Temperature at thermal equilibrium?
Consider a rigid insulated box containing 20.0 g of He(g) at 25.0°C and 1.00 atm in one compartment and 20.0 g of N2(g) at 115.0°C and 2.00 atm in the other compartment. These compartments are connected by a partition which transmits heat. What will be the final temperature in the box at thermal equilibrium? (Cv(He) = 12.5 J/K mol, Cv(N2) = 20.7 J/K mol)
find moles He:
20.0 g @ 4.00g/mol = 5 moles He
heat capacity of He=
(12.5 J/K mol)(5 moles) = 62.5 kJ/K
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find moles of N2:
20.0 g N2 @ 28.01 g/mol = 0.714 moles
find heat capacity of N2 =
(20.7 J/K mol) (0.714 moles) = 14.78 kJ/K
find excess heat within the N2 because it is 90 degrees hotter than the He:
dH = CdT
dH = 14.78 kJ/K (90)
excess heat = 1330kJ
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that 1330kJ is the excess heat within the N2 because it is hotter that the He.
if they were both at 25C …
and the 1330kJ was reintroduced to them mixed together….
they would both rise to that same equilibrium temp…
that they would have arrived at anyway
their combined heat capacities are:
14.78kJ/degree & 62.5 kJ/K = 77.28kJ/mol
find the equilib temp rise above 25C:
dH = CdT
1330 kJ = 77.28 kJ/degree (dT)
dT = 17.2 C
so the final temp is
25.0 plus 17.2 =
42.2 C
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