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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^(0)C. Now the te,nperature of the bigger bulb is raised to 100^(0)C and that of the smaller bulb being kept at 0^(0)C. If the initial pressure was 76 cm of mercury, what is the final pressure?

Answer» <html><body><p></p>Solution :Let P be the initial <a href="https://interviewquestions.tuteehub.com/tag/pressure-1164240" style="font-weight:bold;" target="_blank" title="Click to know more about PRESSURE">PRESSURE</a> of the bulbs and `P^(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)` be the final pressure. Initial temperature of the bulbs be T. The final temperatures of the bulbs of volume `V_(1) and V_(2)` be `T_(1) and T_(2) `respectively. <br/> By ideal <a href="https://interviewquestions.tuteehub.com/tag/gas-1003521" style="font-weight:bold;" target="_blank" title="Click to know more about GAS">GAS</a> equation, as mass of the gas in the bulbs remains constant <br/> `(P_(1)V_(1))/(T ) + (P_(2)V_(2))/(T) = (P^(1)V_(1))/(T_(1)) + (P^(1)V_(2))/(T_(2))` <br/> `(P)/(T) (V_(1) + V_(2)) = P^(1) ((V_(1))/(T_(1)) + (V_(2))/(T_(2)) )(because P_(1) = P_(2) = P ) ` <br/> `(76)/(<a href="https://interviewquestions.tuteehub.com/tag/273-1832932" style="font-weight:bold;" target="_blank" title="Click to know more about 273">273</a>) (100 + 200 ) = P^(1) ((100)/(273) + (200)/(373))` <br/> `(76 xx <a href="https://interviewquestions.tuteehub.com/tag/300-305868" style="font-weight:bold;" target="_blank" title="Click to know more about 300">300</a>)/(273) = p^(1) xx 100 ((919)/(273 xx 373))` <br/> `P^(1) = (76 xx 3 xx 373)/(919) = 92.53` cm</body></html>


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