Question 11.2.1: Use vapor pressure. A student leaving campus for spring brea......
Use vapor pressure.
A student leaving campus for spring break wants to make sure the air in her dorm room has a high water vapor pressure so that her plants are comfortable. The dorm room measures 4.0 m × 3.0 m × 3.0 m and the student places a large pan containing 8.0 L of water in the room. Assuming that the room is airtight, that there is no water vapor in the air when she closes the door, and that the temperature remains a constant 25 °C, will all of the water in the pan evaporate? The vapor pressure of water at 25 °C is 23.8 mm Hg
You are asked to determine whether a liquid will evaporate in a sealed room.
You are given the size of the room and the volume and temperature of the liquid.
To determine whether or not all the water would evaporate, first calculate the vapor pressure the water would exert if all the water were to evaporate. If this pressure is greater than the vapor pressure of water at 25 °C, it implies that not all of the water can evaporate.
n_{H_{2}O} = 8.0 L H_{2}O × \frac{10^{3}\text{ mL}}{1\text{ L}} \times \frac{1.00\text{ g}}{1\text{ mL}} \times \frac{1\text{ mol H}_{2}\text{O}}{18.0\text{ g}} = 440 mol
v_{room} = 4.0 m × 3.0 m × 3.0 m × \left(\frac{10^{2}\text{ cm}}{1\text{ m}} \right)^{3} \times \frac{1\text{ mL}}{1\text{ cm}^{3}} \times \frac{1\text{ L}}{10^{3}\text{ ml}} = 3.6 × 10^{4} L
P = \frac{nRT}{V} =\frac{\left(440\text{ mol}\right) \left(0.082057\text{ L} \cdot \text{atm/K} \cdot \text{mol}\right)(298 \text{ K}) }{3.6 \times 10^{4}\text{ L}} \times \frac{760\text{ mm HG}}{1\text{ atm}} = 230 mm Hg
The equilibrium vapor pressure of water at 25 °C (23.8 mm Hg) is much smaller than the vapor pressure in the room if all of the water evaporates (230 mm Hg). Therefore, not all of the water in the pan will evaporate in an airtight room. (In reality, the room is probably not airtight and more water will evaporate than predicted.)