posted by lindsay on .
1)A pocket of gas is discovered in a deep drilling operation. The gas has a temperature of 480c and is at a pressure of 12.8 atm. What volume of gas is required to provide 18.0 L of gas at the surface where the conditions are 22 C and 1.00 atm?
is this correct for number 1?
480c = 753k, 22c = 295k
p1 x v1/T1 = p2 x v2/T2
12.8 x v1/753 = 1.00 x 18.0/295
.016 = .061
.016 is V1
Hmmm. Put this in your calc or the google search window:
18* 1/12 * 753/295
I get almost 4 liters.
What is the volumeof 45.0 grams of NO at 20 c and a pressure of 740 mm Hg?
2) The molar mass is 16 + 12 = 28.00, so
45.0 g is n = 1.61 moles. That number of moles would occupy
22.4 x 1.61 = 36.06 liters at 1.00 atm and 273 K. At 20 C and 740 mm Hg. Now what?
First, recalcualte the molar mass of NO. I get more like 30 than 28.
Then solve one of two ways.
Use your NEW volume (using the correct molar mass) and P1V1/T2 = P2V2/T2 OR
use PV = nRT where n is the NEW number of mols.
The answer is about 37 L or so.