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DeltaS=(q(rev))/(T), so...

`DeltaS=(q_(rev))/(T)`, so

A

`Delta S` is defined only for reversible process.

B

For irreversible process, `DeltaS` is calculated by considering the irreversible.

C

For irreversible process, `A rarr B` and same process taking place reversible, `DeltaS` is same.

D

`Delta S_(sys)` is always the for irreversible process.

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The correct Answer is:
3
S is state function hence `DeltaS` will be indepent of type of process ( reversible `//` irrersible ) for `A rarr B`
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Assertion:The SI unit of entropy is JK^(-1)mol^(-1) Reason: DeltaS=(q_(rev))/(T)

The entropy change can be calculated by using the expression DeltaS-(q_(rev))/(T) . When water freezes in a glass beaker, choose the correct statement amongst the following:

Work done by the system in isothermal reversible process is w_(rev.)= -2.303 nRT "log"(V_(2))/(V_(1)) . Also in case of adiabatic reversible process work done by the system is given by: w_(rev.) = (nR)/(gamma -1) [T_2 - T_1] . During expansion disorder increases and the increase in disorder is expressed in terms of change in entropy DeltaS = q_(rev.)/T . The entropy changes also occurs during transformation of one state to other end expressed as DeltaS = DeltaH/T . Both entropy and enthalpy changes obtained for a process were taken as a measure of spontaniety of process but finally it was recommended that decrease in free energy is responsible for spontaniety and DeltaG=DeltaH - T DeltaS . The heat of vaporisation and heat of fusion of H_2O are 540 cal//g and 80 cal//g . This ratio of (DeltaS_(vap.))/(DeltaS_("fusion")) for water is:

Work done by the system in isothermal reversible process is w_(rev.)= -2.303 nRT "log"(V_(2))/(V_(1)) . Also in case of adiabatic reversible process work done by the system is given by: w_(rev.) = (nR)/(gamma -1) [T_2 - T_1] . During expansion disorder increases and the increase in disorder is expressed in terms of change in entropy DeltaS = q_(rev.)/T . The entropy changes also occurs during transformation of one state to other end expressed as DeltaS = DeltaH/T . Both entropy and enthalpy changes obtained for a process were taken as a measure of spontaniety of process but finally it was recommended that decrease in free energy is responsible for spontaniety and DeltaG=DeltaH - T DeltaS . A chemical change will definitely be spontaneous if:

Work done by the system in isothermal reversible process is w_(rev.)= -2.303 nRT "log"(V_(2))/(V_(1)) . Also in case of adiabatic reversible process work done by the system is given by: w_(rev.) = (nR)/(gamma -1) [T_2 - T_1] . During expansion disorder increases and the increase in disorder is expressed in terms of change in entropy DeltaS = q_(rev.)/T . The entropy changes also occurs during transformation of one state to other end expressed as DeltaS = DeltaH/T . Both entropy and enthalpy changes obtained for a process were taken as a measure of spontaniety of process but finally it was recommended that decrease in free energy is responsible for spontaniety and DeltaG=DeltaH - T DeltaS . Which statements are correct? (1) The expansion work for a gas into a vacuum is equal to zero. (2) 1 mole of a gas occupying 3 litre volume on expanding to 15 litre at constant pressure of 1atm does expansion work 1.215 kJ . (3) The maximum work done during expansion of 16gO_2 at 300K from 5dm^3 to 25 dm^3 is 2.01 kJ . (4) The DeltaS for S to L is almost negligible in comparision to DeltaS for L to G . (5) DeltaS = 2.303 nR "log"(V_(2))/(V_(1)). (at constant T )

Work done by the system in isothermal reversible process is w_(rev.)= -2.303 nRT "log"(V_(2))/(V_(1)) . Also in case of adiabatic reversible process work done by the system is given by: w_(rev.) = (nR)/(gamma -1) [T_2 - T_1] . During expansion disorder increases and the increase in disorder is expressed in terms of change in entropy DeltaS = q_(rev.)/T . The entropy changes also occurs during transformation of one state to other end expressed as DeltaS = DeltaH/T . Both entropy and enthalpy changes obtained for a process were taken as a measure of spontaniety of process but finally it was recommended that decrease in free energy is responsible for spontaniety and DeltaG=DeltaH - T DeltaS . Ag_2O_((s)) to 2Ag_((s)) + 1/2 O_(2(g)) attains equilibrium at temperature… K is : (The DeltaH and DeltaS for the reaction are 30.5kJ mol^(-1) and 66J mol^(-1) K^(-1) )

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