A 3.00L flask contains 2.33g of argon gas at 312 mm Hg. What is the temperature of the gas?

Question

A 3.00L flask contains 2.33g of argon gas at 312 mm Hg. What is the temperature of the gas?

Answer 1

pV=nRT

p=312 mm Hg/760 mm/ATM

V=3 L

n=2.33 g/40 g/g-mole

R=0.0821 atm-L/mole-degreeK

Now solve.  

Answer 2

2.33g/40g/mol. = 0.058mol. Ar.

1 mol. at STP = 22.4L volume.

0.058mol. x 22.7L/mol. = 1.3L.

Using the Combined Gas Law.

P1 x V1 x T2 = P2 x V2 x T1.

760mmHg x 1.3L x 273K = 312mmHg x 3Lx T1

T1 = (760 x 1.3 x 273) / (312 x 3).

= 269,724 / 936 = 288K - 273 = 15°C.

Yep !, I must be getting old ...

Answer 3

PV = nRT

T = PV / nR

n = 2.33 g x (1 mole / 40 g ) = .05825 mole

R = .0821 L atm / mole K

P = 312 mm Hg x (1 atm / 760 mmHg) = 0.4105 atm

V = 3.00 L

T = [(0.4105 atm) x (3.00 L)] / [(0.05825 mole) x (0.0821 L atm / mole K)]

T = 258 K  {3 sig figs}

****** norrie *******

your T1 and T2 are reversed