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The vapor pressure of water at 25 C is 23.76 torr. If 1.25 g of water is enclosed in a 1.5-L container, will any liquid be present? IF so, what mass of liquid?

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User BruceCui
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To determine if any liquid water will be present in the container, we need to calculate the vapor pressure of water at the temperature inside the container. If the vapor pressure inside the container is less than the vapor pressure of pure water at 25°C, then some liquid water will be present.

We can use the ideal gas law to calculate the pressure inside the container, assuming that the water vapor behaves like an ideal gas. The ideal gas law is given by:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the absolute temperature.

We can rearrange this equation to solve for the number of moles of water vapor:

n = PV/RT

We know the volume of the container is 1.5 L, and the temperature is 25°C = 298 K. The gas constant R is 0.08206 L atm/mol K.

To find the pressure inside the container, we need to know how much water vapor is present. We can calculate the number of moles of water vapor from the mass of water using the molar mass of water (18.015 g/mol):

n = m/M

where m is the mass of water and M is the molar mass of water.

n = 1.25 g / 18.015 g/mol = 0.069 mol

Now we can substitute the values we know into the ideal gas law equation to find the pressure inside the container:

P = nRT/V = (0.069 mol)(0.08206 L atm/mol K)(298 K) / 1.5 L = 1.14 atm

To convert this to torr, we multiply by 760 torr/atm:

P = 1.14 atm × 760 torr/atm = 865.4 torr

Comparing this value to the vapor pressure of water at 25°C (23.76 torr), we can see that the pressure inside the container is much greater than the vapor pressure of pure water. Therefore, no liquid water will be present in the container. All of the water has vaporized to form water vapor inside the container.

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User Spacegoing
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