An ideal monoatomic gas is confined in a horizontal cylinder by a spring loaded piston (as shown in the figure).  Initially the gas is at temperature  T1, pressure P1  and volume  V1 and the spring is in its relaxed state.  The gas is then heated very slowly to temperature T2, pressure P2 and volume  V2.  During this process the piston moves out by a distance  x.Assume that pressure in the chamber containing spring remains the same throughout the motion of the piston. Ignoring the friction between the piston and the cylinder, the correct statement(s) is (are)

Ideal  mono atomic gas   γ=53Spring released state  Fsp=0 Gas heated slowly (external agent)(Process is not adiabatic / isobaric/ isochoric).  Initial conditions  Final conditions P1                               P2V1                               V2T1                                T2 (A)  V2=2V1T2=3T1,P2=? From ideal gas equation P1V1P2V2=nRT1nRT2⇒P1P212=13⇒P2=3P12Equilibrium of pistonP2=P1+Pspring ⇒3P12=P1+kxA⇒P12=KxA∴ Sping energy: U=12Kxsp2=Kxsp22K⇒U=12P12xA(x)2=12P12(Ax)=14P1V2−V1⇒U=P1V14 Ans. A correct  (B) If V2=2V1 and T2=3T1  Change in internal energy of gas ΔU=nCV(ΔT) [valid irrespective of the process ⇒ΔU=P1V1RT13R23T1−T1=3P1V1Ans. A Correct (C) If V2=3V1 and T2=4T1P=P1+kXAwork done by the gas is W=∫PdV=∫P1+kxAdV=∫P1dV+∫kxAAdx⇒W=P1V2−V1+kx22 Using ,P1V1P2V2=nRT1nRT2⇒P1P213=14⇒P2=43P1Using ,P=P1+kxA⇒43P1=P1+kxA⇒kx=P1A3 Also, V2=v1+Ax⇒3V1=V1+Ax⇒x=2V1AThus W=P1V2−V1+kx22=P13V1−V1+12×P1A3×2V1A=73P1V1Option C is correct Heat supplied =Q=W+ΔU⇒Q=73P1V1+32P2V2−P1V1⇒Q=73P1V1+3243P13V1−P1V1=416P1V1Option D is wrong