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Suppose 0.420 mol0.420 \mathrm{~mol} of an ideal gas is isothermally and reversibly expanded in the four situations given below. What is the change in the entropy of the gas for each situation?

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Answered 2 years ago
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Given:\textbf{Given:}

n=0.550moln = 0.550 \: \: \mathrm{mol}

Solution:\textbf{Solution:}

We know that for an isothermal process, the change in internal energy ΔU=0\Delta U = 0, which means that the work WW is just equal to the heat QQ. Moreover, the formula in obtaining the work in this case is:

Q=W=nRTln(V2V1)\begin{align} Q = W = nRT \ln(\dfrac{V_2}{V_1}) \end{align}

The change in entropy ΔS\Delta S can be calculated by:

ΔS=QT\begin{align} \Delta S &= \dfrac{Q}{T} \end{align}

Plugging in Equation 1 to Equation 2, we get:

ΔS=nRTln(V2V1)TΔS=nRln(V2V1)\begin{align} \Delta S &= \dfrac{nRT \ln(\dfrac{V_2}{V_1})}{T} \\ \\ & \boxed{\Delta S = nR \ln(\dfrac{V_2}{V_1})} \end{align}

Now we can proceed in determining the change in entropy for all cases as shown below:

A:\textbf{A:}

ΔS=nRln(V2V1)=0.5508.314ln(0.8000.200)=6.34JK\begin{align*} \Delta S &= nR \ln(\dfrac{V_2}{V_1}) \\ &= 0.550 \cdot 8.314 \cdot \ln(\dfrac{0.800}{0.200}) \\ &= \boxed{6.34 \: \: \mathrm{\dfrac{J}{K}}} \end{align*}

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