Calculate the work done on 16.0 g of oxygen gas at 100 degrees C that is compressed isothermally to half its initial volume.

The given temperature is $100^\circ\text{C}$.

$$T = 100^\circ\text{C} + 273.15 = 373.15 \text{ K}$$

The mass of Oxygen gas is $16.0 \text{ g}$.
The molar mass of Oxygen gas ($\text{O}_2$) is $2 \times 16.00 \text{ g/mol} = 32.00 \text{ g/mol}$.

$$n = \frac{\text{mass}}{\text{molar mass}} = \frac{16.0 \text{ g}}{32.00 \text{ g/mol}} = 0.50 \text{ mol}$$

For an isothermal process, the work done on the gas is given by:

$$W = nRT \ln\left(\frac{V_i}{V_f}\right)$$

Where:
$n$ is the number of moles
$R$ is the ideal gas constant ($8.314 \text{ J/(mol}\cdot\text{K)}$)
$T$ is the temperature in Kelvin
$V_i$ is the initial volume
$V_f$ is the final volume

The problem states that the volume is halved, so $V_f = \frac{1}{2}V_i$.
Therefore, $\frac{V_i}{V_f} = \frac{V_i}{\frac{1}{2}V_i} = 2$.

$$W = (0.50 \text{ mol}) \times (8.314 \text{ J/(mol}\cdot\text{K)}) \times (373.15 \text{ K}) \times \ln(2)$$ $$W = 0.50 \times 8.314 \times 373.15 \times 0.693147$$ $$W \approx 1074.00 \text{ J}$$

Rounding to the nearest whole number, the work done is $1074 \text{ J}$.