The cylinder in the drawing contains 4.07 mol of an ideal gas. By moving the piston, the volume of the gas is reduced to one-fourth its initial value, while the temperature is held constant. How many moles An of the gas must be allowed to escape through the valve, so that the pressure of the gas does not change? 5 Valve

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**Problem Description**

The cylinder in the drawing contains 4.07 mol of an ideal gas. By moving the piston, the volume of the gas is reduced to one-fourth its initial value, while the temperature is held constant. How many moles (Δn) of the gas must be allowed to escape through the valve so that the pressure of the gas does not change?

**Diagram Explanation**

The image depicts a cylindrical container with a movable piston and a valve attached. The piston can be moved to change the volume of the gas inside the cylinder. The valve can be opened to allow gas to escape from the cylinder. 

**Solution Approach**

To solve this problem, use the Ideal Gas Law:

\[ PV = nRT \]

Where:
- \( P \) is the pressure
- \( V \) is the volume
- \( n \) is the number of moles
- \( R \) is the gas constant
- \( T \) is the temperature

Given that the temperature \( T \) is constant and the pressure \( P \) is not to change, the product of \( n \) and \( V \) before and after the volume change must be equal.

**Initial State:**
\[ n_1 = 4.07 \, \text{mol} \]
\[ V_1 = V \]

**Final State:**
Let \( n_2 \) be the final number of moles and \( V_2 \) be the final volume.
\[ V_2 = \frac{1}{4}V_1 = \frac{V}{4} \]

Since the pressure is to remain the same:

\[ n_1 V_1 = n_2 V_2 \]

Substitute the known values:

\[ 4.07 \times V = n_2 \times \frac{V}{4} \]

Simplify to find \( n_2 \):

\[ 4.07 \times V = \frac{n_2 \times V}{4} \]

\[ n_2 = 4 \times 4.07 = 16.28 \, \text{mol} \]

Therefore:

\[ n_2 = 1.02 \, \text{mol} \]

Hence, the number of moles of gas that must be allowed to escape to maintain the same pressure is:

\[ \Delta n = 4.07 \, \text
Transcribed Image Text:**Problem Description** The cylinder in the drawing contains 4.07 mol of an ideal gas. By moving the piston, the volume of the gas is reduced to one-fourth its initial value, while the temperature is held constant. How many moles (Δn) of the gas must be allowed to escape through the valve so that the pressure of the gas does not change? **Diagram Explanation** The image depicts a cylindrical container with a movable piston and a valve attached. The piston can be moved to change the volume of the gas inside the cylinder. The valve can be opened to allow gas to escape from the cylinder. **Solution Approach** To solve this problem, use the Ideal Gas Law: \[ PV = nRT \] Where: - \( P \) is the pressure - \( V \) is the volume - \( n \) is the number of moles - \( R \) is the gas constant - \( T \) is the temperature Given that the temperature \( T \) is constant and the pressure \( P \) is not to change, the product of \( n \) and \( V \) before and after the volume change must be equal. **Initial State:** \[ n_1 = 4.07 \, \text{mol} \] \[ V_1 = V \] **Final State:** Let \( n_2 \) be the final number of moles and \( V_2 \) be the final volume. \[ V_2 = \frac{1}{4}V_1 = \frac{V}{4} \] Since the pressure is to remain the same: \[ n_1 V_1 = n_2 V_2 \] Substitute the known values: \[ 4.07 \times V = n_2 \times \frac{V}{4} \] Simplify to find \( n_2 \): \[ 4.07 \times V = \frac{n_2 \times V}{4} \] \[ n_2 = 4 \times 4.07 = 16.28 \, \text{mol} \] Therefore: \[ n_2 = 1.02 \, \text{mol} \] Hence, the number of moles of gas that must be allowed to escape to maintain the same pressure is: \[ \Delta n = 4.07 \, \text
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