One mole of a monoatomic gas is brought from state 'A' to state 'B' along path ACB. Temperature at 'A' is 'To' Heat absorbed along path ACB is equal To
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a
112RT0
b
92RT0
c
72RT0
d
52RT0
answer is A.
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Detailed Solution
For ACP0V0T0=P02V0TC⇒TC=2T0For CBP02T0=2P0TB⇒TB=4T0For path ACBH=nCPTC−TA+nCVTB−TC=n52RT0+n32R2T0=112nRT0