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

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Solution

At first, at $273 K$
the mole in each bulb $= \frac{n_{t o t a l}}{2} = \frac{1}{2} . \frac{P .2 V}{R T} = \frac{76 V}{273 \times R}$
On heating second bulb, some mole of gas are transferred to another bulb until the pressure of both bulb become same.
$∴ n_{1} = \frac{P \times V}{273 R}$ and $n_{2} = \frac{P V}{R \times 335}$
$n_{t o t a l} = n_{1} + n_{2}$
$\Rightarrow 2 \times \frac{76 V}{273 R} = \frac{P \times V}{273 R} + \frac{P V}{335 \times R} \Rightarrow P = 83.75$ unit of $A g$ .

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

At first, at 273Kthe mole in each bulb =ntotal2=12.P.2VRT=76V273×ROn heating second bulb, some mole of gas are transferred to another bulb until the pressure of both bulb become same.∴n1=P×V273R and n2=PVR×335ntotal=n1+n2⇒2×76V273R=P×V273R+PV335×R⇒P=83.75 unit of Ag.