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Entropy for a Reversible Process

One kg of an ...

Question

One kg of an ideal gas (gas constant, $R = 400$ J/kgK; specific heat at constant volume, $c_{v} = 1000$ J/kgK) at $1$ bar, and $300$ K is contained in sealed rigid cylinder. During an adiabatic process, $100$ kJ of work is done on the system by a stirrer. The increase in entropy of the system is..... J/K
287.682

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Solution

The correct option is A 287.682
$m = 1 k g; R = 400$ j/kg-K

$c_{v} = 1000 j / k g - K$
$T_{i} = 300 k; P_{i} = 10 k P a$
$W = 100 k J$
Applying first law:
$δ Q = Δ U + δ W$
$\Rightarrow$ $0 = m c_{v} (T_{f} - T_{i}) + (- 100) \times 10^{3} J$
$\Rightarrow$ $0 = 1 \times 1000 \times (T_{f} - 300) - 100 \times 10^{3}$
$\Rightarrow$ $T_{f} = 400 K$
Entorpy change for ideal gas
$Δ S = m m [c_{v} I n \frac{T_{f}}{T_{i}} + R I n \frac{V_{f}}{V_{i}}]$
(As volume is constnat $∵ V_{i} = V_{f})$
$= 1 \times [1000 I n \frac{400}{300} + 0]$
$= 287.682 J / K$

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