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Entropy for a Reversible Process

In an isobari...

Question

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

2.303 C_{p} log (T_{2} / T_{1})

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

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

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C_{v} ln (T_{2} / T_{1})

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Solution

The correct option is A $2.303 C_{p} log (T_{2} / T_{1})$
The entropy change for a process, when $T$ and $P$ are the variable is given by

$Δ S = C_{P} ln \frac{T_{2}}{T_{1}} - R ln \frac{P_{2}}{P_{1}}$

For an isobaric process $P_{1} = P_{2}$ .Hence the above equation reduces to

$C_{P} ln \frac{T_{2}}{T_{1}} = Δ S$

or $Δ S = 2.303 C_{p} log (T_{2} / T_{1})$

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