Q5. A cylinder with insulated walls contains a movable insulated piston. On each side of the piston are n moles of an ideal g Show transcribed image text Expert Answerinformation icon Pankaj Singh's Avatar Pankaj Singh answered this Was this answer helpful? Thumbs up inactive0Thumbs down inactive0 760 answers SInce the there is insulation around the piston and walls thus system is closed which means that the no of moles is going to remain same. Now on the side without heater, the whole setup is insulated and no heat is given thus; Q = 0 => that the process is adiabatic which means that PV\gamma= constant Thus;PoVo1.4 = 3PoV1.4 => V = 0.456Vo So, by using ideal gas equation; PoVo/To = 3PoV/T => T = 1.37To Now in adiabatic process; work done on gas is given by W = dU => W = nCv(T-To) => W = 0.37nCvTo b) On the heater side the since the whole volume of the cylinder is constant we saw that on the non-heater side the volume compressed to 0.456Vo which means that the gas expanded on heater side by 2Vo - 0.456Vo = 1.544Vo For piston to be at equilibrium the pressure has to be same on both sides; Thus P = 3Po Now using the ideal gas equation; PoVo/To = PV/T => PoVo/To = 3Po*1.544Vo/T => T = 4.632To Now the work done by the gas here will be equal to the work done on the gas in the non-heater chamber, because of the piston. W = 0.37nCvTo Now the change in internal energy is given as- dU = nCv(4.632To - To) => dU = 3.632nCvTo Now using thermodynamic law; heat flown can be given as dQ = W + dU => dQ = 0.37nCvTo + 3.632nCvTo => dQ = 4nCvTo