Calculate the change in internal energy (δe) for a system that is giving off 25.0 kj of heat and is changing from 12.00 l to 6.00 l in volume at 1.50 atm pressure. (remember that 101.3 j = 1 l∙atm) 14)
a.+25.9 kj
b.-16.0 kj
c.-25.9 kj
d.-24.1 kj
e.937 kj

Since the system itself is giving off heat, this is a reduction in the internal energy.

heat = - 25,000 J

Since work is being done on the system, therefore it is an additional energy to the system. Work is given as:

work = - P dV

work = - 1.50 atm (6 L – 12 L)

work = 9 L atm

Since it is given that 1 L atm is equivalent to 101.3 J, therefore the total energy added is:

energy due to work = 9 L atm (101.3 J / 1 L atm)

energy due to work = 911.7 J

Therefore the total change in internal energy is the sum of heat and energy due to work:

Change in internal energy = - 25,000 J + 911.7 J

Change in internal energy = - 24,088.3 J

Therefore, approximately 24.1 kJ of energy is lost by the system in the total process.

-24.1 kJ

your answer is, orbit around the nucleus