For the process, 1 Ar (300 K, 1 bar) → 1 Ar (200 K, 10 bar), assuming ideal gas behaviour, the change in molar entropy is

Δ S = n C p ln ⁡ T 2 T 1 - n R ln ⁡ p 2 p 1 = 2.303 C p log ⁡ T 2 T 1 - 2.303 R log ⁡ p 2 p 1 For monoatomic gas like Ar, C p = 5 2 R = 5 × 8.314 2 = 20.8 Δ S = 2.303 × 20.8 log ⁡ 200 300 - 2.303 × 8.314 log ⁡ 10 1 = 2.303 × 20.8 log ⁡ 2 3 - 2.303 × 8.314 = 47.9 × - 0.176 - 19.15 = - 8.43 - 19.15 = - 27.58 J/K/mol

Chemistry - Thermodynamics - II Question with Solution | TestHub

ChemistryThermodynamics - IIAdiabatic Process/PolytropicMedium2 minPYQ_2020

ChemistryMediumsingle choice

For the process, 1 Ar (300 K, 1 bar) $\to$ 1 Ar (200 K, 10 bar), assuming ideal gas behaviour, the change in molar entropy is

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Answer:

Solution:

$Δ S = n C_{p} \ln \frac{T_{2}}{T_{1}} - n R \ln \frac{p_{2}}{p_{1}}$

$= 2.303 C_{p} \log \frac{T_{2}}{T_{1}} - 2.303 R \log \frac{p_{2}}{p_{1}}$

For monoatomic gas like Ar, $C_{p} = \frac{5}{2} R$

$= \frac{5 \times 8.314}{2} = 20.8$

$Δ S = 2.303 \times 20.8 \log \frac{200}{300} - 2.303 \times 8.314 \log \frac{10}{1}$

$= 2.303 \times 20.8 \log \frac{2}{3} - 2.303 \times 8.314$

$= 47.9 \times (- 0.176) - 19.15$

$= - 8.43 - 19.15 = - 27.58$ J/K/mol

Stream:NTA_ABHYASSubject:ChemistryTopic:Thermodynamics - IISubtopic:Adiabatic Process/Polytropic

⏱ 2mℹ️ Source: PYQ_2020

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