4.00 g of He, 20.0g F2, and 12.0 g Ar are placed in a 12.0-L container at 18.0 °C. The total pressure
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let ,
Pt = total pressure
no of mole of He = 4/4 = 1
no of mole of F2 = 20/38 = 0.526
no of mole of Ar = 12/40 =3/10 = 0.3
total no of mole of all gases = 1 +0.526 + 0.3 = 1.826
now , use PV= nRT
here P = Pt
V = 12 L
n = 1.826
R = 0.082
T = 18 +273 = 191 K
so,
Pt × 12 = 1.826 × 0.082 × 291
Pt = 1.826 × 0.082 × 291/12
= 3.63 atm
Pt = total pressure
no of mole of He = 4/4 = 1
no of mole of F2 = 20/38 = 0.526
no of mole of Ar = 12/40 =3/10 = 0.3
total no of mole of all gases = 1 +0.526 + 0.3 = 1.826
now , use PV= nRT
here P = Pt
V = 12 L
n = 1.826
R = 0.082
T = 18 +273 = 191 K
so,
Pt × 12 = 1.826 × 0.082 × 291
Pt = 1.826 × 0.082 × 291/12
= 3.63 atm
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