Caculate the change in internal energy (delta e) for a system that is absorbing 35.8 kJ of heat and is expanding from 8...
Caculate the change in internal energy (delta e) for a system
that is absorbing 35.8 kJ of heat and is expanding from 8.00 to
24.0 L in volume at 1.00 atm. (Remember that 101.3 J= 1
l*atm).
I know that the answer is going to be a postive since energy is
going to be absorbed...I just cannot figure out how to get the
right answer. Please help.
Thank you
Homework Answers
?U = change in internal energy
Q = heat added.. + if heat is added... - if heat is removed
W = work done ON the system.. + if done on the system... - if done by the system...
Q is straight forward right? +35.8 kJ
W isn't....
W = P ?V... for the isobaric (constant pressure) expansion of a gas...
W = 1.00 atm x (24.0 L - 8.00 L) = 16.0 L atm...
and notice the units on W? we need to convert that to Joules...let's start with some conversion factors...
101,325 Pa = 1 atm
1 Pa =1 N / m²
1000 L = 1 m³
1 J = 1 N m...
then via dimensional analysis...
16.0 L atm x (101325 Pa / 1 atm) x (1 N / m² / 1 Pa) x (1 m³ / 1000 L) x (1 J / 1 N m) = 1621 J = 1.6 kJ
alternaltely... if you didn't follow all that.. the link in the source states 1 J = 9.869x10^-3 L atm... so that...
16.0 L atm x (1 J / 9.689x10^-3 L atm) = 1.6x10^3 J = 1.6 kJ...either way...
finally...since the volume is EXPANDING, the work is done BY the system on it's environment. Not "on the system". so the work value is "-" for this case..and...
?U = Q + W = +35.8 kJ - 1.6 kJ = 34.2 kJ
?U = change in internal energy
Q = heat added.. + if heat is added... - if heat is removed
W = work done ON the system.. + if done on the system... - if done by the system...
Q is straight forward right? +35.8 kJ
W isn't....
W = P ?V... for the isobaric (constant pressure) expansion of a gas...
W = 1.00 atm x (24.0 L - 8.00 L) = 16.0 L atm...
and notice the units on W? we need to convert that to Joules...let's start with some conversion factors...
101,325 Pa = 1 atm
1 Pa =1 N / m²
1000 L = 1 m³
1 J = 1 N m...
then via dimensional analysis...
16.0 L atm x (101325 Pa / 1 atm) x (1 N / m² / 1 Pa) x (1 m³ / 1000 L) x (1 J / 1 N m) = 1621 J = 1.6 kJ
alternaltely... if you didn't follow all that.. the link in the source states 1 J = 9.869x10^-3 L atm... so that...
16.0 L atm x (1 J / 9.689x10^-3 L atm) = 1.6x10^3 J = 1.6 kJ...either way...
finally...since the volume is EXPANDING, the work is done BY the system on it's environment. Not "on the system". so the work value is "-" for this case..and...
?U = Q + W = +35.8 kJ - 1.6 kJ = 34.2 kJ
Add Answer to:
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