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 $T_{1}$ , pressure $P_{1}$ and volume $V_{1}$ and the spring is in its relaxed state. The gas is then heated very slowly to temperature $T_{2},$ pressure $P_{2}$ and volume $V_{2}$ . 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)

Question

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 $T_{1}$ , pressure $P_{1}$ and volume $V_{1}$ and the spring is in its relaxed state. The gas is then heated very slowly to temperature $T_{2},$ pressure $P_{2}$ and volume $V_{2}$ . 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)

Infinity Learn · Accepted Answer

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 Δ$$U=nCV($$Δ$$T)$$ [valid irrespective of the process ⇒Δ$$U=P1V1RT13R23T1$$−$$T1=3P1V1Ans$$. 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

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