Question

A glass tube, sealed at both ends, is 100 cm long. It lies horizontally with the middle 10 cm containing mercury. The two ends of the tube contain air at ${27}^0$C and at a pressure 76 cm of mercury. The air column on one side is maintained at $0^0$C and the other side is maintained at ${127}^0$C. Calculate the length of the air column on the cooler side. Neglect the changes in the volume of mercury and of the glass.

Answer 1

Here 2L + 10 = 100 cm

$\Rightarrow$ 2L = 90 cm

$\Rightarrow$ L = 45 cm

Applying combined gas equation to part 1 of the tube,

$\frac{\left(45A\right)P_0}{300}=\frac{(45-x)A{P}_1}{273}$

$\Rightarrow P_1=\frac{273\times 45\times P_0}{300(45-x)}$

Applying combined gas equation to part 2 of the tube,

$\frac{\left(45A\right)P_0}{300}=\frac{(45+x)A{P}_2}{400}$

$\Rightarrow P_2\frac{400\times 45P_0}{300(45+x)}$

$P_1=P_2$

$\Rightarrow \frac{273\times 45\times P_0}{300(45-x)}=\frac{400\times 45\times P_0}{300(45+x)}$

$\Rightarrow \frac{45\times 273}{45-x}=\frac{400\times 45}{45+x}$

$\Rightarrow (45-x)400=(45+x)273$

$\Rightarrow 18000-400x=12285+273x$

$\Rightarrow$ (400+273) x = 18000-12285

$\Rightarrow$ x = 8.49

$P_1=\frac{273\times 45\times 76}{300+36.51}$

= 85.25 cm of Hg

Length of air column on the air cooler side

= L-x

= 45 - 8.49 = 36.51 cm.