The entropy change when 2 , text{moles} of an | vedclass

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(C) The formula for entropy change is $\Delta S = nC_V \ln \left( \frac{T_2}{T_1} \right) + nR \ln \left( \frac{V_2}{V_1} \right)$.Since the process i

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$3 R \ln \left( \frac{573}{473} \right)$

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(C) The formula for entropy change is $\Delta S = nC_V \ln \left( \frac{T_2}{T_1} ight) + nR \ln \left( \frac{V_2}{V_1} ight)$.Since the process is isochoric,the volume remains constant,so $\Delta S = nC_V \ln \left( \frac{T_2}{T_1} ight)$.Given:$n = 2 \, 	ext{moles}$$C_V = \frac{3}{2} R$ (for a monoatomic gas)$T_1 = 200 + 273 = 473 \, 	ext{K}$$T_2 = 300 + 273 = 573 \, 	ext{K}$Substituting the values:$\Delta S = 2 	imes \left( \frac{3}{2} R ight) 	imes \ln \left( \frac{573}{473} ight)$$\Delta S = 3 R \ln \left( \frac{573}{473} ight)$.

The entropy change when $2 \, \text{moles}$ of an ideal monoatomic gas is heated from $200 \, ^\circ \text{C}$ to $300 \, ^\circ \text{C}$ isochorically is:

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