To solve the problem it is necessary to consider the concepts and formulas related to the change of ideal gas entropy.
By definition the entropy change would be defined as
[tex]\Delta S = C_p ln(\frac{T_2}{T_1})-Rln(\frac{P_2}{P_1})[/tex]
Using the Boyle equation we have
[tex]\Delta S = C_p ln(\frac{T_2}{T_1})-Rln(\frac{v_1T_2}{v_2T_1})[/tex]
Where,
[tex]C_p[/tex] = Specific heat at constant pressure
[tex]T_1[/tex]= Initial temperature of gas
[tex]T_2[/tex]= Final temprature of gas
R = Universal gas constant
[tex]v_1[/tex]= Initial specific Volume of gas
[tex]v_2[/tex]= Final specific volume of gas
According to the statement, it is an isothermal process and the tank is therefore rigid
[tex]T_1 = T_2, v_2=v_1[/tex]
The equation would turn out as
[tex]\Delta S = C_p ln1-ln1[/tex]
Therefore the entropy change of the ideal gas is 0
Into the surroundings we have that
[tex]\Delta S = \frac{Q}{T}[/tex]
Where,
Q = Heat Exchange
T = Temperature in the surrounding
Replacing with our values we have that
[tex]\Delta S = \frac{230kJ}{(30+273)K}[/tex]
[tex]\Delta S = 0.76 kJ/K[/tex]
Therefore the increase of entropy into the surroundings is 0.76kJ/K
