Questions #11, 12, and 13:

Part 1 of 3: 2.4 cm³ of water is boiled at atmospheric pressure to become 4514.7 cm³ of steam, also at atmospheric pressure. Calculate the work done by the gas during this process. The latent heat of vaporization of water is 2.26 × 10⁶ J/kg. Answer in units of J.

Part 2 of 3: Find the amount of heat added to the water to accomplish this process. Answer in units of J.

Part 3 of 3: Find the change in internal energy. Answer in units of J.

Respuesta :

During the isobaric expansion;

  • work done by the gas is 457.2 J
  • heat absorbed by the water is 4520 J
  • the change in internal energy of a system is 4062.8 J.

What is the work done by a gas in an isolated process?

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:

  • Q = m×ΔHvap

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

ACCESS MORE