Two bulbs of equal volumes connected through a stopcock, contained 0.7 mole of H_(2) gas at 0.5 atm pressure and 27^(@)C (at the open position of the stopcock). IF the first bulb was heated to 127^(@)C keeping the other at the same temperature, i.e, 27^(@)C, what will be the final pressure and moles in each bubl ?

Two bulbs of equal volumes connected through a stopcock, contained 0.7

1.	Two bulbs of equal volumes connected through a stopcock, contained 0.7 mole of H_(2) gas at 0.5 atm pressure and 27^(@)C (at the open position of the stopcock). IF the first bulb was heated to 127^(@)C keeping the other at the same temperature, i.e, 27^(@)C, what will be the final pressure and moles in each bubl ?
Answer» Solution :Let the volume of each bulb be v litres. For the two connected bulbs : p = 0.5 atm, n = 0.7 moles T = 27 + 273 = 300 K and VOL. = 2V. We have, pV = nRT `0.5(2V) = 0.7 XX 0.0821 xx 300` V = 17.22 litres When one of the bulbs is maintained at `127^(@)C`, i.e, 400 K and the other at 300 K, let the moles of `H_(2)` in these bulbes be `n_(1)` and `n_(2)` RESPECTIVELY. `therefore n_(1) + n_(2) = 0.7` .....(1) Since stopcock is OPEN, the pressure in each bulb will be the same. Let it be p atm. Thus for the bulb at 400 K. `p xx V = n_(1) xx 0.0821 xx 400` or `17.22 p = 32.8 n_(1)` or `p = 1.90 n_(1)` ......(1) And for the second bulb at 300 K, `p xx 17.22 = n_(2)xx 0.0821 xx 300` `p = 1.42 n_(2)` .......(3) From equations (1), (2) and (3), we get p = 0.571 atm, `n_(1) = 0.3` mole, `n_(2) = 0.4` mole.