## Related questions with answers

Suppose $0.420 \mathrm{~mol}$ 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?

Solution

Verified$\textbf{Given:}$

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

$\textbf{Solution:}$

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

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

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

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

Plugging in Equation 1 to Equation 2, we get:

$\begin{align} \Delta S &= \dfrac{nRT \ln(\dfrac{V_2}{V_1})}{T} \\ \\ & \boxed{\Delta S = nR \ln(\dfrac{V_2}{V_1})} \end{align}$

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

$\textbf{A:}$

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

## Create an account to view solutions

## Create an account to view solutions

## More related questions

1/4

1/7