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Q)

Two glass bulbs with equal volume are connected by a narrow tube and filled with a gas at $0^{\large\circ}C$ and pressure of $76\;cm\; Hg$ . One of the bulb is then placed in a water bath maintained at $62^{\large\circ}C$. What is the new value of the pressure inside the bulbs? The volume of the connecting tube is negligible.

$(a)\;23.75 cm\; Hg\qquad(b)\;83.75 cm\; Hg\qquad(c)\;29.75 cm\; Hg\qquad(d)\;30.76 cm\; Hg$

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A)
Initially, when both bulbs are at 273 K. the mole in each bulb,
$ = \large\frac{Total \;mole}{2} = \large\frac{P.2V}{2RT} =\large\frac{PV}{RT} = \large\frac{76\times V}{R\times273}$
On heating second bulb some mole of gas are transferred to bulb One till the pressure in two bulbs becomes same.
For bulb 1 $n_1 = \large\frac{P_1 \times V}{R\times273}$
For bulb 2 $n_2 = \large\frac{P_2\times V}{R\times 335}$
Since $n = n_1 + n_2 $
$\therefore \large\frac{76\times V}{R\times273}\times2 = \large\frac{P_1\times V}{R\times273}+\large\frac{P_2\times V}{R\times335}$
also $P_1 = P_2$
$\therefore P_1 = 83.75$ cm Hg
Hence answer is (b)
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