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Enthalpy

An ideal gas ...

Question

An ideal gas undergoes a process from state $1$ $T_{1} = 300 K, P_{1} = 100 k P a$ to state $2$ $T_{2} = 600 K, P_{2} = 500 k P a$ . The specific heats of the ideal gas are: $c_{p} = 1 k J / k g - K$ and $c_{v} = 0.7 k J / k g - K$ . The change in specific entropy of the ideal gas from state $1$ to state $2$ (in kJ/kg-K) is
(correct to two decimal places)
0.21

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Solution

The correct option is A 0.21
Ideal gas
State -1: $T_{1} = 300 K, p_{1} = 100 k P a$
State-2: $T_{2} = 600 K, p_{2} = 500 k P a,$
$c_{p} = 1 k J / k g - K, C_{v} = 0.7 k J / k g - k,$
$c_{p} - c_{v} = R,$
$\Rightarrow c_{p} - c_{v} = 1 - 0.7 = R$
$R = 0.3 k J / k g - K$
Change in specific entropy,
$s_{2} - s_{1} = c_{p} l n \frac{T_{2}}{T_{1}} - R l n \frac{p_{2}}{p_{1}}$
$= 1 \times l n \frac{600}{300} - 0.3 l n \frac{500}{100}$
$= 0.21 k J k g - K$

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