Question in Chemistry

Question

A sample of krypton gas occupies a volume of 8.61 L at 52.0°C and 350. torr.

If the volume of the gas sample is increased to 11.0 L, while its temperature is decreased to –20.0°C, the resulting gas pressure will be__________ torr.

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A mixture of helium and krypton gases is maintained in a 6.98 L flask at a pressure of 2.66 atm and a temperature of 90 °C. If the gas mixture contains 0.699 grams of helium, the number of grams of krypton in the mixture is ________________ g.

Answer 1

Let's start with the first problem. I would recommend using this equation based off of the ideal gas law:

P₁V₁/T₁ = P₂V₂/T₂

Before we start plugging things in, make sure our units are correct. This equation was derived from the ideal gas law containing the gas constant R. There's a whole chart of gas constant forms using different variables. Find the one that has the most similar variables. The form I will use has units of L*torr per K*mole.

Comparing to the units in our problem, liters are good, torr is good, but oh no, we aren't in Kelvin! Let's convert that quickly. Our initial temperature of 52ºC translates to 325.15 K, and our final temperature of -20ºC translates to 253.15 K. Now we can plug in all our values.

We are solving for the new pressure, so let's plug in everything else. Our equation should look like this:

350(8.61)/325.15 = P₂(11)/253.15

Simply solve the equation using order of operations to get P₂=213 torr.

Conceptually, this checks out. We have not only increased the volume, but decreased the temperature significantly, so it's logical that our slower molecules generate a lower pressure in the vessel.

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Problem two also uses the ideal gas law, but actually uses the "original" version, which looks like this:

PV = nRT

Where P is the pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature. We're solving for the number of moles in this case, because the partial pressures of gas mixtures and the moles of those gasses are congruent!

I will be using a gas constant of R = 0.0821 L*atm per K*mole. We plug all our known values in, and don't forget to convert Celsius to Kelvin again!! Our new equation looks like:

2.66(6.98) = n(0.0821)(363.15)

We can solve for the number of moles in the tank by simplifying. We then get

n = 0.6227 moles

This means that there are 0.6227 moles TOTAL in the tank. We know there are 0.699g He in the tank, so let's see how many moles that is. Use stoichiometry - we should get:

0.699 g / 4.0026 g per mole = 0.175 moles He

So, now we know there are 0.175 moles of He, but our equation tells us there should be 0.6227 moles TOTAL. The remaining mystery moles must be the Kr! Solve for the moles of Kr:

0.6227 - 0.175 = 0.448 moles Kr

However they didn't ask for moles of Kr, they asked for g of Kr, so we can do a simple conversion:

0.448 moles Kr * 83.798 g/mole = 37.55 g Kr.