What is the volume, in l, of 0.691 moles of o_2 2 ? gas having a temperature of 66.1 ? c and pressure of 5.91 atm

The volume of gas is 3.26 L.

Explanation:

Given data:

Pressure = P = 5.91 atm

Temperature = T= 66.1 °C = 66.1 +273.15 = 339.25 K

Number of moles of oxygen gas= n = 0.691 mol

Volume = V = ?

Formula:

PV = nRT

R is general gas constant and its value is  0.0821 dm³.atm.K⁻¹.mol⁻¹ .

Now we will put the values in equation.

V = nRT/ P

V = ( 0.691 mol× 0.0821 dm³.atm.K⁻¹.mol⁻¹ × 339.25 K) / 5.91 atm

V = 19.25 dm³. atm/  5.91 atm

V = 3.26 dm³

   or

V = 3.26 L   (dm³= L)

first one is the correct answer