N_{2} gas is heated from 300, K to 600, K thr | vedclass

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(B) The change in entropy for an isobaric process is given by the formula:$\Delta S = n C_{P} \ln \left( \frac{T_{2}}{T_{1}} \right)$For a diatomic ga

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$20$

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(B) The change in entropy for an isobaric process is given by the formula:$\Delta S = n C_{P} \ln \left( \frac{T_{2}}{T_{1}} ight)$For a diatomic gas like $N_{2}$, the molar heat capacity at constant pressure is $C_{P} = \frac{7}{2} R$.Given: $n = 1\, 	ext{mole}$, $T_{1} = 300\, K$, $T_{2} = 600\, K$, and $R \approx 8.314\, J/(mol \cdot K)$.Substituting the values:$\Delta S = 1 	imes \frac{7}{2} 	imes 8.314 	imes \ln \left( \frac{600}{300} ight)$$\Delta S = 3.5 	imes 8.314 	imes \ln(2)$Using $\ln(2) \approx 0.693$:$\Delta S = 3.5 	imes 8.314 	imes 0.693 \approx 20.16\, J/K$.Rounding to the nearest integer, the change in entropy is $20\, J/K$.

$N_{2}$ gas is heated from $300\, K$ to $600\, K$ through an isobaric process. Find the change in entropy of the gas. $(n = 1\, \text{mole})$ (in $J/K$)

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