Question

What is the volume in mL of 3.59 g CO2 of gas at 2.09 atm and -14 oC?

Answer 1

800 ml is the volume in mL of 3.59 g CO2 of gas at 2.09 atm and -14°C.

Explanation:

Data given:

mass of the carbon dioxide gas = 3.59

pressure of the carbon dioxide gas =  2.09 atm

temperature of the carbon dioxide gas  = -14 degrees OR 273.15 +(-14) = 259.15 K

atomic mass of carbon dioxide =44.01 grams/mole

R = 0.0821 L.atm/mole K

volume occupied by the gas =?

number of moles of CO2 = $\frac{mass}{atomic mass of 1 mole}$

number of moles = $\frac{3.59}{44.01}$

number of moles of carbon dioxide gas = 0.081 moles

from ideal gas law

PV = nRT

V = $\frac{0.081 X 0.0821 X 259.15}{2.09}$

V = 0.80 L is the volume of carbon dioxide gas

volume in ml = 0.8 x 1000

volume in ml= 800 ml

Answer 2

The volume in mL will be "800 mL".

Given that:

• Mass of CO₂ = 3.59
• Pressure of CO₂ = 2.09 atm
• Temperature of CO₂ = -14°C

                                           = 259.15 K

• Atomic mass of CO₂ = 44.01 g/mol

Now,

The number of moles of CO₂ will be:

= $\frac{Mass}{Atomic \ mass}$

= $\frac{3.59}{44.01}$

= $0.081 \ moles$

By using the Ideal gas law, we get

→ $PV = nRT$

or,

→    $V = \frac{nRT}{P}$

By substituting the values, we get

         $= \frac{0.081\times 0.0821\times 259.15}{2.09}$

         $= 080 \ L$

or,

         $= 0.8\times 1000$

         $= 800 \ mL$

Thus the above answer is appropriate.

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