Suppose a monatomic ideal gas is contained within a vertical cylinder that is fitted with a movable piston. The piston is frictionless and has a negligible mass. The area of the piston is` 3.14 xx 10^(-2) m^2,` and the pressure outside the cylinder is `1.01 xx 10^5 ` Pa. Heat (2093 J) is removed from the gas. Through what distance does the piston drop?

Options

A. `0.185 m`
B. `0.389 m`
C. `0.539 m`
D. `0.264 m`

Answer

To solve the problem, we need to determine the distance the piston drops when 2093 J of heat is removed from a monatomic ideal gas in a cylinder. We will use the first law of thermodynamics and the relationship between pressure, area, and displacement.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Area of the piston, \( A = 3.14 \times 10^{-2} \, \text{m}^2 \)
- External pressure, \( P = 1.01 \times 10^5 \, \text{Pa} \)
- Heat removed from the gas, \( Q = -2093 \, \text{J} \) (negative because heat is removed)

2. **Understand the First Law of Thermodynamics:**
The first law states:
\[ \Delta U = Q + W \]
where \( \Delta U \) is the change in internal energy, \( Q \) is the heat added to the system, and \( W \) is the work done by the system. Since heat is removed, \( Q \) is negative.

3. **Calculate Work Done by the Gas:**
At constant pressure, the work done by the gas when the piston moves down is given by:
\[ W = P \Delta V \]
where \( \Delta V \) is the change in volume. The change in volume can also be expressed in terms of the area of the piston and the distance it moves down:
\[ \Delta V = A \cdot x \]
where \( x \) is the distance the piston drops.

4. **Substituting Work into the First Law:**
Rearranging the first law gives:
\[ \Delta U = Q + P \Delta V \]
Substituting for \( \Delta V \):
\[ \Delta U = Q + P(A \cdot x) \]

5. **Relate Change in Internal Energy to Heat Removed:**
For a monatomic ideal gas, the change in internal energy can be expressed as:
\[ \Delta U = \frac{3}{2} n R \Delta T \]
However, since we are not given the number of moles or temperature change directly, we will use the heat removed directly in our calculations.

6. **Setting Up the Equation:**
From the first law:
\[ -2093 = P(A \cdot x) \]
Rearranging gives:
\[ x = \frac{-2093}{P \cdot A} \]

7. **Substituting Values:**
Substitute the values for \( P \) and \( A \):
\[ x = \frac{-2093}{(1.01 \times 10^5) \cdot (3.14 \times 10^{-2})} \]

8. **Calculating \( x \):**
\[ x = \frac{-2093}{(1.01 \times 10^5) \cdot (3.14 \times 10^{-2})} \approx \frac{-2093}{3164.14} \approx -0.661 \, \text{m} \]

Since we are interested in the distance the piston drops, we take the absolute value:
\[ x \approx 0.661 \, \text{m} \]

### Final Answer:
The piston drops approximately **0.661 meters**.