Ideal gas law - looks like easy question

[Stacker]

Could someone check my answer.

"If a sample of CO2 gas has a mass of 1.5 grams when the atmospheric pressure is 202 kPa and the temperature is 25 degrees C, what would the density of the gas be at STP?"

Solve for initial volume under conditions 202kpa (1.99 atm), 1.5g CO2, and 25C (298K)
PV = N * R * T is modified to PV = (mass/molar mass) * R * T
(1.99 atm) (V) = (1.5 g /44 g/mol) (0.0821 L*atm/mol*K) (298K)
V = 0.419 L

At STP, conditions are always 1 atm, 273K, and 1 mole gas = 22.4 L
(P1) * (V1)/ (T1) = (P2) * (V2)/ (T2) is modified to V2 = V1 * (P1/P2) * (T2/T1)
V2 = (0.419 L) * (1 atm/ 1.99 atm) * (273K/ 298K) = 0.193 L

Density is (moles/L).
0.034 moles/ 0.193 L = 0.18 moles/L

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[FROGGBUSTER]

looks good to me

 

[Jab1113]

Could someone check my answer.

"If a sample of CO2 gas has a mass of 1.5 grams when the atmospheric pressure is 202 kPa and the temperature is 25 degrees C, what would the density of the gas be at STP?"

Solve for initial volume under conditions 202kpa (1.99 atm), 1.5g CO2, and 25C (298K)
PV = N * R * T is modified to PV = (mass/molar mass) * R * T
(1.99 atm) (V) = (1.5 g /44 g/mol) (0.0821 L*atm/mol*K) (298K)
V = 0.419 L

At STP, conditions are always 1 atm, 273K, and 1 mole gas = 22.4 L
(P1) * (V1)/ (T1) = (P2) * (V2)/ (T2) is modified to V2 = V1 * (P1/P2) * (T2/T1)
V2 = (0.419 L) * (1 atm/ 1.99 atm) * (273K/ 298K) = 0.193 L

Density is (moles/L).
0.034 moles/ 0.193 L = 0.18 moles/L

I dont follow this. What is the correct answer??

Your density equation is wrong its D= (mass/volume)

And why wouldn't you just manipulate the ideal gas law from PV=nRT and make it P(MW) = D*R*T (MW = molecular weight, D = density, R = gas constant, T = Temperature, P = pressure)?

Im a little confused on this question, lemme know what the correct answer is and then we can work from there.

 

[Stacker]

I dont follow this. What is the correct answer??

Your density equation is wrong its D= (mass/volume)

And why wouldn't you just manipulate the ideal gas law from PV=nRT and make it P(MW) = D*R*T (MW = molecular weight, D = density, R = gas constant, T = Temperature, P = pressure)?

Im a little confused on this question, lemme know what the correct answer is and then we can work from there.

I found this question on yahoo answers so I don't know the correct answer.
You're right density is mass/volume, my bad.

 

[Jab1113]

I found this question on yahoo answers so I don't know the correct answer.
You're right density is mass/volume, my bad.

Better to make the mistake on SDN then on the DAT lol

I think the whole calculating it at STP is throwing me off. It gives us regular values where we are not at STP, we know that 1kPa = 7.5 Torr and 1 atm=760 torr so that how you got the 1.99 atm.

So I think you did everything right. Just at the end if you do mass/volume then you should get the answer.

Anybody else wanna help?

 

[vicsin]

I memorized the formula density= PMM/RT
= simple to manipulate but useful to know .

 

[yappy]

Aye, from Kaplan.

Density = (Partial pressure * Molar mass)/ (RT)

I memorized the formula density= PMM/RT
= simple to manipulate but useful to know .

 

[rockclock]

Could someone check my answer.

"If a sample of CO2 gas has a mass of 1.5 grams when the atmospheric pressure is 202 kPa and the temperature is 25 degrees C, what would the density of the gas be at STP?"

Solve for initial volume under conditions 202kpa (1.99 atm), 1.5g CO2, and 25C (298K)
PV = N * R * T is modified to PV = (mass/molar mass) * R * T
(1.99 atm) (V) = (1.5 g /44 g/mol) (0.0821 L*atm/mol*K) (298K)
V = 0.419 L

At STP, conditions are always 1 atm, 273K, and 1 mole gas = 22.4 L
(P1) * (V1)/ (T1) = (P2) * (V2)/ (T2) is modified to V2 = V1 * (P1/P2) * (T2/T1)
V2 = (0.419 L) * (1 atm/ 1.99 atm) * (273K/ 298K) = 0.193 L

Density is (moles/L).
0.034 moles/ 0.193 L = 0.18 moles/L

You guys fell for the trap...density of CO2 at STP is independent of all of the initial conditions they gave you:

(1 mol/22.4 L)*(44 g C/1 mol C) = 2.0 g/L

 

[rockclock]

Classic DAT question btw...right answer is a simple calculation you can do in your head; falling into the trap will cost you time and still get you the wrong answer.

 

[yappy]

Rockclock's mind is so sharp. It's insane

Classic DAT question btw...right answer is a simple calculation you can do in your head; falling into the trap will cost you time and still get you the wrong answer.