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
![\Delta S = C_p ln(\frac{T_2}{T_1})-Rln(\frac{P_2}{P_1})](https://tex.z-dn.net/?f=%5CDelta%20S%20%3D%20C_p%20ln%28%5Cfrac%7BT_2%7D%7BT_1%7D%29-Rln%28%5Cfrac%7BP_2%7D%7BP_1%7D%29)
Using the Boyle equation we have
![\Delta S = C_p ln(\frac{T_2}{T_1})-Rln(\frac{v_1T_2}{v_2T_1})](https://tex.z-dn.net/?f=%5CDelta%20S%20%3D%20C_p%20ln%28%5Cfrac%7BT_2%7D%7BT_1%7D%29-Rln%28%5Cfrac%7Bv_1T_2%7D%7Bv_2T_1%7D%29)
Where,
= Specific heat at constant pressure
= Initial temperature of gas
= Final temprature of gas
R = Universal gas constant
= Initial specific Volume of gas
= Final specific volume of gas
According to the statement, it is an isothermal process and the tank is therefore rigid
![T_1 = T_2, v_2=v_1](https://tex.z-dn.net/?f=T_1%20%3D%20T_2%2C%20v_2%3Dv_1)
The equation would turn out as
![\Delta S = C_p ln1-ln1](https://tex.z-dn.net/?f=%5CDelta%20S%20%3D%20C_p%20ln1-ln1)
<em>Therefore the entropy change of the ideal gas is 0</em>
Into the surroundings we have that
![\Delta S = \frac{Q}{T}](https://tex.z-dn.net/?f=%5CDelta%20S%20%3D%20%5Cfrac%7BQ%7D%7BT%7D)
Where,
Q = Heat Exchange
T = Temperature in the surrounding
Replacing with our values we have that
![\Delta S = \frac{230kJ}{(30+273)K}](https://tex.z-dn.net/?f=%5CDelta%20S%20%3D%20%5Cfrac%7B230kJ%7D%7B%2830%2B273%29K%7D)
![\Delta S = 0.76 kJ/K](https://tex.z-dn.net/?f=%5CDelta%20S%20%3D%200.76%20kJ%2FK)
<em>Therefore the increase of entropy into the surroundings is 0.76kJ/K</em>