Calculate the work done for a reversible isothermal expansion of 1 mole of an ideal gas at 25°C from 10 atm to 5 atm.

Step 1: Identify given values and convert units.
The number of moles of gas is $n = 1 \text{ mol}$.
The process is reversible and isothermal, meaning the temperature is constant.
The temperature is $T = 25^\circ\text{C}$. Convert this to Kelvin:

$$T = 25 + 273.15 = 298.15 \text{ K}$$

The gas constant $R = 8.314 \text{ J mol}^{-1}\text{ K}^{-1}$.

The problem states the gas is "expanded reversibly and isothermally from 5 atm to 10 atm". An expansion implies that the volume increases and the pressure decreases. Therefore, the initial pressure ($P_1$) should be greater than the final pressure ($P_2$). The given pressures "from 5 atm to 10 atm" suggest an increase in pressure, which would be a compression, not an expansion. To be consistent with the term "expanded", we will assume the initial pressure was $P_1 = 10 \text{ atm}$ and the final pressure was $P_2 = 5 \text{ atm}$.

Step 2: Apply the formula for work done in a reversible isothermal process.
For a reversible isothermal process involving an ideal gas, the work done ($W$) is given by:

$$W = -nRT \ln\left(\frac{V_2}{V_1}\right)$$

For an isothermal process, $P_1V_1 = P_2V_2$, which means $\frac{V_2}{V_1} = \frac{P_1}{P_2}$.
Substituting this into the work equation:

$$W = -nRT \ln\left(\frac{P_1}{P_2}\right)$$

Step 3: Substitute the values and calculate the work done.
Using $n = 1 \text{ mol}$, $R = 8.314 \text{ J mol}^{-1}\text{ K}^{-1}$, $T = 298.15 \text{ K}$, $P_1 = 10 \text{ atm}$, and $P_2 = 5 \text{ atm}$:

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