Here Q=0 and W=0. Therefore from first law of thermodynamics ΔU=Q+W=0∴ Internal energy of the system with partition = Internal energy of the system without partition.n1CvT1+n2CvT2=(n1+n2)CvT∴T=n1+n2n1T1+n2T2But n1=RT1P1V1 and n2=RT2P2V2∴T=RR1P1V1+RRR2R2V2RT1P1V1×T1+RR2P2V2×T2=P1V1T2+P2V2T1T1T2(P1V1+P2V2)