Question

At what pressure will 36 grams of CO2 occupy 28.9 L at standard temperature

Answer 1

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

$p= \frac{nRT}{V} \\ = \frac{0.818 \times 0.08206 \times 273}{28.9} = 0.6342 \;atm$

Answer: 0.6342 atm