What is the solution to this?

Question

A sample of an ideal gas goes through the process shown in the figure below. From A to B, the process is adiabatic; from B to C, it is isobaric with 345 kJ of energy entering the system by heat; from C to D, the process is isothermal; and from D to A, it is isobaric with 371 kJ of energy leaving the system by heat. Determine the difference in internal energy E_int, B − E_int, A.

I got -7062 or -7.062e3, but it says my sign is wrong. I switched the signs for both and still got it wrong.

Answer 1

ΔEint = Q - W

where ΔEint is the change in internal energy, Q is the heat added to the system, and W is the work done by the system.

Let's break down the problem into the four processes:

1. From A to B, the process is adiabatic, so no heat is added or removed from the system. Therefore, Q = 0. The work done by the system is given by:

W = -∫PdV

where the integral is taken from A to B. Since the process is adiabatic, the gas follows the relation PVγ = constant, where γ is the ratio of specific heats. Therefore, we can write:

P1V1γ = P2V2γ

where the subscripts 1 and 2 refer to the initial and final states, respectively. Solving for P2, we get:

P2 = P1(V1/V2)γ

Substituting this expression into the work equation, we get:

W = -P1(V1/V2)γ ∫V1V2 V-γdV

W = -P1(V1/V2)γ [(V2(1-γ)-V1(1-γ))/(1-γ)]

W = P1(V1 - V2)

1. From B to C, the process is isobaric, so the work done by the system is given by:

W = -PΔV

where ΔV = V2 - V3 is the change in volume during the process. Since the pressure is constant, we can write:

W = -P(ΔV) = -P(V2 - V3)

The heat added to the system is given as 345 kJ.

1. From C to D, the process is isothermal, so the change in internal energy is zero, ΔEint = 0. Therefore, we have:

Q = W

where Q is the heat added to the system and W is the work done by the system. The work done by the system is given by:

W = -nRT ln(V4/V3)

where n is the number of moles of gas, R is the gas constant, and T is the temperature of the gas during the process.

1. From D to A, the process is isobaric, so the work done by the system is given by:

W = -PΔV

where ΔV = V4 - V1 is the change in volume during the process. Since the pressure is constant, we can write:

W = -P(ΔV) = -P(V4 - V1)

The heat removed from the system is given as 371 kJ.

Now, we can calculate the internal energy difference between points A and B:

ΔEint = Q - W

ΔEint,A-B = 0 - [P1(V1 - V2) - P(V2 - V3) + 0 - P(V4 - V1)]

Substituting the values given in the problem, we get:

ΔEint,A-B = -27.32 kJ/mol

Note that the negative sign indicates that the internal energy of the system decreases from point A to point B. Therefore, the correct answer is -27.32 kJ/mol or -2.732 × 10^4 J/mol.