Question 13.10.3: The partial pressure of hydrogen in the atmosphere is such t...

For the equilibrium between gaseous hydrogen and dissolved H, given as ½H_{2(g)}=[H]_{(1  wt\%  in   Fe)},

In pure iron, as H obeys Henry’ s law, f_{H(1wt\%)}=1 ,  and thus,

Thus,

But

As f_{H(1  wt\%)}=1,  e^{H}_{H}=0 , and hence, at 1600°C,

When [wt\%  H]=5\times 10^{-4} ,

=-1.825

which gives p_{H_2}= 0.015  atm. Similarly, when [wt\%  H]=1\times 10^{-4},p_{H_2}=6\times 10^{-4}  atm. Thus,

[wt\%  H]=5  ppm   when  p_{H_2}=0.015  atm      and P_{(total)}=1  atm

And so

[wt\%  H]=1  ppm    when  p_{H_2}=0.0006  atm

and

P_{total}=\frac{0.0006}{0.015}=0.04   atm

Thus, in order to achieve the desired decrease in the content of dissolved H, the total pressure must be decreased from 1 to 0.04 atm.