During the isobaric expansion;
The work done by a gas in an isolated expansion is the product of pressure and change in volume.
[tex]W = p\times \Delta V[/tex]
1 atm = 101325 Pa
2.4 cm³ = 2.4 × 10^-6 m³
4514.7 cm³ = 4514.7 × 10^-6 m³
[tex]W = 101325 Pa \times (4514.7\times 10^{-6} m^3 - 2.4\times10^{-6} m^3) = 457.2 J \\ [/tex]
Thus, work done by the gas is 457.2 J
Tje Heat absorbed by the water during vaporization process is calculated using the formul:
mass of water = density × volume
mass of water = 1 g/cm³ × 2.4 cm³ = 2 g = 0.002 kg
Q = 0.002 kg × 2.26 × 10⁶ J/kg
Q = 4520 J
Thus, heat absorbed by the water is 4520 J
The change in internal energy, ΔU of a system is given as the heat absorbed minus the work done by the system.
[tex]\Delta U = Q-W [/tex]
ΔU = 4520 J − 457.2 J = 4062.8 J
Therefore, the change in internal energy of a system is 4062.8 J.
Learn more about change in internal energy at: https://brainly.com/question/14126477