A horizontal uniform glass tube of 100 cm le | vedclass

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(C) Initial state: The tube length is $100 \ cm$, with a $10 \ cm$ mercury column in the middle. The length of air on each side is $(100 - 10) / 2 = 4

Vedclass · Accepted Answer

$102.4$

Vedclass · Answer

(C) Initial state: The tube length is $100 \ cm$, with a $10 \ cm$ mercury column in the middle. The length of air on each side is $(100 - 10) / 2 = 45 \ cm$. Initial pressure $P_0 = 76 \ cm$ of $Hg$, initial temperature $T_0 = 31 + 273 = 304 \ K$.Applying the ideal gas law $\frac{PV}{T} = 	ext{constant}$, for the initial state: $\frac{P_0 V_0}{T_0} = \frac{76 	imes 45}{304}$.Final state: Let the new length of the air column at $0^{\circ} C$ $(273 \ K)$ be $l$, and the length of the air column at $273^{\circ} C$ $(546 \ K)$ be $(90 - l)$. Let the new pressure be $P'$.Since the mercury column is at rest, the pressure on both sides must be equal: $P_2 = P_3 = P'$.Using the gas law for both sides:$\frac{P' l}{273} = \frac{P' (90 - l)}{546} = \frac{76 	imes 45}{304}$.From $\frac{P' l}{273} = \frac{P' (90 - l)}{546}$, we get $2l = 90 - l$, which implies $3l = 90$, so $l = 30 \ cm$.Now, substitute $l = 30 \ cm$ into the equation: $\frac{P' 	imes 30}{273} = \frac{76 	imes 45}{304}$.$P' = \frac{76 	imes 45 	imes 273}{304 	imes 30} = \frac{76 	imes 45 	imes 273}{9120} = 102.375 \approx 102.4 \ cm$ of $Hg$.

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