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

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)

Ask Question

Tag:MathPhyChemBioOther

Take Test

...