A bubble of 5.00 mol helium is submerged at a certain depth in liquid water when
ID: 1479344 • Letter: A
Question
A bubble of 5.00 mol helium is submerged at a certain depth in liquid water when
the water ( and thus the helium ) undergoes a temperature increase T of 16C 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
Cp -Cv =R
Cp = specific heat at constant pressure
Cv = specific heat at constant volume = (3/2)R
R =8.314 J/mol.K
dT =16 oC
(a) At constant pressure
Q = nCpdT =n(Cv+R)dT
Q = 5*(5/2)*8.314*16
Q= 1662.8 J
(b) Change internal energy Eint = nCvdT
Eint = 5*(3/2)*8.314*16 = 997.68 J
(c) From first law of thermodynamics
Q = Eint +W
W =Q -Eint = 1662.8 -997.68
W = 665.12 J
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.