Mixed Gas Question 4

Question

1.If I have 91 liters of gas held at a pressure of 1.6 atm and a temperature of 816 K, how many moles of gas do I have? Answer with correct significant figures and unit spelled out completely.

R = 0.0821 atm. L.mole^-1 .K^-1

2.If some neon gas at 158 kPa were allowed to change from 15 dm³ to 70 dm³ under constant temperature, what pressure, in kPa, would the neon gas exert under these new conditions?  Answer with correct significant figures and unit in abbreviation

Answer 1

Hello, Jahnine,

a) Use the Ideal Gas Law to calculate the number of moles under these conditions.

PV=nRT, rearrange to solve for n

n = PV/RT

n = (1.6atm)(91L)/(0.0821atm*L*mole^-1K^-1)(816K)

n = 2.17 moles 2.2 moles for two sig figs.

b) Use the Combined Gas Law P1V1/T1 = P2V2/T2, where the subscripts 1 and 2 represent the initial and final states, respectively.

We want tio find P2, the new pressure. Rearrange the equation for P2:

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

Note how I arranged the volume and temperature variables - as ratios. This makes it easier to enter the date, cancel units, and underswtand what is happening. We are increasing volume at constant temperature. The temperature ratio cancels out, and we can now see that since (V1/V2) is smaller than 1 (15dm³/70dm³), we'd expect a significant drop in pressure.

P2 = P2(V1/V2), where (V1/V2) = (15dm³/70dm³) [That's around a 75% drop, so we should expect that order of magnitude in the final result]

Since the temperature and volume units cancel, we can stick with kPa, the desired unit. I calculate 33.86 kPa. 34 kPa for 2 sig figs. That's about the magnitude we expected.

Bob