Calculate the work done by an ideal gas during an irreversible expansion from 5 atm to 1 atm at 25°C with n=1 mol and P_ext=1 atm.

The number of moles of ideal gas is $n = 1 \text{ mol}$.
The initial temperature is $T = 25^\circ\text{C}$. Convert this to Kelvin:

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

The initial internal pressure of the gas is $P_1 = 5 \text{ atm}$.
The constant outside pressure against which the gas expands is $P_{ext} = 1 \text{ atm}$.
The final internal pressure of the gas is $P_2 = 1 \text{ atm}$.
The gas constant $R = 0.08206 \text{ L atm mol}^{-1}\text{ K}^{-1}$ will be used for volume calculations, and $1 \text{ L atm} = 101.325 \text{ J}$ for work conversion.

For an irreversible expansion against a constant external pressure, the work done ($W$) is given by:

$$W = -P_{ext} \Delta V$$

where $\Delta V = V_2 - V_1$ is the change in volume. We assume the temperature remains constant during the expansion for volume calculations, as no information about temperature change is provided.

The ideal gas law is $PV = nRT$. So, $V = \frac{nRT}{P}$.

$$V_1 = \frac{nRT}{P_1}$$ $$V_1 = \frac{(1 \text{ mol})(0.08206 \text{ L atm mol}^{-1}\text{ K}^{-1})(298.15 \text{ K})}{5 \text{ atm}}$$ $$V_1 = \frac{24.466 \text{ L atm}}{5 \text{ atm}}$$ $$V_1 = 4.8932 \text{ L}$$

$$V_2 = \frac{nRT}{P_2}$$ $$V_2 = \frac{(1 \text{ mol})(0.08206 \text{ L atm mol}^{-1}\text{ K}^{-1})(298.15 \text{ K})}{1 \text{ atm}}$$ $$V_2 = \frac{24.466 \text{ L atm}}{1 \text{ atm}}$$ $$V_2 = 24.466 \text{ L}$$

$$\Delta V = V_2 - V_1$$ $$\Delta V = 24.466 \text{ L} - 4.8932 \text{ L}$$ $$\Delta V = 19.5728 \text{ L}$$

$$W = -P_{ext} \Delta V$$ $$W = -(1 \text{ atm})(19.5728 \text{ L})$$ $$W = -19.5728 \text{ L atm}$$

Using the conversion factor $1 \text{ L atm} = 101.325 \text{ J}$:

$$W = -19.5728 \text{ L atm} \times \frac{101.325 \text{ J}}{1 \text{ L atm}}$$ $$W = -1983.10 \text{ J}$$

The work done is:

$$-1983.10 \text{ J}$$