A 25-liter sample of steam at 100°C and 1.0 atm is cooled to 25°C and expanded until the pressure is 19.71 mm Hg. If no water condenses, calculate the final volume of the water vapor.240 L770 L1.0 L1.3 x 10 -3 LNEXT QUESTION

To calculate the final volume of the water vapor, we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

First, let's convert the initial and final temperatures from Celsius to Kelvin. The initial temperature is 100°C, so we add 273.15 to get 373.15 K. The final temperature is 25°C, so we add 273.15 to get 298.15 K.

Next, let's calculate the initial number of moles of water vapor. We can use the ideal gas law equation to solve for n. Rearranging the equation, we have n = PV / RT.

For the initial state, the pressure is 1.0 atm and the volume is 25 liters. The ideal gas constant R is 0.0821 L·atm/(mol·K). Plugging in these values, we get n = (1.0 atm * 25 L) / (0.0821 L·atm/(mol·K) * 373.15 K).

Calculating this, we find that the initial number of moles is approximately 0.798 moles.

Since no water condenses, the number of moles remains constant during the process. Therefore, the final number of moles is also 0.798 moles.

Now, let's calculate the final volume of the water vapor. We can rearrange the ideal gas law equation to solve for V. Plugging in the values, we have V = nRT / P.

For the final state, the pressure is 19.71 mm Hg, which we need to convert to atm by dividing by 760 (since 1 atm = 760 mm Hg). The pressure is then approximately 0.0259 atm.

Plugging in the values, we get V = (0.798 moles * 0.0821 L·atm/(mol·K) * 298.15 K) / 0.0259 atm.

Calculating this, we find that the final volume of the water vapor is approximately 770 liters.

Therefore, the correct answer is 770 L.