Liquid brasses conform to the following relationship: 2 RTInγ^ =-5000xr , Zn Cu where R and T are in cal/deg/mole and K respectively. The vapour pres­ sures of pure copperand pure zinc (in mm Hg) are given as l0= 9 PCu -^τ^- 1.21 log T+ 13.21, and log pZn = - £252. - 0.755 log T + 11.24. Calculate (i) the vapour pressure of copper over a brass containing 40 atom% zinc at 1227°C (1500 K), and ( i i ) the vapour pressure of zinc over the above brass at the same temperature.

To solve this problem, we need to use the given equations and plug in the given values.

(i) The vapour pressure of copper over a brass containing 40 atom% zinc at 1227°C (1500 K)

First, we need to calculate the activity coefficient (γ) using the given equation:

2 RTInγ = -5000x_Zn

where R = 1.987 cal/deg/mole, T = 1500 K, and x_Zn = 0.4 (40 atom% zinc)

Solving for γ, we get:

Inγ = -50000.4 / (21.987*1500) γ = e^(Inγ)

Next, we use the equation for the vapour pressure of pure copper (p_Cu):

log p_Cu = - 1.21 log T + 13.21

Substitute T = 1500 K into the equation to get p_Cu.

Finally, the vapour pressure of copper over the brass is given by:

p_Cu_brass = γ * p_Cu

(ii) The vapour pressure of zinc over the above brass at the same temperature

We use the same process as above, but this time we use the equation for the vapour pressure of pure zinc (p_Zn):

log p_Zn = - 252. - 0.755 log T + 11.24

Substitute T = 1500 K into the equation to get p_Zn.

Finally, the vapour pressure of zinc over the brass is given by:

p_Zn_brass = γ * p_Zn

Remember to convert the final answers from log scale to normal scale by taking the antilog.