If4 310 dm− of water is introduced into a31 dm flaskat 300 K, how many moles of water are in thevapour phase when equilibrium is established(Given vapour pressure of2H O at 300K is 3170 Pa
Question
If4 310 dm− of water is introduced into a31 dm flaskat 300 K, how many moles of water are in thevapour phase when equilibrium is established(Given vapour pressure of2H O at 300K is 3170 Pa
Solution
To determine the number of moles of water in the vapor phase when equilibrium is established, we can use the ideal gas law and the given vapor pressure of water at 300 K.
Step 1: Convert the given volume of water from cubic decimeters (dm^3) to liters (L). 4 310 dm^3 = 4 310 L
Step 2: 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.
Since we are given the pressure (vapor pressure of water at 300 K) and the volume, we can rearrange the equation to solve for the number of moles (n): n = PV / RT
Step 3: Substitute the given values into the equation. P = 3170 Pa (vapor pressure of water at 300 K) V = 4 310 L (volume of water) R = 0.0821 L·atm/(mol·K) (ideal gas constant) T = 300 K (temperature)
n = (3170 Pa * 4 310 L) / (0.0821 L·atm/(mol·K) * 300 K)
Step 4: Calculate the number of moles of water in the vapor phase. n ≈ 55.6 moles
Therefore, when equilibrium is established, there are approximately 55.6 moles of water in the vapor phase.
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