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Ch.12 - Liquids, Solids & Intermolecular Forces

Chapter 12, Problem 100

A sealed flask contains 0.55 g of water at 28 °C. The vapor pressure of water at this temperature is 28.35 mmHg. What is the minimum volume of the flask in order that no liquid water be present in the flask?

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everyone in this example, it says that a vessel at 35 degrees Celsius consists of 350.155 g of water at 35 degrees Celsius, the vapor pressure of water is 42.2 millimeters of mercury. And were asked that in order to make sure that no liquid water is present in the vessel, what should be the minimum volume of the vessel? So we're going to be solving for volume here. So based on the prompt were given info for pressure, were given a pressure of 42.2 millimeters of mercury and we actually want pressure to be in a. T. M. So we're going to recall the conversion factor that for 1 80 M, we have 760 millimeters of mercury. So this allows us to cancel out units of millimeters of mercury were left with a T. M. And this gives us a pressure equal to 0.555 to a T. M's. We're also given information on temperature from the prompt as 35°C. However, we want temperature to be in units of Kelvin. So we're going to add to 73.15 to get to Kelvin and we should have a kelvin temperature of 308.15 Kelvin when doing so. Were also given a mass of water. And so we want to go ahead and recognize that since we need to find volume and were given mass, we're going to recall upon the following formula where we take pressure times volume and set that equal to the molds of our substance, multiplied by the gas constant R. And then multiplied by temperature. So even though we need to find volume, we're going to actually need to find our moles which is in here first. So we're going to find our moles of water First by taking our massive water given from the prompt as .155 g of water. And we're going to multiply it by the molar mass of water, which we should recall from our periodic tables is 18.016g of water for one mole of water. This allows us to cancel out gramps, leaving us with molds, which is what we want. And we're going to get a value for moles of water equal to 8.6035 times 10 to the negative third power and units of moles. So now that we have molds of our water, we can go ahead and sulfur volume. So solving for volume, what we should have by reorganizing our formula above, we're going to take the moles of our gaseous water, multiplied by the gas constant. R multiplied by temperature and then divided all this by the pressure. So what we will have is our moles of our water, which we just determined is 8.6035 times 10 to the negative third power moles. We're going to plug in R. Gas constant R. So recall that R is eight point sorry, that should be 0.8206 Leaders times a T. M's, divided by moles, times kelvin. And then we're going to multiply by our temperature, which we converted to Kelvin as 308.15 Kelvin. We're then going to divide all of this by our pressure, which we converted to a T. M. S above as 0.555 to 80 M. So canceling out our units, we're going to get rid of a T. M. S. We're going to get rid of moles and we're also going to get rid of kelvin leaving us with leaders as our unit of volume, which is what we want here. And once we do the math, we should get that. Our volume is equal to a value of 3.9-80 m. So this would be our final answer here for the volume of our vessel or the minimum volume rather of our vessel. So I hope that everything I reviewed was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video