At what pressure will 36 grams of CO2 occupy 28.9 L at standard temperature
MM(CO2) = 44 g/mol
n(CO2) "= \\frac{36}{44}=0.818 \\;mol"
V=28.9 L
T= 273 K
Ideal Gas Law:
pV=nRT
R = 0.08206 L×atm/mol×K
"p= \\frac{nRT}{V} \\\\\n\n= \\frac{0.818 \\times 0.08206 \\times 273}{28.9} = 0.6342 \\;atm"
Answer: 0.6342 atm
Comments
Leave a comment