Question

What is the volume occupied by 0.254 mol of helium gas at 1.32 atm and 308 K ?

Answer 1

Given information about He gas is-

Moles of He = n = 0.254 mols

Pressure of He = P=1.32 atm

Temperature= T=308 K

R (ideal gas constant) = 0.0821 L atm K​​​​​-1 mol​​-1

The ideal gas equation is-

PV = nRT

Putting the given values we have -

1.32 × V = n × R × T

1.32 × V = 0.254 × 0.0821 × 308

V = 6.423 ÷ 1.32 = 4.829 L.