You visited us 1 times! Enjoying our articles? Unlock Full Access!

Entropy Change in Isochoric Process

One kg of an ...

Question

One $k g$ of an idela gas (gas constnat $R = 287 J / k g K$ ) undergoes a nirreversible process from state $- 1$ ( $1$ bar, $300 K$ ) to state $- 2$ ( $2$ bar, $300 K$ . The change in specific entropy $(s_{2} - s_{1})$ of the gas (in $K / k g K$ ) in the process is
198.93

Open in App

Solution

The correct option is A 198.93
Change in specific entropy of idela gas
$s_{f} - s_{i} = c_{p} I n \frac{T_{f}}{T_{i}} - R I n \frac{P_{f}}{P_{i}}$
$= 1005 I n \frac{300}{300} - 287 I n \frac{2}{1}$
$0 - 287 I n 2 = - 198.93 J / k g - K$

Suggest Corrections

Similar questions

p_{1} = 1

T_{1} = 27^{o} C

p_{2} = 3

T_{2} = 127^{o} C

287

γ = 1.4

1

T_{1} = 300 K, P_{1} = 100 k P a

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

1

2

(R = 287 J / k g K)

1 (20^{o} C, 0.9 m^{3})

2 (20^{o} C, 0.6 m^{3})

s_{2} - s_{1}

c_{p} = 1 k J / k g K, R = 0.30 k J / k g K

R = 400

c_{v} = 1000

1

300

100

Join BYJU'S Learning Program