Question

A horizontal uniform glass tube of 100 cm length sealed at both ends contains a 10 cm mercury column in the middle. The temperature and pressure of air on either side of the mercury column are respectively \[0{}^\circ C\]and 80 cm of mercury. If the air column at one end is kept at \[0{}^\circ C\]and the other end at\[273{}^\circ C\], the pressure of air which is \[0{}^\circ C\]is (in cm of Hg)

A. 76
B. 88.2
C. 102
D. 132

Answer 1

Hint: Using the gas law equation the calculation should be carried out. In order to make the mercury column to be equilibrium, the new pressure on both the ends of the glass tube must be equal to each other.

Formula used:
\[\dfrac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}}=\dfrac{{{P}_{2}}{{V}_{2}}}{{{T}_{2}}}\]

Complete step by step answer:
From given, we have the data,
The length of the glass tube = 100 cm
The length of the mercury column = 10 cm
The temperature of the air at either side of the glass tube = \[0{}^\circ C\]
The pressure of air at either side of the glass tube = 80 cm of mercury.

We will make use of the ideal gas law equation and it is given as follows.
\[\dfrac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}}=\dfrac{{{P}_{2}}{{V}_{2}}}{{{T}_{2}}}\]
Where \[{{P}_{1}},{{P}_{2}}\]are the initial and final pressure values, \[{{T}_{1}},{{T}_{2}}\] are the initial and final temperature values and \[{{V}_{1}},{{V}_{2}}\] are the initial and final volume values

Let the displacement of the mercury column on the side whose temperature is changed to \[0{}^\circ C\] be represented by “x”. The constant of the air column is equal to \[\dfrac{PV}{T}\].
In order to make the mercury column to be equilibrium, the new pressure on both the ends of the glass tube must be equal to each other.
So, we have,
\[\dfrac{{{P}_{0}}\times 45}{273+31}=\dfrac{{{P}_{1}}\times (45-x)}{273+0}\] …… (1)
And
\[\dfrac{{{P}_{0}}\times 45}{273+31}=\dfrac{{{P}_{1}}\times (45+x)}{273+273}\] …… (2)
As the LHS parts both the equations (1) and (2) are the same, so we can equate the equations.
So, we get,
\[\begin{align}
  & \dfrac{{{P}_{1}}\times (45-x)}{273+0}=\dfrac{{{P}_{1}}\times (45+x)}{273+273} \\
 & \Rightarrow \dfrac{45-x}{273}=\dfrac{45+x}{546} \\
\end{align}\]
Upon further solving the equation, we get,
\[\begin{align}
  & 546(45-x)=273(45+x) \\
 & \Rightarrow 24570-546x=12285+273x \\
 & \Rightarrow x=15 \\
\end{align}\]
Upon further solving, we get the value of the displacement as, x = 15 cm and thus, the value of the pressure to be 102.4 cm of Hg.
As the pressure of air is obtained to be equal to 102 cm of Hg, thus, the option (C) is correct.

Note:
The things to be on your finger-tips for further information on solving these types of problems are: The units of the given parameters should be taken into consideration while solving the problem. If in question, asked to determine the final value in mm unit, then write the final answer in terms of that unit only, even if while calculating, if you make use of other length units. The unit of the temperature should also be converted to Kelvin.