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Standard XII

Chemistry

Thermodynamic Processes

One kg of air...

Question

One kg of air in a closed system undergoes an irreversible process from an initial state of $p_{1} = 1$ bar (absolute) and $T_{1} = 27^{o} C$ , to a final state of $p_{2} = 3$ bar (absolute) and $T_{2} = 127^{o} C$ . If the gas constant of air is $287$ J/kgK and the ratio of the specific heats $γ = 1.4$ , then the change in the specific entropy (in J/kgK) of the air in the process is

28.4

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172.0

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indeterminate, as the process is irreversible

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- 26.3

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Solution

The correct option is D $- 26.3$
$P_{1} = 1 b a r, P_{2} = 3 b a r, T_{1} = 300 K, T_{2} = 400$
K,
$c_{p} = \frac{γ R}{γ - 1} = \frac{1.4 \times 287}{0.4}$
$= 1005 J / k g K$
$Δ S = c_{p} l n \frac{T_{2}}{T_{1}} - R l n \frac{P_{2}}{P_{1}}$
$= 1005 ln \frac{400}{300} - 287 l n \frac{3}{1}$
$- 26.32 J / k g K$

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The correct option is D −26.3 P1=1bar,P2=3bar,T1=300K,T2=400 K, cp=γRγ−1=1.4×2870.4 =1005J/kgK ΔS=cplnT2T1−RlnP2P1 =1005ln400300−287ln31 −26.32J/kgK