Ideal Gases

Problem : Molly admires her red balloon, which has a volume of 2.0 liters at sea level (1.0 atm). A clown catches her eye, and she lets go of the balloon. The red balloon goes up and up until the pressure around it is 0.80 atm. Assuming isothermal conditions, what is the new volume of Molly's red balloon?

Solution for Problem 1 >>

This problem is a straightforward application of Boyle's law. P ₁ V ₁ = P ₂ V ₂ rearranges to = V ₂ . After plugging in values, we find that the balloon's volume V ₂ = 2.5 liters.

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Problem : Graph the pressure vs. volume relationship dictated by Boyle's law. If Boyle's law stated that P = aV , where a < 0 , what would the graph of P vs. V look like?

Solution for Problem 2 >>

A graph of P vs. V according to Boyle's law is shown below:

Boyle's Law

When P = aV , the graph of P vs. V will look like this:

P = aV

Note that P = aV cannot be true. The equation predicts negative volumes as P increases.

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Problem : Initially the volume and pressure of a sample of gas are 1 dm ³ and 10 smoots, respectively. The volume is raised isothermally to 10 dm ³ . What is the pressure of the gas in smoots under these conditions?

Solution for Problem 3 >>

Don't let the unfamiliar units of dm ³ and smoots confuse you. Your first reaction may to be convert to SI units. In this case you can't; smoots are completely imaginary. Instead realize that the equation P ₁ V ₁ = P ₂ V ₂ works as long as the units of P ₁ P ₂ and V ₁ V ₂ are the same. The actual units of pressure or volume don't matter. So rearrange the equation to = P ₂ . Plugging in values, we find that P ₂ = 1 smoot.

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Problem : One end of a mercury filled manometer is open to the atmosphere, while the other is closed and contains a vacuum. What does the height difference h of the Hg columns measure?

Problem : The pressure of gas A ( P _A ) is 3.0 atm. The height of the mercury column h is 1140 mm. What is the pressure of gas B ( P _B ) in atmospheres? Assume that 1 mm Hg = 1/760 atm.