If the process carried out on one mole of monatomic ideal gas is shown in the figure PV-diagram with P0V0=13RT0, the correct match is, [2019]
(1)
I. W1→2→3=W1→2+W2→3 [W2→3=0, ΔV=0]
=P0×V0+0
=P0V0=RT03 [∵P0V0=13RT0 given]
∴ I→Q
II. ΔU1→2→3=ΔU1→2+ΔU2→3
=nCvΔT1→2+nCvΔT2→3
=1×3R2(Tf-Ti)1→2+1×3R2(Tf-Ti)2→3
=32[2P0V0-P0V0]+32[3P02×2V0-P0×2V0]
=3P0V0=3×13RT0=RT0
II→R
III. Q1→2→3=Q1→2+Q2→3=nCpΔT1→2+nCvΔT2→3
=52P0V0+1×32[3P02×2V0-P0(2V0)]
=82P0V0=82×RT03=43RT0
III→S
IV. Q1→2=nCpΔT1→2=nCp(Tf-Ti)
=1×5R2[P0(2V0)R-P0V0R] [∵PV=nRT]
=52P0V0=52(RT03)=5RT06
∴ IV→U