# (Dry) Volume Ideal Gas Law Calculation

• sandmanvgc
In summary: So, in summary, for calculating the dry volume you subtract the pressure of water vapour, and for calculating the wet volume you add the volume of water vapour.

#### sandmanvgc

Homework Statement
If the question was asking for (dry) volume, how would you do that?
Relevant Equations
In photos below
If the question was asking for (dry) volume, how would you do that?

There is a container with a mix of water vapor and oxygen.
With no vapor of water present in same container, pressure of oxygen would be 1 atm at 35° F.

Lnewqban said:
With no vapor of water present, pressure of oxygen would be 1 atm.
Why wouldn’t you subtract 0.555 due to water vapw from there?

Perhaps I didn't understand your original question.
Sorry.
Let's wait for other members to post.

sandmanvgc said:
Why wouldn’t you subtract 0.555 due to water vapw from there?

That's what they did, no? They calculated volume of the gas collected using partial pressure of the oxygen (so the volume calculated is the wet gas volume, as water @ 0.0555 atm occupies exactly the same volume, just increasing the total pressure). Dry volume would be closer to C.

Sadly, looks like whoever wrote the calculations down made a classical beginner's mistake and rounded down all intermediate numbers, so the errors accumulate and answers given are difficult to reproduce. Dry volume should be 436 mL, wet volume should be 462 mL. That's assuming 8.00 g of Ag2O, as technically 8 g has only one significant digit.

Lnewqban
Another way of doing this is to determine the total number of moles of gas first. The mole fraction of water is 0.0555, so the number of moles of water is (0.017)(0.0555)/0.9445=0.0010. So the total number of moles of gas is 0.018. This number of moles of gas is at 1 atm. See what you get for the volume of this wet gas mixture from the ideal gas law.

chemisttree and Lnewqban
Borek said:
That's what they did, no? They calculated volume of the gas collected using partial pressure of the oxygen (so the volume calculated is the wet gas volume, as water @ 0.0555 atm occupies exactly the same volume, just increasing the total pressure). Dry volume would be closer to C.

Sadly, looks like whoever wrote the calculations down made a classical beginner's mistake and rounded down all intermediate numbers, so the errors accumulate and answers given are difficult to reproduce. Dry volume should be 436 mL, wet volume should be 462 mL. That's assuming 8.00 g of Ag2O, as technically 8 g has only one significant digit.
Why don't you subtract the pressure due to water vapor when calculating the dry volume?

sandmanvgc said:
Why don't you subtract the pressure due to water vapor when calculating the dry volume?

You are asking the same question for the second time. They DID subtract. Total pressure is 1 atm, dry oxygen pressure is 0.945 atm

Borek said:
You are asking the same question for the second time. They DID subtract. Total pressure is 1 atm, dry oxygen pressure is 0.945 atm
In the question they solve for (Wet) Volume

My original question in the topic was if instead they were asking for the (Dry) volume, what would be the difference?

Dry volume should be 436 mL, wet volume should be 462 mL

Going by your answers, it seems when you calc'ed the dry volume, difference is that you don't subtract the pressure due to water vapor from the barometric reading given in the question. I'm confused on this concept and thought if calculating the dry volume you would subtract the pressure due to H20 Vap. Going back to original question is why do you subtract for finding Wet volume but not for the dry?

If you measured the wet volume, you would have to subtract the volume of water vapour to get the dry volume. If, as in the question, you want to calculate the wet volume, you have to add to the dry volume the volume of water vapour (which is equivalent to subtracting the vapour pressure of water).

## 1. What is the ideal gas law?

The ideal gas law is a mathematical equation that describes the relationship between the pressure, volume, temperature, and number of moles of an ideal gas. It is expressed as 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.

## 2. How do you calculate the volume of an ideal gas using the ideal gas law?

To calculate the volume of an ideal gas using the ideal gas law, you need to rearrange the equation to solve for V. This can be done by dividing both sides of the equation by P, giving you V = nRT/P. Then, you can plug in the values for n, R, T, and P to solve for V.

## 3. What is the difference between dry volume and volume of an ideal gas?

Dry volume refers to the volume of a gas without taking into account the water vapor present in the gas. On the other hand, the volume of an ideal gas takes into account the water vapor present in the gas. This is important because water vapor can affect the pressure and volume of a gas, and therefore, the ideal gas law calculation.

## 4. What units should be used when plugging in values for the ideal gas law?

The ideal gas law equation requires that pressure be measured in atmospheres (atm), volume in liters (L), temperature in Kelvin (K), and the gas constant (R) in units of L*atm/mol*K. It is important to make sure all units are consistent when plugging in values for the ideal gas law calculation.

## 5. Can the ideal gas law be used for all gases?

The ideal gas law is most accurate for ideal gases, which are gases that follow the kinetic theory of gases and have no intermolecular forces. However, it can also be used for real gases under certain conditions, such as low pressure and high temperature. In these cases, the ideal gas law may not be as accurate, but it can still provide a reasonable estimate.