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.
I don't know what you did.
you have the vapor pressure, you have the moles of water (convert mass), you have the temp, solve for V
I can check your work.
To solve this problem, you'll need to use the ideal gas law equation, PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.
To find the minimum volume of the flask, we need to calculate the number of moles of water vapor in the flask and then convert it into volume.
1. Convert the temperature from Celsius to Kelvin: 28 degrees Celsius + 273.15 = 301.15 K.
2. Use the ideal gas law to calculate the number of moles (n) of water vapor. Rearranging the equation, we have n = PV / RT.
n = (28.36 mmHg / 760 mmHg/atm) * (0.60 g / 18.015 g/mol) / (0.0821 L.atm/mol.K * 301.15 K)
n = 0.00137 moles
3. Convert the number of moles into volume using the ideal gas law again. Rearranging the equation, we have V = nRT / P.
V = (0.00137 moles * 0.0821 L.atm/mol.K * 301.15 K) / (28.36 mmHg / 760 mmHg/atm)
V ≈ 0.038 liters or 38 milliliters
Therefore, the minimum volume of the flask in order for there to be no liquid water present is approximately 38 milliliters.