A horizontal uniform glass tube of 100cm length, sealed at both ends, contains 10cm mercury column in the middle. The temperature and pressure of air on either side of mercury column are respectively 31 C ∘ and 76 cm of

Question

Quantrex Academy · Accepted Answer

Correct answer: C. 102.4

On keeping the temperature of the ends of tube at

\text{0} \text{C} ∘

and

\text{273} \text{C} ∘

.

A horizontal uniform glass tube of 100cm length, sealed at both ends, contains 10cm mercury column in the middle. The

Applying ideal gas equation

\dfrac{{p}_{1}{V}_{1}}{{T}_{1}} = \dfrac{{p}_{2}{V}_{2}}{{T}_{2}} = \dfrac{{p}_{3}{V}_{3}}{{T}_{3}}

\dfrac{76 \times 45}{( 273 + 31 )} = \dfrac{{p}_{2} \times l}{( 273 + 0 )} = \dfrac{{p}_{3} ( 90 - l )}{273 + 273}

\dfrac{76 \times 45}{304} = \dfrac{{p}_{2} \times l}{273} = \dfrac{{p}_{3} ( 90 - l )}{546}

\dfrac{{p}_{2} \times l}{273} = \dfrac{{p}_{3} ( 90 - l )}{546}

(Mercury column is at rest, so pressure

difference

{p}_{2} - {p}_{3} = 0 ⇒ {p}_{2} = {p}_{3} )

∴\dfrac{{p}_{2} \times l}{273} = \dfrac{{p}_{2} ( 90 - l )}{546}

⇒ 2l = 90 - l ⇒ l = 30

cm

\dfrac{76 \times 45}{304} = \dfrac{{p}_{2} \times 30}{273}

⇒ {p}_{2} = \dfrac{76 \times 45 \times 273}{30 \times 304}

{p}_{2} = 102.4

A horizontal uniform glass tube of 100cm length, sealed at both...

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