What is the entropy change in $2 m o l N_{2}$ when its temperature is taken from $400 K$ to $800 K$ , at constant pressure?

30 J / K

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60 J / K

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40 J / K

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20 J / K

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Solution

The correct option is C $40 J / K$
$Δ S = n C_{P} l n \frac{T_{2}}{T_{1}}$

$C_{v} (for diatomic) = \frac{5}{2} R$
$C_{p} = C_{v} + R$
$C_{P} = \frac{7}{2} R$
$∴$
$Δ S = 2 \times \frac{7}{2} R \times l n \frac{800}{400} = 40.33 \approx 40 J / K$

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Pressure of 10 moles of an ideal gas is changed from 2 atm to 1 atm against constant external pressure without change in temperature. If surrounding temperature (300 K) and pressure (1 atm) always remains constant then calculate total entropy change ( $Δ S_{s y s t e m} + Δ S_{s u r r o u n d i n g})$ for given process. [Given: ℓn2 = 0.70 amd R=8.0 J/mol/K]

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Q. One mole of an ideal monoatomic gas expands isothermally against constant external pressure of 1 atm from initial volume of 1L to a state where its final pressure becomes equal to external pressure. If initial temperature of gas is 300 K then total entropy change of system in the above process is :
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What is the entropy change in 2 mol N2 when its temperature is taken from 400 K to 800 K, at constant pressure?

The correct option is C 40 J/KΔS=nCPlnT2T1 Cv (for diatomic) =52R Cp=Cv+R CP=72R ∴ ΔS=2×72R×ln800400=40.33≈40 J/K

What is the entropy change in $2 m o l N_{2}$ when its temperature is taken from $400 K$ to $800 K$ , at constant pressure?

The correct option is C $40 J / K$
$Δ S = n C_{P} l n \frac{T_{2}}{T_{1}}$

$C_{v} (for diatomic) = \frac{5}{2} R$
$C_{p} = C_{v} + R$
$C_{P} = \frac{7}{2} R$
$∴$
$Δ S = 2 \times \frac{7}{2} R \times l n \frac{800}{400} = 40.33 \approx 40 J / K$