A sealed flask contains 0.60 g of water at 28 degrees celcius. The vapor pressure of water at this temperature is 28.36 mmHg.
What is the minimum volume of the flask in order that no liquid water be present in the flask?
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I tried to solve this problem earlier using PV = nRT, but I kept getting the wrong answer.
To find the minimum volume of the flask in order for no liquid water to be present, we need to determine the pressure at which the water will completely evaporate. This pressure is known as the vapor pressure.
The ideal gas equation, PV = nRT, can be used to solve this problem. However, to apply this equation, we need to make some assumptions:
1. The water can be treated as an ideal gas.
2. The volume of the water vapor is negligible compared to the volume of the flask.
3. The temperature remains constant at 28 degrees Celsius.
Let's go step by step:
Step 1: Convert the given temperature from Celsius to Kelvin.
- Kelvin = Celsius + 273.15
- T (in Kelvin) = 28 + 273.15 = 301.15 K
Step 2: Calculate the number of moles (n) of water present in the flask.
- n = mass / molar mass
- The molar mass of water (H2O) is approximately 18 g/mol.
- n = 0.60 g / 18 g/mol = 0.0333 mol
Step 3: Rearrange the ideal gas equation to solve for pressure (P).
- P = nRT / V
- Since we want to find the minimum volume V, we can rearrange the equation as:
- V = nRT / P
Step 4: Substitute the given values into the equation.
- R = 0.0821 L·atm/mol·K (universal gas constant)
- P = 28.36 mmHg (vapor pressure of water)
V = (0.0333 mol) * (0.0821 L·atm/mol·K) * (301.15 K) / (28.36 mmHg)
Step 5: Convert the pressure from mmHg to atm.
- 1 atm = 760 mmHg
- P (in atm) = 28.36 mmHg / 760 mmHg/atm = 0.0373 atm
V = (0.0333 mol) * (0.0821 L·atm/mol·K) * (301.15 K) / (0.0373 atm)
Step 6: Calculate the volume.
V ≈ 7.68 L
Therefore, the minimum volume of the flask in order for no liquid water to be present is approximately 7.68 liters.