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Elden [556K]
3 years ago
9

A total of 25.6 kJ of heat energy is added to a 5.46 L sample of helium at 0.991 atm. The gas is allowed to expand against a fix

ed external pressure to a volume of 18.7 L .
Required:
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?
Physics
1 answer:
bagirrra123 [75]3 years ago
3 0

Answer:

(a) W = 1329.5 J = 1.33 KJ

(b) ΔU = 24.27 KJ

Explanation:

(a)

Work done by the gas can be found by the following formula:

W = P\Delta V

where,

W = Work = ?

P = constant pressure = (0.991 atm)(\frac{101325\ Pa}{1\ atm}) = 100413 Pa

ΔV = Change in Volume = 18.7 L - 5.46 L = (13.24 L)(\frac{0.001\ m^3}{1\ L}) = 0.01324 m³

Therefore,

W = (100413 Pa)(0.01324 m³)

<u>W = 1329.5 J = 1.33 KJ</u>

<u></u>

(b)

Using the first law of thermodynamics:

ΔU = ΔQ - W (negative W for the work done by the system)

where,

ΔU = change in internal energy of the gas = ?

ΔQ = heat added to the system = 25.6 KJ

Therefore,

ΔU = 25.6 KJ - 1.33 KJ

<u>ΔU = 24.27 KJ</u>

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