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'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 $\mathrm{Hg}$ )
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Verified Answer
The correct answer is:
102.4
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,
$\frac{76 \times 45}{304}=\frac{p_2 \times 30}{273}$
$\Rightarrow \quad p_2=\frac{76 \times 45 \times 273}{30 \times 304}$
$p_2=102.4$
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,
$\frac{76 \times 45}{304}=\frac{p_2 \times 30}{273}$
$\Rightarrow \quad p_2=\frac{76 \times 45 \times 273}{30 \times 304}$
$p_2=102.4$
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