Home
Class 12
CHEMISTRY
Entropy is a state function and its valu...

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

Text Solution

Verified by Experts

The correct Answer is:
B
`DeltaS= 2.303 xx 1 xx C_(p)"log"((600)/(300)) = 23.03 implies " "C_(p) =(10)/("log"2)`
`C_(v) = C_(P)-R (10)/("log"2) -8.3`
Promotional Banner

Topper's Solved these Questions

  • THERMODYNAMICS

    RESONANCE|Exercise exercise-3 Part:(I)|23 Videos

  • THERMODYNAMICS

    RESONANCE|Exercise exercise-3 part-1 Advanced level Solutions|30 Videos

  • THERMODYNAMICS

    RESONANCE|Exercise exercise-2 Part:(III): One or more than one options correct|10 Videos

  • TEST SERIES

    RESONANCE|Exercise CHEMISTRY|50 Videos

Similar Questions

Explore conceptually related problems

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]

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.

If K_(c) is not numerically equal to K_(p) , how can both of the following equations be valid? DeltaG^(ɵ)=-2.303 RT log K_(c), DeltaG^(ɵ)=-2.303 RT log K_(p)

Examine the following plots and predict whether in (i) P_(1) lt P_(2) "and" T_(1) gt T_(2) , in (ii) T_(1) = T_(2)ltT_(3) , in (iii) V_(1) gt V_(2), "in" (iv) P_(1) gt P_(2) "or" P_(2) gt P_(1)