Let there are n1 moles of hydrogen and n2 moles of helium in the given mixture. As Pv=nRT
Then the pressure of the mixture
P=n1RTV+n2RTV=(n1+n2)RTV
⇒2×101.3×103=(n1+n2)×(8.3×300)20×10−3
or, (n1+n2)=2×101.30×103×20×10−3(8.3)(300)
or, n1+n2=1.62....(1)
The mass of the mixture is (in grams)
n1×2+n2×4=5
⇒(n1+2n2)=2.5.....(2)
Solving the eqns. (1) and (2), we get
n1=0.74 and n2=0.88
Hence, mHmHe=0.74×20.88×4=1.483.52=25.