What is the entropy change for the isothermal expansion of $0.75 g$ $"0.75 g"$ of $He$ $"He"$ from $5.0 L$ $"5.0 L"$ to $12.5 L$ $"12.5 L"$ ?

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What is the entropy change for the isothermal expansion of $0.75 g$ $"0.75 g"$ of $He$ $"He"$ from $5.0 L$ $"5.0 L"$ to $12.5 L$ $"12.5 L"$ ?

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Answer 1 · 2017-03-30T16:42:44

I got $1.428 J/K$ $"1.428 J/K"$ .

I assume you aren't given an equation and have to derive it.

If we treat entropy as a function of the temperature and volume (that is, $S = S (T, V)$ $S = S(T,V)$ ), and consider a constant-temperature expansion, then the total derivative is:

$d S (T, V) = {(\frac{\partial S}{\partial T})}_{V} d T + {(\frac{\partial S}{\partial V})}_{T} d V$ $dS(T,V) = ((delS)/(delT))_VdT + ((delS)/(delV))_TdV$

Since we are looking at the change in entropy due to a change in volume, we only consider ${(\frac{\partial S}{\partial V})}_{T}$ $((delS)/(delV))_T$ . Take the time to digest this step. This is the key step to understanding where to begin.

Another key step is to relate this to a derivative we are more familiar with. Recall that the Helmholtz free energy $A$ $A$ is a function of temperature and volume, and that its Maxwell relation is:

$d A = - S d T - P d V = - (S d T + P d V)$ $dA = -SdT - PdV = -(SdT + PdV)$

Since $A$ $A$ is a state function, its second derivatives are continuous, so that its cross derivatives are equal.

${(\frac{\partial S}{\partial V})}_{T} = {(\frac{\partial P}{\partial T})}_{V}$ $((delS)/(delV))_T = ((delP)/(delT))_V$

You should get to know this pretty well. The righthand side is a derivative we are familiar with. The ideal gas law can then be used.

$P V = n R T$ $PV = nRT$

$\Rightarrow P = \frac{n R T}{V}$ $=> P = (nRT)/V$

Now, the derivative is feasible in terms of an equation we know:

${(\frac{\partial P}{\partial T})}_{V} = \frac{\partial}{\partial T} {[\frac{n R T}{V}]}_{V}$ $((delP)/(delT))_V = del/(delT)[(nRT)/V]_V$

$= \frac{n R}{V} \frac{d T}{d T}$ $= (nR)/V(dT)/(dT)$

$= \frac{n R}{V}$ $= (nR)/V$

This therefore gives:

${(\frac{\partial S}{\partial V})}_{T} = {(\frac{\partial P}{\partial T})}_{V} = \frac{n R}{V}$ $((delS)/(delV))_T = ((delP)/(delT))_V = (nR)/V$

Next, we can integrate this over the two volumes to get the change in entropy. By moving the $d V$ $dV$ in the ${(\frac{\partial S}{\partial V})}_{T}$ $((delS)/(delV))_T$ derivative onto the other side:

$int_(S_1)^(S_2) dS = DeltaS = int_(V_1)^(V_2) (nR)/VdV$

The integral of $1/V$ is $ln|V|$ , so we have:

$DeltaS = nRint_(V_1)^(V_2) 1/VdV = nR|[ln|V|]|_(V_1)^(V_2)$

$= nR(lnV_2 - lnV_1)$

$=> color(blue)(DeltaS = nRln(V_2/V_1))$

So, given $"0.75 g"$ of helium, we can get the mols of gas and thus find the change in entropy.

$color(blue)(DeltaS) = 0.75 cancel"g He" xx cancel("1 mol")/(4.0026 cancel"g He") xx "8.314472 J/"cancel"mol"cdot"K" xx ln("12.5 L"/"5.0 L")$

$=$ $color(blue)("1.428 J/K")$

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What is the entropy change for the isothermal expansion of $0.75 g$ $"0.75 g"$ of $He$ $"He"$ from $5.0 L$ $"5.0 L"$ to $12.5 L$ $"12.5 L"$ ?

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What is the entropy change for the isothermal expansion of $0.75 g$ $"0.75 g"$ of $He$ $"He"$ from $5.0 L$ $"5.0 L"$ to $12.5 L$ $"12.5 L"$ ?