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 dа ta:

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)