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

I would use PV = nRT and solve for P. If pressure is greater than 23.76 torr, there will be liquid present.

Using PV = nRT, plug in 23.76 converted to atm and solve for n = moles H2O in vapor. Subtract from moles in 1.25 initially. The difference is amount water liquid left.

Well, let me calculate that for you.

First, we need to convert the water vapor pressure from torr to atm, otherwise known as "the scientist's unit of endless confusion." Alright, just kidding. It's actually quite straightforward.

So, 1 atm is equal to 760 torr. Therefore, 23.76 torr is approximately 0.031 atm.

Now, let's use some mathematical magic to figure out if any liquid water will be present. We can do this using the ideal gas law equation:

PV = nRT

Where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature in Kelvin.

We can rearrange the equation to solve for n (moles):

n = PV / RT

Given that the pressure is 0.031 atm, volume is 1.5 L, the ideal gas constant is 0.0821 L·atm/(mol·K), and the temperature is 25°C + 273.15 K (because Kelvin is cooler than Celsius), we can calculate the number of moles of water vapor.

Now, let's bring in the molar mass of water (H2O) which is approximately 18 g/mol. This will allow us to determine the mass of water vapor in grams.

So, the mass of water vapor is:

mass = number of moles x molar mass

Now, we can calculate the number of moles and the mass of water vapor.

Finally, if you subtract the mass of water vapor from the initial mass of water in the container, you'll get the mass of liquid water, if any.

Voila! Just plug in the numbers, and you'll have an answer with a side of mathematical humor.

To determine if any liquid will be present, we need to compare the vapor pressure of water at 25 degrees Celsius to the actual water vapor pressure inside the container.

Step 1: Convert the given mass of water from grams to moles.
The molar mass of water (H2O) is approximately 18.01528 g/mol.
Number of moles = mass (in grams) / molar mass
Number of moles = 1.25 g / 18.01528 g/mol = 0.069 moles (rounded to three decimal places)

Step 2: Use the Ideal Gas Law equation to find the pressure of the water vapor.
The Ideal Gas Law equation is: PV = nRT
Where:
P = pressure (in atmospheres)
V = volume of the container (in liters)
n = number of moles
R = ideal gas constant (0.0821 L.atm/(mol.K))
T = temperature (in Kelvin)

Given:
V = 1.5 L (the volume of the container)
T = 25 + 273.15 = 298.15 K (temperature converted to Kelvin)

Rearranging the equation, we can solve for pressure:
P = nRT / V

P = (0.069 mol * 0.0821 L.atm/(mol.K) * 298.15 K) / 1.5 L
P ≈ 1.411 atm (rounded to three decimal places)

Step 3: Convert the pressure to torr.
1 atm ≈ 760 torr
1.411 atm * 760 torr/atm ≈ 1073.536 torr (rounded to three decimal places)

Step 4: Compare the water vapor pressure to the actual water vapor pressure.
The water vapor pressure at 25 degrees Celsius is given as 23.76 torr. This is much less than 1073.536 torr. Therefore, some liquid water will be present in the container.

Step 5: Calculate the mass of liquid water.
Since all the water is not evaporated, we can find the mass of liquid water by subtracting the vaporized water mass from the total mass.

Mass of liquid water = Total mass of water - Mass of water vaporized
Mass of water vaporized = Number of moles of water vaporized * Molar mass of water

Number of moles of water vaporized = Number of moles - Number of moles of water remaining
Number of moles of water remaining = Number of moles of water vaporized = 0.069 moles (since all the water will not convert into vapor)

Mass of liquid water = (0.069 moles - 0.069 moles) * 18.01528 g/mol (molar mass of water)
Mass of liquid water = 0 grams

Therefore, there will not be any mass of liquid water present in the container.

To determine if any liquid will be present, we need to compare the actual vapor pressure of the water at a given temperature with the water vapor pressure at that temperature.

First, we need to calculate the actual vapor pressure of water at 25 degrees Celsius using the given information. The vapor pressure of water increases with temperature, so we need to find the vapor pressure at 25 degrees Celsius.

Next, we can compare the calculated vapor pressure with the water vapor pressure at 25 degrees Celsius. If the calculated vapor pressure is higher, then no liquid water will be present. If the calculated vapor pressure is lower, then there will be liquid water present.

To calculate the actual vapor pressure of water at 25 degrees Celsius, you can use a formula called the Antoine equation:

log10(P) = A - (B / (T + C))

where P is the vapor pressure, T is the temperature in Celsius, and A, B, and C are constants specific to the substance (water in this case).

For water, the Antoine equation constants are:
A = 8.07131
B = 1730.63
C = 233.426

Plugging in the values:
T = 25 degrees Celsius

log10(P) = 8.07131 - (1730.63 / (25 + 233.426))

Now, we need to solve this equation to find the value of log10(P). Taking the antilog (10 raised to the power of both sides of the equation), we get:

P = 10^(8.07131 - (1730.63 / (25 + 233.426)))

Using a scientific calculator or computer software, we can find the value of P:

P ≈ 23.76 torr

Comparing this calculated vapor pressure (23.76 torr) with the given water vapor pressure at 25 degrees Celsius (also 23.76 torr), we see that they are equal. Therefore, the system is at equilibrium, and both liquid and vapor phases of water will be present in the container.

To find the mass of liquid water in the system, we need to determine how much water vapor is in equilibrium with the given mass of water.

To do this, we can use the ideal gas law:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin.

Rearranging the equation to solve for n:

n = PV / (RT)

Given:
P = 23.76 torr
V = 1.5 L
T = 25 + 273.15 = 298.15 K (convert Celsius to Kelvin)

n = (23.76 torr * 1.5 L) / (0.0821 L·atm/mol·K * 298.15 K)

Now, we can calculate the value of n in moles.

Next, since 1 mole of water has a mass of approximately 18 grams, we can determine the mass of liquid water using the molar mass.

Mass of liquid water = n * molar mass of water

Substituting the value of n:

Mass of liquid water = n * 18 grams

Now, calculate the mass of liquid water using this formula.

1.204 g