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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?

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

hail is a solid form of precipitation composed of large, layered ice particles called pellets that often have spherical shapes. this phenomenon occurs during thunderstorms in the could type known as cumulus

hope this you out ^w^

what is the actual question though?