Gas Laws in Chemistry

Question

Which sample contains more atoms: 1,340 mL of neon at 35.5 ˚C and 840 mmHg or 1.48 L of argon at 48.9 ˚C and 0.790 atm? 

moles Ne = _____mol

moles Ar =  _____mol

Which sample contains more:

1. Neon
2. Argon
3. Need more Information
4. Equal Amounts

Answer 1

PV = nRT

nR = PV/T (R can does need to be included in the calculations for comparisons)

Ne

given :

P = 840 mmHg x 1 atm/760 = 1.105 atm

V = 1.340 L

T = 35.5 ˚C = 318.65 K

nR = 1.105 x 1.34 ÷ 318 = 0.0047 R

Ar

given :

P = 0.790 atm

V = 1.48 L

T = 48.9 ˚C = 322.05 K

nR = 0.79 x 1.48 ÷ 322.05 = 0.0036 R

Answer :

There is more Ne

NOTES:

1. I left units out as this was just a comparison. To calculate the actual mols, use units and multiply by R (0.0821 L*atm/(mol*K)
2. Technically, 840 mmHg is 2 significant figures. (840. mmHg would be 3)