As illustrated in the figure, a cylindrical container holds oxygen...

The pressure of the gas inside the cylinder is the sum of the atmospheric pressure and the pressure due to the weight of the piston.

$P = P_{\text{atm}} + \dfrac{mg}{A}$

Substituting the given values:

$P = 100 \times 10^3 + \dfrac{50 \times 10}{100 \times 10^{-4}}$

$P = 10^5 + 0.5 \times 10^5 = 1.5 \times 10^5 \text{ Pa}$

Since the cylinder is heated slowly, the process is isobaric. The change in volume of the gas is:

$\Delta V = A \times d = 100 \times 10^{-4} \times 0.2 = 2 \times 10^{-3} \text{ m}^3$

The work done by the gas during this expansion is:

$W = P \Delta V = (1.5 \times 10^5) \times (2 \times 10^{-3}) = 300 \text{ J}$

For an isobaric process, the heat supplied $Q$ is related to the work done $W$ by:

$Q = n C_p \Delta T = \dfrac{C_p}{R} (n R \Delta T) = \dfrac{\gamma}{\gamma - 1} W$

Given $\gamma = 1.4$, we get:

$Q = \dfrac{1.4}{1.4 - 1} \times 300$

$Q = \dfrac{1.4}{0.4} \times 300 = \dfrac{7}{2} \times 300 = 1050 \text{ J}$