Solution:

Using the ideal gas law \( PV = nRT \), we find the moles \( n \) in each vessel:

1. For vessel \( A \) with \( O_2 \):
\[
n_{\text{O}_2} = \frac{PV}{RT}
\]

2. For vessel \( B \) with \( H_2 \) at \( 2P \):
\[
n_{\text{H}_2} = \frac{2PV}{RT} = 2 \times \frac{PV}{RT}
\]

Now, the mass \( m = n \times \text{molar mass} \):

- Mass of \( O_2 \) in \( A = n_{\text{O}_2} \times 32 = \frac{PV}{RT} \times 32 \)
- Mass of \( H_2 \) in \( B = n_{\text{H}_2} \times 2 = 2 \times \frac{PV}{RT} \times 2 \)

So, the mass ratio is:
\[
\frac{\text{mass of } O_2}{\text{mass of } H_2} = \frac{\frac{PV}{RT} \times 32}{2 \times \frac{PV}{RT} \times 2} = \frac{32}{4} = 8:1
\]