Entropy is a state function and its value depends on two or three variable temperature (T), Pressure(P) and volume (V). Entropy change for an ideal gas having number of moles(n) can be determined by the following equation.
`DeltaS=2.303 “nC”_(v)”log”((T_(2))/(T_(1))) + 2.303 “nR log” ((V_(2))/(V_(1)))`
`DeltaS=2.303 “nC”_(P) “log”((T_(2))/(T_(1))) + 2.303 “nR log” ((P_(1))/(P_(2)))`
Since free energy change for a process or a chemical equation is a deciding factor of spontaneity , which can be obtained by using entropy change `(DeltaS)` according to the expression, `DeltaG = DeltaH -TDeltaS` at a temperature `T`
An isobaric process having one mole of ideal gas has entropy change `23.03` J/K for the temperature range `27^(@)` C to `327^(@)` C . What would be the molar specific heat capacity `(C_(v))`?
A. `(10)/(“log”2) J//K mol`
B. `(10)/(“log”2) – 8.3 J//K` mol
C. `10 xx “log”2 J//K mol`
D. `10 log2 + 8.3 J//k` mol

Entropy is a state function and its value depends on two or three variable temperature (T), Pressure(P) and volume (V). Entropy change for an ideal gas having number of moles(n) can be determined by the following equation.
`DeltaS=2.303 “nC”_(v)”log”((T_(2))/(T_(1))) + 2.303 “nR log” ((V_(2))/(V_(1)))`
`DeltaS=2.303 “nC”_(P) “log”((T_(2))/(T_(1))) + 2.303 “nR log” ((P_(1))/(P_(2)))`
Since free energy change for a process or a chemical equation is a deciding factor of spontaneity , which can be obtained by using entropy change `(DeltaS)` according to the expression, `DeltaG = DeltaH -TDeltaS` at a temperature `T`
What would be the entropy change involved in thermodynamic expansion of `2` moles of a gas from a volume of `5` L to a volume of `50` L at `25^(@)` C [Given `R=8.3 J//”mole”-K]`
A. `38.23` J/K mol
B. `26.76` J/K
C. `20` J/k
D. `28.23` J/K

Entropy is a state function and its value depends on two or three variable temperature (T), Pressure(P) and volume (V). Entropy change for an ideal gas having number of moles(n) can be determined by the following equation.
`DeltaS=2.303 “nC”_(v)”log”((T_(2))/(T_(1))) + 2.303 “nR log” ((V_(2))/(V_(1)))`
`DeltaS=2.303 “nC”_(P) “log”((T_(2))/(T_(1))) + 2.303 “nR log” ((P_(1))/(P_(2)))`
Since free energy change for a process or a chemical equation is a deciding factor of spontaneity , which can be obtained by using entropy change `(DeltaS)` according to the expression, `DeltaG = DeltaH -TDeltaS` at a temperature `T`
For a reaction `M_(2)O(s) rarr 2M(s) + (1)/(2) O_(2)(g) , DeltaH = 30 KJ//mol` and `DeltaS=0.07 KJ/mole “at” 1` atm. Calculate upto which temperature the reaction would not be spontaneous.
A. `T gt 428.6` K
B. `T gt 300.8` K
C. `T lt 300.8` K
D. `T lt 428.6` K