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Standard XII

Chemistry

Law of Mass Action

One kg of air...

Question

One kg of air $(R = 287 J / k g K)$ undergoes an irreversible process between equilibrium state $1 (20^{o} C, 0.9 m^{3})$ and equilibrium state $2 (20^{o} C, 0.6 m^{3})$ . The change in entroy $s_{2} - s_{1}$ (in J/kgK) is
-116.368

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Solution

The correct option is A -116.368
Change in entropy,
$s_{2} - s_{1} = c_{v}$ ln $\frac{T_{2}}{T_{1}} + R l n \frac{V_{2}}{V_{1}} = R l n \frac{V_{2}}{V_{1}}$
Since internal energy,
$d u = 0$
So, $T_{1} = T_{2}$
$∴ s_{2} - s_{1} = 287 l n \frac{0.6}{0.9} = - 116.368$ J\kgK

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One kg of air (R=287J/kgK) undergoes an irreversible process between equilibrium state 1(20oC,0.9m3) and equilibrium state 2(20oC,0.6m3). The change in entroy s2−s1 (in J/kgK) is -116.368

The correct option is A -116.368Change in entropy, s2−s1=cv ln T2T1+RlnV2V1=RlnV2V1 Since internal energy, du=0 So, T1=T2 ∴s2−s1=287ln0.60.9=−116.368J\kgK

One kg of air $(R = 287 J / k g K)$ undergoes an irreversible process between equilibrium state $1 (20^{o} C, 0.9 m^{3})$ and equilibrium state $2 (20^{o} C, 0.6 m^{3})$ . The change in entroy $s_{2} - s_{1}$ (in J/kgK) is
-116.368

The correct option is A -116.368
Change in entropy,
$s_{2} - s_{1} = c_{v}$ ln $\frac{T_{2}}{T_{1}} + R l n \frac{V_{2}}{V_{1}} = R l n \frac{V_{2}}{V_{1}}$
Since internal energy,
$d u = 0$
So, $T_{1} = T_{2}$
$∴ s_{2} - s_{1} = 287 l n \frac{0.6}{0.9} = - 116.368$ J\kgK