Two vessels A and B contain same mass of Oxygen and Hydrogen respectively at the same temperature. The volume of B is twice that of A. The ratio of gas pressure in A to that in B is

Using the ideal gas equation, $PV = nRT = \dfrac{m}{M}RT$, where $m$ is the given mass and $M$ is the molar mass.

For vessel A containing Oxygen ($O_2$, $M_A = 32$):
$P_A = \dfrac{mRT}{32 V_A}$

For vessel B containing Hydrogen ($H_2$, $M_B = 2$):
$P_B = \dfrac{mRT}{2 V_B}$

Given that the volume of B is twice that of A ($V_B = 2V_A$), we get:
$P_B = \dfrac{mRT}{2(2V_A)} = \dfrac{mRT}{4V_A}$

Taking the ratio of the pressures:
$\dfrac{P_A}{P_B} = \dfrac{\dfrac{mRT}{32 V_A}}{\dfrac{mRT}{4 V_A}} = \dfrac{4}{32} = \dfrac{1}{8}$

The ratio of gas pressure in A to that in B is $1:8$.

Answer: $1:8$