QuestionAugust 4, 2025

34. An open flask contains 0.200 mol of air. Atmospheric pressure is 745 mmHg and room temperature is 68^circ F. How many moles are present in the flask when the pressure is 1.10 atm and the temperature is 33^circ C

34. An open flask contains 0.200 mol of air. Atmospheric pressure is 745 mmHg and room temperature is 68^circ F. How many moles are present in the flask when the pressure is 1.10 atm and the temperature is 33^circ C
34. An open flask contains 0.200 mol of air. Atmospheric pressure is 745 mmHg and room temperature is 68^circ F. How many moles are present in the flask when the pressure is 1.10 atm and the temperature is 33^circ C

Solution
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Answer

n_2 \approx 0.197 \text{ mol} Explanation 1. Convert Units Convert 68^{\circ}F to Celsius: C = \frac{5}{9}(F - 32) = \frac{5}{9}(68 - 32) = 20^{\circ}C. Convert mmHg to atm: 745 \text{ mmHg} \times \frac{1 \text{ atm}}{760 \text{ mmHg}} = 0.980 \text{ atm}. 2. Apply Ideal Gas Law Use the formula PV = nRT. For initial conditions: P_1 = 0.980 \text{ atm}, T_1 = 20 + 273 = 293 \text{ K}, n_1 = 0.200 \text{ mol}. 3. Calculate Final Moles Rearrange to find n_2: \frac{P_1V}{T_1} = \frac{P_2V}{T_2} \Rightarrow n_2 = n_1 \cdot \frac{P_2}{P_1} \cdot \frac{T_1}{T_2}. Substitute P_2 = 1.10 \text{ atm}, T_2 = 33 + 273 = 306 \text{ K}. n_2 = 0.200 \cdot \frac{1.10}{0.980} \cdot \frac{293}{306}.

Explanation

1. Convert Units<br /> Convert $68^{\circ}F$ to Celsius: $C = \frac{5}{9}(F - 32) = \frac{5}{9}(68 - 32) = 20^{\circ}C$. Convert mmHg to atm: $745 \text{ mmHg} \times \frac{1 \text{ atm}}{760 \text{ mmHg}} = 0.980 \text{ atm}$.<br />2. Apply Ideal Gas Law<br /> Use the formula $PV = nRT$. For initial conditions: $P_1 = 0.980 \text{ atm}$, $T_1 = 20 + 273 = 293 \text{ K}$, $n_1 = 0.200 \text{ mol}$.<br />3. Calculate Final Moles<br /> Rearrange to find $n_2$: $\frac{P_1V}{T_1} = \frac{P_2V}{T_2} \Rightarrow n_2 = n_1 \cdot \frac{P_2}{P_1} \cdot \frac{T_1}{T_2}$. Substitute $P_2 = 1.10 \text{ atm}$, $T_2 = 33 + 273 = 306 \text{ K}$.<br /> $n_2 = 0.200 \cdot \frac{1.10}{0.980} \cdot \frac{293}{306}$.
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