Question

Two glass bulbs of equal volumes are connected by a narrow tube and are filled with an ideal gas at $0^{\circ }C$ and a pressure of 76 cm of mercury. One of the bulbs is then placed in a water bath maintained at ${62}^{\circ }C$. What is the new value of the pressure inside the bulbs?

Answer 1

Initially, when both bulbs are at 273 K. The mole in each bulb,
$=\frac{Totalmole}{2}=\frac{P.2V}{2RT}=\frac{PV}{RT}=\frac{76\times V}{R\times 273}$
On heating second bulb some mole of gas is transferred to the bulb.
For bulb $1n_1=\frac{P_1\times V}{R\times 273}$
For bulb $2n_1=\frac{P_2\times V}{R\times 335}$
Since $n=n_1+n_2$
$\therefore \frac{76\times V}{R\times 273}\times 2=\frac{P_1\times V}{R\times 273}+\frac{P_2\times V}{R\times 335}$
Also $P_1=P_2$
$\therefore P_1=83.75cmHg$