Final answer:
The work done by the gas when its volume increases from 1.9 m^3 to 4.1 m^3 at atmospheric pressure is 222,915 J. The change in internal energy of the gas, after 5.80 × 10^5 J of heat is added, is 357,085 J.
Step-by-step explanation:
When 5.80 × 10^5 J of heat is added to a gas in a cylinder with a light frictionless piston at atmospheric pressure, and its volume increases from 1.9 m^3 to 4.1 m^3, we can calculate the work done by the gas and the change in internal energy of the gas.
To calculate the work done (W) by the gas, we use the equation for work in an isobaric (constant pressure) process: W = PΔV. Atmospheric pressure (P) can be taken as 101,325 Pa (Pascal), and the change in volume (ΔV) is 4.1 m^3 - 1.9 m^3 = 2.2 m^3. Hence, the work done by the gas is:
W = PΔV = (101,325 Pa) × (2.2 m^3) = 222,915 J
The change in internal energy (U) of the gas can be determined by using the first law of thermodynamics, which states that the change in internal energy is equal to the heat added to the system (Q) minus the work done by the system: ΔU = Q - W. Therefore, the change in internal energy is:
ΔU = Q - W = (5.80 × 10^5 J) - (222,915 J) = 357,085 J