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

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Solution

We have $P V = n R T$
As total no of gas molecules in the bulbs would remain constant,
we have $n_{1} + n_{2} = n_{1}^{'} + n_{2}^{'}$
$\frac{P_{0} V}{R T_{0}} + \frac{P_{0} V}{R T_{0}} = \frac{P . V}{R T} + \frac{P V}{R T_{0}}$
i.e. $\frac{2 P_{0}}{T_{0}} = P (\frac{1}{T} + \frac{1}{T_{0}})$
Hence new pressure $P = (\frac{2 T}{T + T_{0}}) \cdot P_{0}$
Here $P_{0} = 76 c m$ of $H g$ (initial pressure), $T_{0} = 0^{\circ} = 273 K$
$T = 62^{\circ} C = 335 K$
$∴ P = \frac{2 \times 335}{(273 + 335)} \times 76 = 83.75 c m$ of $H g$ .

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

We have PV=nRTAs total no of gas molecules in the bulbs would remain constant,we have n1+n2=n′1+n′2P0VRT0+P0VRT0=P.VRT+PVRT0i.e. 2P0T0=P(1T+1T0)Hence new pressure P=(2TT+T0)⋅P0Here P0=76 cm of Hg (initial pressure), T0=0∘=273 KT=62∘C=335 K∴P=2×335(273+335)×76=83.75 cm of Hg.