What volume of “wet” methane would have to be collected at 20.0 oC and 760.0 torr to be sure the sample contained 2.00 x 102 ml of dry methane at the same temperature and pressure ?
My Work: All I have so far is that I use P1V1=P2V2 but I don't know the difference between wet and dry methane?
The terms "wet" and "dry" methane refer to the presence or absence of other gases or impurities in the methane sample. In this context, "dry" methane means it only contains pure methane without any other gases or impurities, while "wet" methane may contain some amount of water vapor or other impurities.
To solve the problem, we can use the ideal gas law equation, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
Step 1: Convert the given temperature from Celsius to Kelvin.
Adding 273.15 to the temperature in Celsius gives us:
T = 20.0 oC + 273.15 = 293.15 K
Step 2: Calculate the number of moles of dry methane using the ideal gas law.
Rearranging the equation, we have:
n = PV / RT
Given:
P = 760.0 torr
V = 2.00 x 10^2 mL = 2.00 x 10^2 cm^3 = 2.00 x 10^2 / 1000 L = 0.200 L
R = 0.0821 L atm/(mol K) (ideal gas constant)
T = 293.15 K
Substituting the values into the equation:
n = (760.0 torr) * (0.200 L) / ((0.0821 L atm/(mol K)) * (293.15 K))
Calculating this will give you the number of moles of dry methane.
Step 3: Convert the number of moles of dry methane to volume of wet methane.
Since methane is an ideal gas, the number of moles of wet methane will be the same as that of dry methane.
Using the equation from Step 2, we have:
n = PV / RT
Rearranging the equation to solve for V, we get:
V = nRT / P
Now substitute the values:
n = (number of moles of dry methane, calculated in Step 2)
R = 0.0821 L atm/(mol K)
T = 293.15 K
P = 760.0 torr
Calculating this will give you the volume of wet methane that needs to be collected at the specified temperature and pressure to ensure you have 2.00 x 10^2 mL of dry methane.
To determine the volume of wet methane required to ensure a sample contains a specific volume of dry methane at the same temperature and pressure, you'll need to understand the difference between wet and dry methane.
Wet methane refers to methane gas that contains water vapor, while dry methane refers to pure, water-free methane gas. The presence of water vapor affects the total volume and pressure of the gas sample.
To solve this problem, you can use the ideal gas law equation, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
Here's the step-by-step process to solve the problem:
1. Convert the given temperature of 20.0 °C to Kelvin by adding 273 to it. The temperature in Kelvin would be 293 K.
2. Calculate the number of moles of dry methane in the given volume (2.00 x 102 ml). To do this, divide the volume by the molar volume of an ideal gas at standard temperature and pressure (STP). The molar volume at STP is approximately 22.4 L/mol. Therefore, the number of moles of dry methane would be (2.00 x 102 ml) / (22.4 L/mol).
3. Apply the ideal gas law equation to the wet methane sample. Assuming the volume of wet methane is V1 and the volume of dry methane is V2, the equation becomes: (760.0 torr) * V1 = (2.00 x 102 ml) * (760.0 torr).
4. Solve for V1 by dividing both sides of the equation by 760.0 torr.
V1 = [(2.00 x 102 ml) * (760.0 torr)] / (760.0 torr)
5. Finally, substitute the temperature, pressure, and volume values into the equation and calculate the volume of wet methane.
Remember to express the volume in liters and ensure that the pressure and volume have consistent units (e.g., convert torr to atmospheres if necessary).
By following these steps, you should be able to determine the volume of wet methane required to contain 2.00 x 102 ml of dry methane at the given temperature and pressure.
So this is what I did
706torr-17.54=742.46
Then plugged into P1V1=P2V2
(742.46)V1=(760)(200)
Solved for V1, and I got 205ml (3 sig figs)
Is this right? Thanks for all your help!
The dry vs wet means that when you collect a gas over water, the total pressure is what you read on the room manometer and that total pressure is the pressure of the CH4 + the vapor pressure of H2O at that temperature. You can look up the vapor pressure of water at the temperature used.
Ptotal = pCH4 + pH2O. Then plug into that equation and solve for pCH4 which will be the pressure of the dry gas.
So you want 200 mL at 760 mm and you will have x mL at 760-v.p. H2O. Solve for x mL.
Post your work if you get stuck.