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Nadusha1986 [10]
3 years ago
15

What is the volume, in l, of 0.682 moles of o'what 2 ​ gas having a temperature of 68.2 ºc and pressure of 5.82 atm?

Chemistry
1 answer:
klemol [59]3 years ago
5 0

We will use Ideal gas equation. PV=nRT where P= pressure(in atm)=5.82atm V=Volume=? n= moles of gases=0.682 R = gas constant= 0.0821L atm / mol K T = temperature= 68.2 C= 68.2 + 273.15 = 341.35 K V= nRT/ P= 0.682 X 0.0821 X 341.35/ 5.82 V= 3.284 L.  

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Remark

The question with these kind of problems is "Which R do you use?" That's where dimensional analysis is so handy. You must look at the units of the givens and choose your R accordingly. You'll see how that works in a moment.

You need to list the givens along with their units and in this case the property you want to solve for. You need all that to determine the R value

Givens

n = 0.25 moles

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Which R

The units of the R you want has to have units of moles, kPa, °K and liters

The R that  you want is 8.314

<em><u>Formula</u></em>

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P 6.23 = 0.25 * 8.314 * 308.15  Combine the left

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P = 640.5/6.23 = 102.81  The answer should be 100 kpA of 1.0 * 10^2 kPa

because the number of moles has only 2 sig digs.

But if sig digs are not a problem 102.8 is likely close enough.

Second Question

You are going to have to clean up the numbers. I think I've got only 1 chance at this. The partial pressures of the 2 gases will add up to the total pressure. So the total pressure was 100 approx and the water vapor was 3.36 kPa. The difference is

Total = air + water vapor

100.18 = air + 3.36 about  Subtract 3.36 from both sides.

100.18 - 3.36 = 96.82 about. Pick the answer that is closest to that. I'll clean up the numbers if I can.

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