A 25-liter sample of steam at 100^circ C and 1.0 atm is cooled to 25^circ C and expanded until the pressure is 19.71 mm Hg. If no water condenses, calculate the final volume of the water vapor. 1.3times 10^-3L 1.0 L 240L 770 L

To solve this problem, we can use the Ideal Gas Law, which is given by:<br /><br />\[ PV = nRT \]<br /><br />where:<br />- \( P \) is the pressure,<br />- \( V \) is the volume,<br />- \( n \) is the number of moles,<br />- \( R \) is the ideal gas constant (0.0821 L·atm/(mol·K)),<br />- \( T \) is the temperature in Kelvin.<br /><br />First, let's convert all the given values to appropriate units and calculate the initial number of moles of steam.<br /><br />Initial conditions:<br />- Volume (\( V_1 \)) = 25 L<br />- Temperature (\( T_1 \)) = \( 100^{\circ}C \) = \( 100 + 273.15 = 373.15 \) K<br />- Pressure (\( P_1 \)) = 1.0 atm<br /><br />Using the Ideal Gas Law to find the number of moles (\( n \)):<br /><br />\[ n = \frac{P_1 V_1}{R T_1} \]<br />\[ n = \frac{1.0 \text{ atm} \times 25 \text{ L}}{0.0821 \text{ L·atm/(mol·K)} \times 373.15 \text{ K}} \]<br />\[ n = \frac{25}{30.635215} \]<br />\[ n \approx 0.816 \text{ moles} \]<br /><br />Now, let's use the final conditions to find the final volume (\( V_2 \)):<br /><br />Final conditions:<br />- Temperature (\( T_2 \)) = \( 25^{\circ}C \) = \( 25 + 273.15 = 298.15 \) K<br />- Pressure (\( P_2 \)) = 19.71 mm Hg = \( \frac{19.71}{760} \) atm = 0.02594 atm<br /><br />Using the Ideal Gas Law again to find the final volume (\( V_2 \)):<br /><br />\[ V_2 = \frac{nRT_2}{P_2} \]<br />\[ V_2 = \frac{0.816 \text{ moles} \times 0.0821 \text{ L·atm/(mol·K)} \times 298.15 \text{ K}}{0.02594 \text{ atm}} \]<br />\[ V_2 = \frac{19.947}{0.02594} \]<br />\[ V_2 \approx 769.3 \text{ L} \]<br /><br />Therefore, the final volume of the water vapor is approximately 770 L.<br /><br />The correct answer is:<br />770 L