Entropy after removing partition separating two gases system 
An ideal gas is seperated by a piston in such a way that the entropy of one prat is$ S_1$ and that of the other part is $S_2$. Given that $S_1>S_2$, if the piston is removed then the total entropy of the system will be..?
Ans: $S_1 +S_2$
JEE Mains March 18th evening attempt 2021

I know that for an ideal gas, the entropy is given as:
$$ S = nC_v \ln T -nR \ln V+C$$
Let the variables of one compartment be $(V,T)$ and the other be $(V',T')$ the combined variables be $(\tilde{V},\tilde{T})$. Now according to the question:
$$ nC_v \ln \tilde{T} - nR \ln \tilde{V} + C = nC_v \ln TT' - nR \ln VV' +2C$$
I simply can't see how the above equation is true! Is there some concept which can be used to simplify understanding the above?
P.S: We aren't taught statistical mechanics for this test
Note: C is an arbitrary constant.
 A: The only way that the final entropy will be $S_1+S_2$ is if the initial temperatures and the initial pressures in the two chambers are equal.
This is for @Buraian:
To get the final temperature and pressure, we use the conditions that the change in internal energy of the combined system is zero, and the final volume is equal to the initial combined volume.  This leads to:
$$n_1C_v(T_F-T_1)+n_2C_v(T_F-T_2)=0$$and$$\frac{n_1RT_1}{P_1}+\frac{n_2RT_2}{P_2}=\frac{(n_1+n_2)RT_F}{P_F}$$The solutions to these equations for $T_F$ and $P_F$ are:
$$T_F=\frac{n_1T_1+n_2T_2}{(n_1+n_2)}$$$$\frac{1}{P_F}=\left(\frac{n_1T_1}{n_1T_1+n_2T_2}\right)\frac{1}{P_1}+\left(\frac{n_2T_2}{n_1T_1+n_2T_2}\right)\frac{1}{P_2}$$The entropy change for the system is given by:
$$\Delta S=n_1C_p\ln{\left(\frac{T_F}{T_1}\right)}+n_2C_p\ln{\left(\frac{T_F}{T_2}\right)}+n_1R\ln{\left(\frac{P_1}{P_F}\right)}+n_2R\ln{\left(\frac{P_2}{P_F}\right)}$$The next step is to show that, unless the initial temperatures and pressures are equal, this entropy change is positive definite.
I'm going to stop here for now. If anyone is interested in my continuing this development to prove the above, please indicate so in a comment and I will continue.
