Answer Check...

A 3.00L sample of gas was collected over water at 45 degrees Celsius. The wet gas exerts pressure of 1.50atm. When dried the sample occupies 1.80L and exerts a pressure of 1.77 atm at 80.0 degrees Celsius. What is the vapor pressure of water at 45 degrees Celsius?

My answer = 0.54atm

I wonder if this is a made up problem. I looked up the vapor pressure of H2O at 45C and found 71.9 mm Hg which is 0.0946 atm.

To find the vapor pressure of water at 45 degrees Celsius, you need to use the information provided about the wet gas and the dried gas.

First, let's consider the ideal gas law equation:

PV = nRT

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

We know the initial volume (V1) of the wet gas is 3.00 L and the final volume (V2) of the dried gas is 1.80 L. We can assume that the number of moles of gas remains constant during the drying process.

Now let's focus on the two different scenarios:

1. Wet gas:
Pressure (P1) = 1.50 atm
Temperature (T1) = 45 degrees Celsius

Convert the temperature to Kelvin:
T1 = 45 + 273 = 318 K

2. Dried gas:
Pressure (P2) = 1.77 atm
Temperature (T2) = 80 degrees Celsius

Convert the temperature to Kelvin:
T2 = 80 + 273 = 353 K

Now we can set up two equations using the ideal gas law:

Equation 1: P1V1 = nRT1
Equation 2: P2V2 = nRT2

Dividing Equation 1 by Equation 2, we get:

(P1V1) / (P2V2) = (nRT1) / (nRT2)

Simplifying the equation, the number of moles (n) cancels out:

(P1V1) / (P2V2) = T1 / T2

Plugging in the given values:

(1.50 atm * 3.00 L) / (1.77 atm * 1.80 L) = 318 K / 353 K

Solving for the unknown, we get:

(4.50 atm L) / (3.186 atm L) = 0.895

Now, we can calculate the vapor pressure of water at 45 degrees Celsius by subtracting the known pressure (P1) from the calculated pressure (P2):

Vapor pressure of water = P2 - P1
= 1.77 atm - 1.50 atm
= 0.27 atm

Therefore, the vapor pressure of water at 45 degrees Celsius is approximately 0.27 atm, not 0.54 atm as you previously stated.