Answer:
P = 4.0794 atm or 413,345.2 Pa
Explanation:
Given
Mass m = 14.0 g
Volume 3.5 L
Temperature T = 75° C = 75 + 273 = 348 K
Known Values
Molar mass of CO = 28 g/mol
Universal gas constant R = [tex]0.082057 L.atm/mol.K[/tex]
Solution
Number of moles in 14 g of CO is
[tex]\\n=\frac{mass}{molar.mass} \\\\n = \frac{14}{28} \\\\n = 0.5 mole[/tex]
[tex]PV = nRT\\\\P = \frac{nRT}{V} \\\\P = \frac{0.5 \times 0.082057 \times 348}{3.5} \\\\P = 4.0794 atm\\\\P = 4.0794 \times 101325\\\\P = 413,345.2 Pa[/tex]
