A bubble of 6.00 mol helium is submerged at a certain depth in liquid water when
ID: 1476142 • Letter: A
Question
A bubble of 6.00 mol helium is submerged at a certain depth in liquid water when the water ( and thus the helium ) undergoes a temperature increase T of 200C at constant pressure. As a result, the bubble expands.
(a) How much heat Q is added to the helium during the expansion and temperature increase? (b) What is the change Eint in the internal energy of the helium during the temperature increase? (c) How much work W is done by the helium as it expands against the pressure of the surrounding water during the temperature increase?
Explanation / Answer
Given that
The number of moles of helium is (n) =6.00mol
Given that the pressure is constant, then heat is given y
Q =nCpDeltaT
Where Cp =Cv+R for the ideal gas
Now we have Q =n(Cv+R)DeltaT
=6(6/2*8.314)(20K)
=2993.04J
Therefore the change in internal energy is DletaEint =nCvDeltaT =6mol(3/2*8.314)20K =1496.52J
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