Ram joined Kota test series, where he was asked to calculate $Δ S_{syB}$ for a process as described below.
A diathermic container (containing an ideal gas) fitted with a piston at equilibrium (without any stopper) and has initial volume 600 litre. Now the external pressure is suddenly reduced to 1 bar and allowed the piston to move upward isothermally. In this process, system absorbs 600 Kj heat.
He calculated $Δ S_{Bув} = 2 kJ / K$ but was awarded zero marks. To identify his mistake he contacted his Kota friend Shyam and explained his problem. When Shyam asked how Ram calculated $Δ S_{sys,}$ . Ram
said simple ! by q/T. If you are Shyam , then help Ram in getting correct answer (in kJ/K). $[In 11 = 2.4]$

Question

Accepted Answer

Heat of system,qsys =−w=1(V−600)×1001000=600⇒ V=6600L;n=1×66000.083×300T=6002=300KΔsysS=nRln⁡V2V1=66000.083×300×8.3ln⁡6600600=5280J/K=5.28kJ/K

Detailed Solution