\]
Calculating the discriminant:
\[
(-130)^2 - 4 \cdot 1 \cdot 3496 = 16900 - 13984 = 2916
\]
Thus,
\[
x = \frac{130 \pm \sqrt{2916}}{2}
\]
Calculating \( \sqrt{2916} = 54 \):
\[
x = \frac{130 \pm 54}{2}
\]
This gives two possible solutions:
1. \( x = \frac{184}{2} = 92 \, \text{cm} \)
2. \( x = \frac{76}{2} = 38 \, \text{cm} \)
### Step 9: Determine the Length of the Air Column
Using \( x = 38 \, \text{cm} \):
\[
\text{Length of air column} = 54 - x = 54 - 38 = 16 \, \text{cm}
\]
### Final Answer
The length of the air column above the mercury in the tube after raising it is:
\[
\boxed{16 \, \text{cm}}
\]
