QuestionJuly 15, 2025

A glass container was initially charged with 3.30 mol of a gas sample at 4.00 atm and 21.7^circ C . Some of the gas was released as the temperature was increased to 30.1^circ C so the final pressure in the container was reduced to 1.25 atm. How many moles of the gas sample are present at the end?

A glass container was initially charged with 3.30 mol of a gas sample at 4.00 atm and 21.7^circ C . Some of the gas was released as the temperature was increased to 30.1^circ C so the final pressure in the container was reduced to 1.25 atm. How many moles of the gas sample are present at the end?
A glass container was initially charged with 3.30 mol of a gas sample at 4.00 atm and 21.7^circ C . Some of the gas was released as the temperature was increased to 30.1^circ C so the final pressure in the container was reduced to 1.25 atm. How many moles of the gas sample are present at the end?

Solution
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Answer

n_2 \approx 0.80 \, \text{mol} Explanation 1. Use the Ideal Gas Law The ideal gas law is given by **PV = nRT**. We will use it to find the initial and final moles of gas. 2. Calculate Initial Moles Initial conditions: P_1 = 4.00 \, \text{atm}, V = \text{constant}, T_1 = 21.7^{\circ}C = 294.85 \, \text{K}, n_1 = 3.30 \, \text{mol}. 3. Calculate Final Moles Final conditions: P_2 = 1.25 \, \text{atm}, T_2 = 30.1^{\circ}C = 303.25 \, \text{K}. Using the relation \frac{P_1}{n_1T_1} = \frac{P_2}{n_2T_2}, solve for n_2: n_2 = \frac{P_2 \cdot n_1 \cdot T_1}{P_1 \cdot T_2} 4. Substitute Values Substitute the known values into the equation: n_2 = \frac{1.25 \times 3.30 \times 294.85}{4.00 \times 303.25} 5. Perform Calculation Calculate n_2 using the above expression.

Explanation

1. Use the Ideal Gas Law<br /> The ideal gas law is given by **$PV = nRT$**. We will use it to find the initial and final moles of gas.<br /><br />2. Calculate Initial Moles<br /> Initial conditions: $P_1 = 4.00 \, \text{atm}$, $V = \text{constant}$, $T_1 = 21.7^{\circ}C = 294.85 \, \text{K}$, $n_1 = 3.30 \, \text{mol}$. <br /><br />3. Calculate Final Moles<br /> Final conditions: $P_2 = 1.25 \, \text{atm}$, $T_2 = 30.1^{\circ}C = 303.25 \, \text{K}$.<br /> Using the relation $\frac{P_1}{n_1T_1} = \frac{P_2}{n_2T_2}$, solve for $n_2$: <br />$$ n_2 = \frac{P_2 \cdot n_1 \cdot T_1}{P_1 \cdot T_2} $$<br /><br />4. Substitute Values<br /> Substitute the known values into the equation:<br />$$ n_2 = \frac{1.25 \times 3.30 \times 294.85}{4.00 \times 303.25} $$<br /><br />5. Perform Calculation<br /> Calculate $n_2$ using the above expression.
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