$\omega=-P_2\left(V_2-V_1\right)$
or we can write
$\omega =-\pi R T\left(1-\frac{p_2}{p_1}\right)$
$\omega =-5 \times 8.314 \times 300\left(1-\frac{1}{10}\right)$
$=-5 \times 8.314 \times 300\left(\frac{9}{10}\right)$
$\omega =-11.224 \,kJ$
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$Ca_3(PO_4)_2 + SiO_2 + C + O_2 + H_2O \to CaSiO_3 + CO_2 + H_3PO_4$
If $1.0\ kg$ each of calcium phosphate and silica are used with excess of $C$ , $O_2$ and $H_2O$ , what is maximum quantity of phosphoric acid that can be produced. $(Ca = 40, P = 31, Si = 28)$
