231k views
4 votes
A total of 25.6 kj of hear energy is added to a 5.18l sample of helium at .987atm The gas is allowed to expand against a fixed external pressure to a volume of 18.4l

User Arshin
by
5.3k points

1 Answer

6 votes

Answer:

A total of 25.6 kj of hear energy is added to a 5.18l sample of helium at .987atm The gas is allowed to expand against a fixed external pressure to a volume of 18.4l

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?

A. Workdone by helium, W = -1.322KJ

B. Internal energy, DE = 24.278KJ

Step-by-step 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

= 18.4 - 5.18

= 13.22L

W = -(0.987 * 13.22)

= -13.0481L.atm

Converting L.atm to joule,

= -13.0481 * 101.325

= -1322.099J

= -1.322KJ

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 = 25.6KJ

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

DE = Q + W

DE = 25.6 + (-1.322)

= 24.278KJ

User Fault
by
5.6k points