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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°C and 76cm of mercury , if the air column at one end is kept at 0°C and the other end at 273°C then pressure of air which is at 0°C is
(in cm of Hg )

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88.2

122

102.4

answer is C.

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Detailed Solution

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}} \Rightarrow \frac{P_{1} {Al}_{1}}{T_{1}} = \frac{P_{2} {Al}_{2}}{T_{2}}$

$\Rightarrow \frac{P_{1} l_{1}}{T_{1}} = \frac{P_{2} l_{2}}{T_{2}} area remains constant$

Whenever temperature of the glass tube changes piston moves through a distance x. So that pressure remains constant Before heating

$\begin{array}{l} \frac{l_{1}}{T_{1}} = \frac{l_{2}}{T_{2}}; \frac{45 - x}{273} = \frac{45 + x}{546} \\ ∴ x = 15 cm \end{array}$

compare left hand side volumes in both cases

$\frac{P_{1}^{1} l_{1}^{1}}{T_{1}^{1}} = \frac{P_{1} l_{1}}{T_{1}}$

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