4
For simplicity, let's assign numerical indices to the compounds of interest — all gaseous products participating in equilibrium:
$$\ce{ZnO(s) + \underset{1}{CO(g)} <=> \underset{2}{Zn(g)} + \underset{3}{CO2(g)}}$$
Partial pressure of carbon monoxide can be found via its mole fraction $x_1$ and given total pressure $p$:
$$p_1 = x_1p\tag{1}$$
To find ...
3
Yes, they are related. The first comes directly from the conservation of number of moles of the solute in a dilution,
$$n_1 = n_2 $$
Since $n = MV$,
$$M_1 V_1 = M_2 V_2$$
The second is related to the conservation of the total number of moles in an isothermal compression or expansion,
$$n_1 = n_2$$
Using the ideal gas law $n = pV/RT$,
$$\frac{P_1 V_1}{...
2
The van der Waals equation is
$$\left(p+\frac{a}{V_m^2}\right)\left(V_m-b\right)=RT$$
which for $p \gg a/V_m^2$ $^\ast$ can be rewritten as
$$pV_m=RT+bp$$
or
$$pV=nRT+nbp \tag{1}$$
The problem says that the tangent of this curve evaluated at the same $T$ has an intercept (the value of $pV$ when $p$ goes to zero) of $\pu{40 Latm}$ when $n=2$, which means ...
2
Since at equilibrium, Moles of CO= Moles of Zn= moles of CO2 (as no information about initial moles are given)
If no information is given, you should just give the three amounts names and treat them as unknowns.
ZnO is exposed to pure CO at 1300 K
This means that initially there is no carbon dioxide and no elemental zinc. Carbon dioxide and elemental ...
1
Just complementing the first answer of @Buck Thorn :
Full rewriting of the van Der Waals equation in terms of $Z=f(p,V_\mathrm{m})$:
$$\left(p+\frac{a}{{V_\mathrm{m}}^2}\right)\left(V_\mathrm{m}-b\right)=RT$$
$$pV_\mathrm{m} - pb + a/V_\mathrm{m} - ab/{V_\mathrm{m}}^2=RT$$
$$pV_\mathrm{m} \left( 1 - \frac{b}{V_\mathrm{m}} + \frac{ a}{ p{V_\mathrm{m}}^2} -...
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