Question

This problem has been solved!
See the answer
A total of 27.10 KJ of heat is added to a 5.30-L sample of helium at 0.981 atm. The gas is allowed to expand against a fixed external pressure to a volume of 23.70 L.
A) calculate the work done on or by the helium gas units of joules, J.
B) what is the change in the helium a internal energy in units kilojoules, KJ?

Answer 1

Answer:

A. Workdone by helium, W = -1.829KJ

B. Internal energy, DE = 25.271KJ

Explanation:

Workdone can be defined as the force moving through a distance. For a gaseous system, when the volume of the gas expands, the system is losing energy. Therefore,

W = -P*DV

Where P is the pressure in pascal

DV is the change in volume in m3

DV = Vfinal - Vinitial

= 23.70 - 5.30

= 18.4L

W = -(0.981 * 18.4)

= -18.0504L.atm

Converting L.atm to joule,

= -18.0504 * 101.325

= -1828.97J

= -1.829KJ

If the system loses heat, Q the rection occurring is Exothermic.

Heat is the transfer of energy from one system to another.

Q = mcDT

Where m is the mass of the system

C is the specific heat capacity

Q = 27.20KJ

Internal energy is the summation of the heat supplied to a system and the workdone by the system

DE = Q + W

DE = 27.10 + (-1.829)

= 25.27KJ