10.0 kJ of heat is absorbed by a gas dissociation reaction occurring reversibly and isothermally at 298 K in an elastic balloon (a closed system) which, together with atmospheric pressure, exerts a constant pressure of 1.20 bar on the gas. During the reaction, the gas mixture expands by 14.1 L. Calculate the change in (a) internal energy, (b) enthalpy, (c) entropy, (d) Gibbs free energy of the gas mixture, and (e) the change of the entropy of the universe due to this process. The gas mixture is not an ideal gas.

a) ΔU = 8.308 kJ

 

b) ΔH = 10 kJ

 

c) ΔS = 33.557 J/K

 

d) ΔG  = 0

 

e) ΔS_Universe = 0

Step-by-step explanation

Given:

Heat absorbed by the gas, Q = 10.0 kJ 

Temperature, T = 298 K

Pressure, P = 1.20 bar(1.20 bar = 1.20 bar × 1 atm/1.10325 bar = 1.1843 atm )

ΔV = 14.1 L

 

 

a) Change in internal energy

 

ΔU = Q + W

 

where:

Q is the heat absorbed or lost by the system

W is work done on or by the system

 

In this system, the gas expands. When a gas expands against external pressure, the gas transfers energy to the surroundings. Therefore, work done is said to be negative because the overall energy of the gas reduces.

 

Thus, the equation for calculating the change in internal energy becomes:

ΔU = Q + (- W)

 

ΔU = Q - W

 

But the work done as the gas expands or is compressed is calculated using the formula:

W = PΔV

 

Therefore:

ΔU = Q - PΔV

 

= 10,000 J -(1.1843 atm × 14.1 L)

 

= 10000 J - (16.6987 L.atm)

 

(In order to carry out the subtraction above, convert L.atm to J).

1 L.atm = 101.325 J

 

16.6987 L.atm = 16.6987 L.atm × 101.325 J/1 L.atm

 

= 1692 J

 

Thus:

ΔU =  10000 J - 1692 J 

 

= 8308 J

 

 

 

b) Change in enthalpy enthalpy(ΔH)

 

where:

 ΔU  is a change in internal energy

PΔV is the work done by expanding or compressing a gas.

 

ΔH = ΔU +  PΔV

 

= 8308 J + 1692 J

 

= 10000 J

 

= 10 kJ

 

 

c) Change in entropy

 

The change in entropy(ΔS) for a reversible process is given by the formula:

 

ΔS = (Q/T)_rev

 

Where:

Q is heat absorbed or lost by a gas

T is temperature

 

Thus:

ΔS = 10kJ/298 K

 

= 10000/298 K

 

= 33.557 J/K

 

 

 

d) Change in Gibbs free energy of the gas mixture

 

ΔG = ΔH - TΔS

 

= 10 kJ - (298 K × 33.557 JK^-1)

 

= 10000 J - 10000 J

 

= 0

 

 

e) Change of the entropy of the universe

 

In this question, the gas absorbed heat(Q). This implies that the value of Q is positive. The gas is the system.

 

Thus:

ΔS_system = Q/T

 

 

However, the surrounding loses heat to the gas. This implies that the value of Q is negative. Thus, the change in entropy of the surrounding is:

ΔS_surrounding = -Q/T

 

Therefore:

 

ΔS_Universe = ΔS_system + ΔS_surrounding

 

=  Q/T + (-Q/T)

 

= Q/T - Q/T

 

= 0

 

Therefore, the change of the entropy of the universe due to this process is zero(ΔS_Universe = 0)