Question

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 fixed 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?

Answer 1

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)

W = 1329.5 J = 1.33 KJ

(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

ΔU = 24.27 KJ