Combined Gas Law

Question

1. A 42.0-g sample of krypton gas has a volume of 11.2 L at STP. The sample is heated to 112ºC and compressed to a volume of 6.83 L. What is the resulting pressure?

 

 2.A gas takes up a volume of 17.0 liters, has a pressure of 2.30 atm, and a temperature of 299 K. If I raise the

    temperature to 350. K and lower the pressure to 1.50 atm, what is the new volume of the gas

Answer 1

Hello, Emma,

1. We can use a simplified version of the ideal gas law for this question. Since there is no change in the moles of gas between the two states, we can write:

P1V1/T1 = P2V2/T2

This is a result of two ideal gas law equations for the two states 1 and 2, in which both the moles, n, and the gas constant, R, both cancel. Rearrange this equation to solve for the unknown, P2, the final pressure:

P2 = P1(V1/V2)(T2/T1)

Note how I've grouped the volume and temperature values, into their own ratios. This helps us 1) keep track of what is happening, 2) allows easy cancelling of units, and 3) reduces transcription errors when entering data.

The gas laws require Kelvin, the absolute temperature scale. Add 273.15 to each ^oC value. Standard temperature and pressure are 0^oC (273.15K) and 1 atm.

Now enter the data and cancel the units. We can see that both temperature (^oC) and volume (L) units will cancel, leaving just pressure, in atm, from the P1 term.

P2 = (1 atm)*(11.2L/6.83L)*(385.2K/273.2K)

P2 = 2.31 atm

===

The second part of the question is solved in a similar manner. The moles of gas do not change, so we can use the same equation P1V1/T1 = P2V2/T2. This time we rearrange for V2, the final volume.

V2 = V1(T2/T1)(P1/P2)

V2 = (1.7L)*(350K/299K)*(2.3atm/1.50atm)

V2 = 30.5 liters

Bob