'A horizontal uniform glass tube of $100 \mathrm{~cm}$, length sealed at both ends contains $10…

'A horizontal uniform glass tube of $100 \mathrm{~cm}$, length sealed at both ends contains $10 \mathrm{~cm}$ mercury column in the middle. The temperature and pressure of air on either side of mercury column are respectively $31^{\circ} \mathrm{C}$ and $76 \mathrm{~cm}$ of mercury. If the air column at one end is kept at $0^{\circ} \mathrm{C}$ and the other end at $273^{\circ} \mathrm{C}$, then the pressure of air which is at $0^{\circ} \mathrm{C}$ is : (in $\mathrm{cm}$ of $\mathrm{Hg}$ )
  1. 76
  2. 88.2
  3. 102.4
  4. 122

Solution

On keeping the temperature of the ends of tube at $0^{\circ} \mathrm{C}$ and $273^{\circ} \mathrm{C}$

Applying ideal gas equation $\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}=\frac{p_3 V_3}{T_3}$ $\frac{76 \times 45}{(273+31)}=\frac{p_2 \times l}{(273+0)}=\frac{p_3(90-l)}{273+273}$ $\frac{76 \times 45}{304}=\frac{p_2 \times l}{273}=\frac{p_3(90-l)}{546}$ $\begin{array}{lll}&\text { I } &&\text { II } && \text { III }\end{array}$ From Part II and III, $\frac{p_2 \times l}{273}=\frac{p_3(90-l)}{546}$ (Mercury column is at rest, so pressure difference, $p_2-p_3=0 \Rightarrow p_2=p_3$ $\therefore \quad \frac{p_2 \times l}{273}=\frac{p_2(90-l)}{546}$ $2 l=90-l$ $\Rightarrow \quad l=30 \mathrm{~cm}$ From I and II, $\frac{76 \times 45}{304}=\frac{p_2 \times 30}{273}$ $\Rightarrow \quad p_2=\frac{76 \times 45 \times 273}{30 \times 304}$ $p_2=102.4$

Asked in: AP EAMCET 2003

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