Question

A 10.45-m³ sample of gas (310K, 745mmHg pressure) was passed through a series of traps that removed teh water from the gas. The gas that came out was found to be at 780mmHg pressure and 273K. Its volume was 8.73-m³. What was the percentage of water vapor (v/v) in the original sample of gas? Assume ideal behavior for all gasous compounds.

Answer 1

Solution :-

lets first calculate moles of the gas initially

PV= nRT

10 45 m3 * 1000 L / 1 m3 = 10450 L

745 mm Hg * 1 atm / 760 mmHg = 0.9803 atm

n= PV/RT

n = 0.9803 atm * 10450 L / 0.08206 L atm per mol K * 310 K

n = 402.7 mol

Now lets calculate the moles of the gas in the final state

8.73m3 * 1000 L / 1 m3 = 8730 L

780 mmHg * 1 atm /760 mmHg = 1.026 atm

n= PV/RT

n= 1.026 atm * 8730 L / 0.08206 L atm per mol K * 273 K

n = 399.82 mol

now lets find the number of the water wapor moles lost

number of water vapor moles = 402.7 mol - 399.82 mol = 2.88 mol

now lets calculate the volume of the 2.88 mol at the initial conditions

402.7 mol = 10.45m3

2.88 mol = ?

2.88 mol * 10.45 m3 / 402.7 mol = 0.0747 m3

so now lets calculate the percent of water vapor in the original sample

% of water vapor = (volume of water vapor / total volume)*100%

                             = (0.0747m3/10.45m3)*100%

                             = 0.715 %

So the percent of the water vapor = 0.715%