Question

A horizontal uniform glass tube of 100cm length is sealed at both ends contains 10 cm mercury column in the middle the temperature and pressure of air on either side of mercury column are respectively 31${}^0$C and 76cm of mercury if the air column at one end is kept at 0${}^0$C and the other end at 273${}^0$C the pressure of air which is at 0${}^0$C is (in cm of Hg )

• A. $76$
• B. $88.2$
• C. $102.4$
• D. $122$

Answer 1

Answer: (C)

$102.4$

Let $x$ be the displacement of the mercury column on the side whose temperature is changed to $0^{\circ }C$. New pressure on both sides must be same so that the mercury column remains in equilibrium.

$\frac{PV}{T}$ is constant for air column.

$⟹\frac{P_0\left(45\right)}{273+31}=\frac{P_1(45-x)}{273+0}$

and $\frac{P_0\left(45\right)}{273+31}=\frac{P_1(45+x)}{273+273}$

Therefore, $x=15cm$ and thus $P=102.4cmofHg$