Final answer:
The work done by the gas in the second situation, where it was heated at constant pressure, is the difference in heat required between constant volume and constant pressure heating. As the gas requires 50 J at constant volume and 46 J at constant pressure, the work done by the gas is 4 J.
Step-by-step explanation:
The question regards the work done by a gas during a heating process at constant pressure in comparison to constant volume. According to the first law of thermodynamics, when a gas is heated at constant volume, all the heat supplied (Q) increases the internal energy (U), and as there is no work done (W), the heat input equals the change in internal energy. However, when a gas is heated at constant pressure the gas does work on its surroundings as it expands and therefore the heat added is shared between raising the internal energy and doing work.
In the given situation, for a temperature increase ΔT1, the gas needs 50 J at constant volume but only 46 J at constant pressure. The difference between these amounts of heat directly gives us the work done by the gas since at constant pressure Q = ΔU + W, and at constant volume Q = ΔU, leading to ΔU being the same. Therefore, the work done W by the gas at constant pressure is the difference between the heat supplied at constant volume and constant pressure, which is 50 J - 46 J = 4 J.