Question

How many moles are contained in a 4.67-l sample of gas at 30°c and 199 kpa?

Answer 1

pv=nRT
where,p=199, R(constant)=8.314, V=4.67 T=30C=293K
n=pv/RT=0.38 moles

Answer 2

Answer is: 0.37 moles are contained.

p = 199 kPa; pressure of the gas.

T = 30°C = 303.15 K; temperature of the gas.

V = 4.67 L; volume of the gas.

R = 8.314 J/K·mol; gas constant..

Use ideal gas law: p·V = n·R·T.

n = p · V ÷ (R · T).

n = 199 kPa · 4.67 L ÷ (8.314 J/K· mol · 303.15 K).

n = 0.37 mol; amount of the gas.