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Question

(C) The total length of the tube is $100 \ cm$. The mercury column is $10 \ cm$ long, so the total length of the air columns is $100 - 10 = 90 \ cm$.

Vedclass · Accepted Answer

$102.4$

Vedclass · Answer

(C) The total length of the tube is $100 \ cm$. The mercury column is $10 \ cm$ long, so the total length of the air columns is $100 - 10 = 90 \ cm$. Initially, the air columns on both sides are equal, each being $45 \ cm$. Initial state: $P_0 = 76 \ cm$ of $Hg$, $T_0 = 31 + 273 = 304 \ K$, $V_0 = 45 \ A$ (where $A$ is the cross-sectional area). Final state: Let the length of the air column at $0^{\circ} C$ $(273 \ K)$ be $l$. The pressure is $P$. The length of the other air column at $273^{\circ} C$ $(546 \ K)$ is $(90 - l)$. Since the mercury column is at rest, the pressure on both sides must be equal, so $P_1 = P_2 = P$. Using the ideal gas law $\frac{PV}{T} = 	ext{constant}$: For the first side: $\frac{76 	imes 45}{304} = \frac{P 	imes l}{273}$ --- $(1)$ For the second side: $\frac{76 	imes 45}{304} = \frac{P 	imes (90 - l)}{546}$ --- $(2)$ Equating $(1)$ and $(2)$: $\frac{P 	imes l}{273} = \frac{P 	imes (90 - l)}{546}$ $2l = 90 - l \Rightarrow 3l = 90 \Rightarrow l = 30 \ cm$. Substituting $l = 30 \ cm$ into equation $(1)$: $\frac{76 	imes 45}{304} = \frac{P 	imes 30}{273}$ $P = \frac{76 	imes 45 	imes 273}{30 	imes 304} = 102.4 \ cm$ of $Hg$.

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