Suppose 4.00 mol of an ideal gas undergo a reversible isothermal expansion from volume $V_{1}$ to volume $V_{2} = 2.00 V_{1}$ at temperature T = 400 K. Find the entropy change of the gas. If the expansion is reversible and adiabatic instead of isothermal,what is the entropy change of the gas?

23.05 \frac{J}{K}, 0 \frac{J}{K}

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24.05 \frac{J}{K}, 0 \frac{J}{K}

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0 \frac{J}{K}, 23.05 \frac{J}{K}

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0 \frac{J}{K}, 0 \frac{J}{K}

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Solution

The correct option is A $23.05 \frac{J}{K}, 0 \frac{J}{K}$
The correct answer is $23.05 \frac{J}{K}, 0 \frac{J}{K}$
$W = n R T ln \frac{V_{2}}{V_{1}} = 4 \times 8.31 \times 400 \times ln \frac{2.0 V_{1}}{V_{1}} = 9216 J$
$Δ S = \int_{i}^{f} \frac{d Q}{T}$
For isothermal process: T=constant
$Δ S = \frac{Q}{T}; Δ E_{i n t} = Q - W = 0 ⟹ Q = W$
Hence,
$Δ S = \frac{W}{T} = \frac{9216}{400} = 23.05 J K^{- 1}$
If the expansion is reversible and adiabatic,
$Q = 0 ⟹ Δ S = 0 J K^{- 1}$

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Suppose 4.00 mol of an ideal gas undergo a reversible isothermal expansion from volume V1 to volume V2=2.00V1 at temperature T = 400 K. Find the entropy change of the gas. If the expansion is reversible and adiabatic instead of isothermal,what is the entropy change of the gas?

The correct option is A 23.05JK,0JKThe correct answer is 23.05JK,0JKW=nRTlnV2V1=4×8.31×400×ln2.0V1V1=9216JΔS=∫fidQTFor isothermal process: T=constantΔS=QT;ΔEint=Q−W=0⟹Q=WHence,ΔS=WT=9216400=23.05JK−1If the expansion is reversible and adiabatic,Q=0⟹ΔS=0JK−1

Suppose 4.00 mol of an ideal gas undergo a reversible isothermal expansion from volume $V_{1}$ to volume $V_{2} = 2.00 V_{1}$ at temperature T = 400 K. Find the entropy change of the gas. If the expansion is reversible and adiabatic instead of isothermal,what is the entropy change of the gas?