One mole of ideal monatomic gas was taken through isochoric heating from `100` K to `1000` K. Calculate `DeltaS_("system"), Delta_("surr")` and `Delta

Question

One mole of ideal monatomic gas was taken through isochoric heating from `100` K to `1000` K. Calculate `DeltaS_("system"), Delta_("surr")` and `Delta

One mole of ideal monatomic gas was taken through isochoric heating from `100` K to `1000` K. Calculate `DeltaS_("system"), Delta_("surr")` and `DeltaS_("total")` in
`(i)` When the process carried out reversibly `" "(ii)` When the process carried out irreversibly (one step)

Monjur Alahi

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2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - (i) Rev. Process `DeltaS_("system") =(3)/(2) R"In" 10; DeltaS_("surr") =-(3)/(2)R in 10 " " DeltaS_("total")=0`
Irr. Process `DeltaS_("system") =(3)/(2) R"In" 10; DeltaS_("surr") =-(3)/(2)R (0.9); DeltaS_("total")=(3)/(2)R(1.403)`
`(i)" "DeltaS_("system") = nC_(v) "In" (T_(2))/(T_(1)) = (3)/(2)"R" "In"10` and
`DeltaS_("surr") =int-(dq_("sys"))/(T) = underset(T_(1))overset(T_(2))int -(nC_(v)dT)/(T) =- nC_(v)"In" (T_(2))/(T_(1)) =-(3)/(2) "Rin" 10`
`(ii)` entropy is a state function.
`DeltaS_("system")=(3)/(2)"R" "In" 10`
`DeltaS_("surr") = (q)/(T) =-(nC_(v)(T_(2)-T_(1)))/(T_(2)) =- (3)/(2) "R" (0.9).`
`DeltaS_("Total") =(3)/(2)"RIn" 10 -(3)/(2) R(0.9)=(3)/(2)"R"(1.403)`