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?

Two bulbs of 100 c.c and 200 c.c capacity, contain the same gas at sam

1.	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» Solution :Let P be the initial PRESSURE of the bulbs and `P^(1)` 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. By ideal GAS equation, as mass of the gas in the bulbs remains constant `(P_(1)V_(1))/(T ) + (P_(2)V_(2))/(T) = (P^(1)V_(1))/(T_(1)) + (P^(1)V_(2))/(T_(2))` `(P)/(T) (V_(1) + V_(2)) = P^(1) ((V_(1))/(T_(1)) + (V_(2))/(T_(2)) )(because P_(1) = P_(2) = P ) ` `(76)/(273) (100 + 200 ) = P^(1) ((100)/(273) + (200)/(373))` `(76 xx 300)/(273) = p^(1) xx 100 ((919)/(273 xx 373))` `P^(1) = (76 xx 3 xx 373)/(919) = 92.53` cm