Question

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the cylinder in the drawing contains 3.02 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?
(image of a cylinder with a valve)
δn = number
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Explanation:

Step1: Recall Ideal Gas Law (PV = nRT)

For constant pressure (P) and temperature (T), \( \frac{V_1}{n_1} = \frac{V_2}{n_2} \) (since \( P = \frac{nRT}{V} \), so \( \frac{n}{V} = \frac{P}{RT} \) is constant).
Initial volume \( V_1 \), initial moles \( n_1 = 3.02 \, \text{mol} \). Final volume \( V_2 = \frac{1}{4}V_1 \).

Step2: Solve for final moles \( n_2 \)

From \( \frac{V_1}{n_1} = \frac{V_2}{n_2} \), substitute \( V_2 = \frac{1}{4}V_1 \):
\( \frac{V_1}{3.02} = \frac{\frac{1}{4}V_1}{n_2} \). Cancel \( V_1 \) (non - zero):
\( \frac{1}{3.02} = \frac{1}{4n_2} \) → \( 4n_2 = 3.02 \) → \( n_2 = \frac{3.02}{4} = 0.755 \, \text{mol} \).

Step3: Calculate moles escaped \( \Delta n \)

\( \Delta n = n_1 - n_2 = 3.02 - 0.755 = 2.265 \, \text{mol} \).

Answer:

\( 2.27 \) (rounded to two decimal places, or \( 2.265 \))