A sample of air was compressed to a volume of 18.0 L. The temperature was 290.0 K, and the pressure was 6.80 atm. How many moles of gas were in the sample?

If the sample had been collected from air at P = 2.0 atm., what was the original volume of the gas (in liters)?

How do I approach this?

#1. PV = nRT and solve for n = mols.

#2. Using n from #1, use PV = nRT and solve for V.

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To approach this problem, you can use the Ideal Gas Law equation, which states:

PV = nRT

Where:
P = pressure of the gas (in atm)
V = volume of the gas (in liters)
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/(K·mol))
T = temperature of the gas (in Kelvin)

To calculate the number of moles of gas in the sample, we need to rearrange the Ideal Gas Law equation to solve for n:

n = PV / (RT)

Now let's calculate the number of moles of gas in the sample given the provided values:

P = 6.80 atm
V = 18.0 L
T = 290.0 K
R = 0.0821 L·atm/(K·mol)

n = (6.80 atm) * (18.0 L) / (0.0821 L·atm/(K·mol) * 290.0 K)

n ≈ 4.10 moles

Therefore, there are approximately 4.10 moles of gas in the sample.

Now, let's move on to the second part of the question. If the sample had been collected from air at P = 2.0 atm and we want to find the original volume of the gas, we can rearrange the Ideal Gas Law equation:

V = nRT / P

Given that we already know the following values:
P = 2.0 atm
n = 4.10 moles
R = 0.0821 L·atm/(K·mol)
T = 290.0 K

V = (4.10 moles) * (0.0821 L·atm/(K·mol)) * (290.0 K) / (2.0 atm)

V ≈ 23.85 L

Therefore, the original volume of the gas collected from air at P = 2.0 atm is approximately 23.85 liters.