A sample of gas collected over water at 42 C occupies a volume of one liter. The wet gas has a pressure of 0.986 atm. The gas is dried, and the dry gas occupies 1.04 L with a pressure of 1.00 atm at 90 C. Using this information, calculate the vapor pressure of water at 42 C.

What I have so far:
PV=nrt

n=PV/RT = (1.00 atm)(1.04L)/(0.0821)(363.15K) = 1.04/29.81 = 0.035 mol

I think you are right so far although I wouldn't round to 2 s.f. as you did.

I think we know Ptotal = Pdry gas + PH2O = 0.986 atm.

Then we should be able top calculate the P of dry gas at 42 C by using n = 0.0349 moles and substituting the conditions at 42 C. Substitute this Pgas into the Dalton equation above and calculate PH2O. I get something like 63 mm. I found a site on the Internet that listed PH2O @ 42 C as 61.5 mm. Check my thinking.

Well, we have a gas that's gone from being wet to being dry. You could say it's gone from being a damp gas to a dry-humor gas. Quite the transformation!

To calculate the vapor pressure of water at 42°C, we need to consider the fact that the total pressure of the wet gas (0.986 atm) is a combination of the pressure exerted by the gas itself and the partial pressure of water vapor.

Since the dry gas no longer contains any water vapor, the pressure exerted by the gas alone is now equal to 1.00 atm. So, the change in pressure due to the removal of water vapor is 0.986 atm - 1.00 atm = -0.014 atm.

Now, we need to use this change in pressure to calculate the vapor pressure of water at 42°C. This is where the ideal gas law comes in handy.

PV = nRT

We already know the number of moles (0.035 mol), the volume (1.04 L), and the temperature (90°C = 363.15 K). Let's plug everything in and solve for P, which represents the vapor pressure of water at 42°C.

P = (nRT - ΔP) / V
P = (0.035 mol)(0.0821 L·atm/(mol·K))(363.15 K) - (-0.014 atm) / 1.04 L

Calculating this out will give you the vapor pressure of water at 42°C. Just remember to keep your answers high in spirits and filled with laughter, even when dealing with science!

To calculate the vapor pressure of water at 42°C, we can use the information given about the wet gas.

Step 1: Find the moles of gas collected over water.
We already have the equation: n = (PV)/(RT).
Using the given values, we have n = (0.986 atm)(1.00 L)/(0.0821 L⋅atm/(K⋅mol))(315 K).

Step 2: Find the moles of water vapor in the gas.
Since the gas was collected over water, the total pressure measured includes the vapor pressure of water. We need to subtract the pressure due to the dry gas to find the pressure due to water vapor.
The pressure due to the dry gas is 0.986 atm - 1.00 atm = -0.014 atm.

Step 3: Find the vapor pressure of water at 42°C.
The total pressure measured is due to both the dry gas and the water vapor. So, the vapor pressure of water at 42°C is equal to the pressure due to water vapor, which is -0.014 atm.

Therefore, the vapor pressure of water at 42°C is -0.014 atm.

To calculate the vapor pressure of water at 42°C, we can use the ideal gas law equation, PV = nRT. But first, we need to determine the number of moles of gas present in the dried sample.

Given:
Initial volume of wet gas, V₁ = 1 liter
Pressure of wet gas, P₁ = 0.986 atm
Dried gas volume, V₂ = 1.04 liters
Pressure of dried gas, P₂ = 1.00 atm
Temperature of dried gas, T₂ = 90°C = (90 + 273.15) K

Step 1: Calculate the number of moles of dried gas
Using the ideal gas law equation PV = nRT, we can rearrange it to solve for n (number of moles):
n₂ = (P₂ * V₂) / (R * T₂)

R is the ideal gas constant and is equal to 0.0821 L·atm/(mol·K).

Substituting the known values, we get:
n₂ = (1.00 atm * 1.04 L) / (0.0821 L·atm/(mol·K) * 363.15 K)

Simplifying the equation:
n₂ ≈ 0.035 mol

Therefore, the number of moles of dried gas is approximately 0.035 mol.

Step 2: Determine the number of moles of water vapor present in the wet gas
Since the wet gas was collected over water, it contains water vapor. We need to determine the number of moles of water vapor in the initial wet gas sample, considering that the volume of the dried gas was greater.

The number of moles of water vapor in the initial wet gas sample is the difference in moles between the number of moles in the dried gas and the number of moles that would occupy the original wet gas volume at the same conditions.

n₁ - n₂ = (P₁ * V₁) / (R * T₁) - (P₂ * V₂) / (R * T₂)

T₁ = 42°C = (42 + 273.15) K

Substituting the known values, we get:
n₁ - 0.035 = (0.986 atm * 1 L) / (0.0821 L·atm/(mol·K) * 315.15 K) - (1.00 atm * 1 L) / (0.0821 L·atm/(mol·K) * 363.15 K)

Simplifying the equation:
n₁ ≈ 0.040 mol

Therefore, the number of moles of water vapor in the initial wet gas sample is approximately 0.040 mol.

Step 3: Calculate the vapor pressure of water at 42°C
Now that we know the number of moles of water vapor in the initial wet gas sample (n₁ ≈ 0.040 mol), we can determine the vapor pressure of water at 42°C.

The vapor pressure of water is the partial pressure of water vapor in the sample. Since the total pressure of the wet gas (P₁) is equal to the sum of the partial pressures of the gas and the water vapor, the vapor pressure of water can be calculated as follows:

Vapor Pressure = P₁ - (P₂ * (n₂ / n₁))

Substituting the known values, we get:
Vapor Pressure = 0.986 atm - (1.00 atm * (0.035 mol / 0.040 mol))

Simplifying the equation:
Vapor Pressure ≈ 0.958 atm

Therefore, the vapor pressure of water at 42°C is approximately 0.958 atm.