Physical Chemistry (1)

Consider one of the flasks at the original condition (both flasks at 27oC) :

P = 0.5 atm
n = 0.7 mol
T = 27 + 273 = 300 K
R = ? atm dm3 mol-1 K-1

Apply PV = nRT
Volume V = nRT / P
Volume V = 0.7 x R x 300 / (0.5)
Volume V = 420R dm3
(The volume of the flask is almost unchanged when heated to 127oC.)

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Consider the case when one of the flasks is put in hot water.

Let Pf = final pressure in the system (both flasks)
Let nh = number of moles of hydrogen in the 127oC flask.
Let nc = number of moles of hydrogen in the 27oC flask.

Consider the flask at 127oC (400 K) :
Since n = PV / RT
nh = Pf x 420R / (R x 400)
Hence, nh = 1.05Pf ...... (*)

Consider the flask at 27oC (300 K) :
Since n = PV / RT
nc = Pf x 420R / (R x 300)
Hence, nc = 1.4Pf ...... (**)

(*) + (**) :
Total number of moles of hydrogen remains unchanged.
nh + nc = 1.05Pf + 1.4Pf = 0.7 x 2
2.45Pf = 1.4
Final pressure of each flask, Pf = 0.571 atm

(*) :
Number of moles of hydrogen in the 127oC flask, nh
= 1.4Pf
= 1.05 x 0.571
= 0.6 mol

(**) :
Number of moles of hydrogen in the 27oC flask, nc
= 1.4Pf
= 1.4 x 0.571
= 0.8 mol