In a cylinder, whose cross–sectional area is $20 c{m}^2$ , a frictionless piston of mass 7.2 kg encloses a 33 cm high air column at  $0^{0}C$ so that there is a 7 cm high empty part above the piston as shown. The atmospheric pressure is $10N/c{m}^2$, the densities of mercury and air in its initial state are $13.6 g/c{m}^3$  and $1.8 g/d{m}^3$  respectively, the specific heat of the air at constant volume is $0.7 J/(g K)$. Use $g = 10m/s^2$. Mercury is poured into the empty part above the piston until the cylinder is full. (Assume constant temperature). If length of mercury column is x cm. Find x/10 = ?
 

The initial pressure of the enclosed air is $p_1=p_0+mg/A$ ,while its final (maximum) pressure is  $p_2=[p_0+(m+m_{Hg})g/A]=p_0+mg/A+Q{H}_{g}gx$
According to Boyle’s law:  $V_2=\frac{p_1}{p_2}{V}_1$

Substituting the expressions for $p_1,p_2,V_1$ and  $V_2$, we get: 
  $(h+h_1-x)A=\frac{p_0+\frac{mg}{A}}{p_0+\frac{mg}{A}+Q{H}_{g}gx}.Ah.$
Let us divide the equation by A and multiply by the denominator of the fraction:
 $(p_0+\frac{mg}{A}+Q{H}_{g}gx)(h+h_1-x)=(p_0+\frac{mg}{A})h$
After rearranging this according to the powers of x, we get:

$$\begin{gathered} Q{H}_{g}g{x}^2-[Q{H}_{g}g(h_1+h)-(p_0+mg/A)]x-(p_0+mg/A)h_1=0 \\ {p}_0=10\frac{N}{c{m}^2}; \frac{mg}{A}=\frac{72}{20}\frac{N}{c{m}^2}=3.6\frac{N}{c{m}^2}; \\ Q{H}_{g}g={\mathrm{13.6.10}}^{-3}\frac{kg}{c{m}^3}.10\frac{m}{s^2} = 0.136\frac{N}{c{m}^3} \end{gathered}$$

Inserting these into the equation, we obtain:
$0.136\frac{N}{c{m}^3}.x^2-(0.136\frac{N}{c{m}^3}.40cm-13.6\frac{N}{c{m}^2}).x-13.6\frac{N}{c{m}^2}.7cm=0$
Dividing the equation by unit  $N/c{m}^2$, we find:
  $0.136\frac{1}{cm}.x^2+\mathrm{8.16.}x-95.2cm=0$
The solution is :
$x=\frac{-8.16+\sqrt{{8.16}^2+\mathrm{4.95.2.0.136}}}{\mathrm{2.0.136}}cm=10cm$