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Properties of Enthalpy

An ideal gas ...

Question

An ideal gas of mass $m$ and temperature $T_{1}$ undergoes a reversible isothermal process from an initial pressure $P_{1}$ to final pressure $P_{2}$ . The heat loss during the process is $Q$ . The entropy change $Δ S$ of the gas is

A

m R I n (\frac{P_{2}}{P_{1}}) - \frac{Q}{T_{1}}

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B

m R I n (\frac{P_{1}}{P_{2}})

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C

m R I n (\frac{P_{2}}{P_{1}})

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D

zero

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Solution

The correct option is B $m R I n (\frac{P_{1}}{P_{2}})$
We know that
$d S = \frac{δ Q}{T}$
Intergrating between states $1$ and $2$ gives
$\int_{1}^{2} Δ S = \int_{1}^{2} \frac{δ Q}{T}$
$S_{2} - S_{1} = \frac{1}{T} \int_{1}^{2} δ Q$ $∵ T = C$
$= \frac{1}{T} \times Q_{1 - 2} = \frac{m R T I n \frac{V_{2}}{V_{2}}}{T}$
$m R I n \frac{V_{2}}{V_{1}} = m R I n \frac{p_{1}}{p_{2}}$
where $\frac{p_{1}}{p_{2}} = \frac{V_{2}}{V_{1}}$ for isothermal process

Alternatively:
We know that the entropy change in terms of temperautre and pressure rations.
$S_{2} - S_{1} = m c_{p} I n \frac{T_{2}}{T_{1}} - S_{2} - S_{1} = m R I n \frac{p_{2}}{p_{1}}$
where $T_{1} = T_{2}$ for isothermal process
$∴ S_{2} - S - 1 = - m R I n \frac{p_{2}}{p_{1}} = m R I n \frac{p_{1}}{p_{2}}$

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