In an isobaric process, when temperature changes from T1 to T2, Δ S is equal to

Question

In an isobaric process, when temperature changes from T1 to T2, Δ S is equal to (A) 2.303 CP log (T2 / T1) (B) 2.303 CP log (T2 / T1) (C) Cp ln (T1 / T2) (D) CV ln (T2 / T1)

Tardigrade Admin · Accepted Answer

The entropy change for a process, when T and P are the variables is given by Δ S=CP In (T2/T1)-R In (P2/P1) For an isobaric process P1 = P2. Hence the above equation reduces to CPIn (T2/T1)=Δ S. or Δ S=2.303CP log (T2/T1).

Q. In an isobaric process, when temperature changes from $T_{1}$ to $T_{2}$ , $Δ S$ is equal to

$2.303 C_{P} lo g (T_{2} / T_{1})$

$2.303 C_{P} lo g (T_{2} / T_{1})$

$C_{p} ln (T_{1} / T_{2})$

$C_{V} ln (T_{2} / T_{1})$

Q. In an isobaric process, when temperature changes from T1​ to T2​, ΔS is equal to

2.303CP​log(T2​/T1​)

2.303CP​log(T2​/T1​)

Cp​ln(T1​/T2​)

CV​ln(T2​/T1​)

The entropy change for a process, when T and P are the variables is given by ΔS=CP​InT1​T2​​−RInP1​P2​​ For an isobaric process P1​=P2​. Hence the above equation reduces to CP​InT1​T2​​=ΔS. or ΔS=2.303CP​logT1​T2​​.

Q. In an isobaric process, when temperature changes from $T_{1}$ to $T_{2}$ , $Δ S$ is equal to

$2.303 C_{P} lo g (T_{2} / T_{1})$

$2.303 C_{P} lo g (T_{2} / T_{1})$

$C_{p} ln (T_{1} / T_{2})$

$C_{V} ln (T_{2} / T_{1})$