Question

Two bulbs of 100 c.c and 200 c.c capacity, contain the same gas at same temperature and pressure are connected by a capillary tube of negligible volume. Initially both were at 0°C. Now the temperature of the bigger bulb is raised to 100°C and that of the smaller bulb being kept at 0°C. If the initial pressure was 76 cm of mercury, what is the final pressure ?​

Answer 1

$\Large\bf {\orange {\underline { \underline {Solutíon}} : - }}$

Let 'P' be the initial pressure of the bulbs and 'P¹' be the final pressure. Initial temperature of the bulbs be 'T' .

The final temperature of the bulbs of volume $\sf V_{1}$ and $\sf V_{2}$ be $\sf T_{1}$ and$\sf T_{2}$ respectively.

By ideal gas equation, as mass of the gas in the bulbs remains constant

$\Longrightarrow \sf \frac{ P_{1} V_{1}}{T} + \frac{ P_{2} V_{2} }{t} = \frac{ {P}^{1} V_{1} }{ T_{1} } + \frac{ {P}^{1} V_{2} }{ T_{2}}$

$\Longrightarrow \sf \frac{P}{T} ( V_{1} + V_{2}) = {P}^{1} \bigg( \frac{ V_{1} }{ T_{1} } + \frac{ V_{2} }{ T_{2}} \bigg) \: \: \: ( \because P_{1} = P_{2} = P)$

$\Longrightarrow \sf \frac{76}{273} (100 + 200) = {P}^{1} \bigg( \frac{100}{273} + \frac{200}{373} \bigg)$

$\Longrightarrow \sf \frac{76 \times 300}{273} = {P}^{1} \times 100 \bigg( \frac{919}{273 \times 373 } \bigg)$

$\Longrightarrow \sf {P}^{1} = \frac{76 \times 3 \times 373}{919}$

$\Longrightarrow \sf {P}^{1} = 92.53cm$

Thus, final pressure is 92.53cm