A gas expands from I to F in Figure .The energy added to the gas by heat is 465 J when the gas goes from I to F along the diagonal path.
(a) What is the change in internal energy of the gas?
(b) How much energy must be added to the gas by heat along the indirect path IAF ?


A gas expands from I to F in Figure The energy added to the gas by heat is 465 J when the gas goes from I to F along the diagonal path a What is the change in i class=


Answer :

Answer:

Part 1: 85 J

Part 2: 718 J

Explanation:

For a gas, work done is equal to -1 times the area under the curve on a P-v diagram. For part 1, the area is equal to the area of a trapezoid. For part 2, the area is a rectangle. The first law of thermodynamics says that the sum of the heat and work that goes into a system must equal its change in internal energy.

Part 1 of 2

Work is equal to -1 times the area under the graph. For a trapezoid:

W = -½ (V₂ − V₁) (P₁ + P₂)

Plug in values (convert L to m³ and atm to Pa).

W = -½ (3.5 L − 1.0 L) (1 m³ / 1000 L) (0.5 atm + 2.5 atm) (101,325 Pa/atm)

W = -380 J

Notice the work is negative, meaning work is going out of the system. Given that 465 J of heat are going in, the change in internal energy is:

Q + W = ΔU

465 J − 380 J = ΔU

ΔU = 85 J

Part 2 of 2

This time, the area under the graph is a rectangle. The work done is therefore:

W = -(V₂ − V₁) P₁

W = -(3.5 L − 1.0 L) (1 m³ / 1000 L) (2.5 atm) (101,325 Pa/atm)

W = -633 J

For the same change in internal energy, the heat required is:

Q + W = ΔU

Q − 633 J = 85 J

Q = 718 J