Answer
The heavier gas has a smaller rms speed. So, the root-mean-square speed of gas A will be less. We know force $F$ is nothing but the change of momentum per unit time. Here momentum remains constant for both gases as well as area. So, there is no difference in pressure and hence no contradiction.
Work Step by Step
At a specific temperature, all gases have the same average kinetic energy. So, for gas $A$ and $B$ we have $3/2 RT=N⟨ϵ_A ⟩$ and $3/2 RT=N⟨ϵ_B ⟩$ where $N$ is Avogadro constant. then we can easily write
$⟨ϵ_A ⟩=⟨ϵ_B ⟩$
or $½m_A ⟨c_A^2 ⟩=½m_B ⟨c_B^2 ⟩$
The root-mean-square speed, $c_rms$ is define by $c_rms=√(⟨c^2 ⟩ $
The ratio of the root-mean-square speeds of two molecules of different masses is equal to the square root of the inverse ratio of the masses
$(c_rms )_1/(c_rms )_2 =√(m_2/m_1 )=√(M_2/M_1 )$
Where $M=N_A m$ is the molar mass. The heavier gas has a smaller rms speed. So, the root-mean-square speed of gas $A$ will be less. We know force $F$ is nothing but the change of momentum per unit time. Here momentum remains constant for both gases as well as area. So, there is no difference in pressure and hence no contradiction.
