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## Homework Statement

Consider the following PV diagram of one mole of a monatomic gas:

The units are in atm and litres.

Find the change in internal energy, heat, and work for all 3 processes and for the cycle as well.

## Homework Equations

PV = nRT

ΔU = Q - W

ΔU = (3/2)*R*ΔT

Q = (5/2)*R*ΔT (for constant pressure)

## The Attempt at a Solution

For the work, I just found the areas:

A -> B: 0 J

B -> C: +3.5 J

C -> A: -1 J

For the cycle, the work done is just the total work added:

3.5 - 1 = 2.5 J

For the cycle, the Q is the same as the Work:

2.5 J

For the cycle, the ΔU = 0.

So far I have this (unless I did something wrong):

Since for a constant volume process the ΔU = Q, I decided to do that A -> B first.

T_A = (0.5 atm)(1 L) = (1)(0.0821)T, T = 6.09 K

T_B = (0.5 atm)(1 L) = (1)(0.0821)T, T = 36.5 K

ΔT = 30.41 K

ΔU = (3/2)*R*ΔT = (3/2)*0.0821*30.41 = 3.75 J = Q

So:

The constant pressure process looks easier so I'll do C -> A next:

T_C = (0.5 atm)(3 L) = (1)(0.0821)T, T = 18.27 K

T_A = (0.5 atm)(1 L) = (1)(0.0821)T, T = 6.09 K

ΔT = -12.18

Q = (5/2)*R*ΔT = (5/2)*0.0821*(-12.18) = -2.5 J

ΔU = Q - W = (-2.5) - (-1) = -1.5

So:

Now I can just add up the unknown columns:

ΔU for B -> C: 3.75 + (-1.5) + x = 0, x = -2.25

Q for B -> C: 3.75 + (-2.5) + x = 2.5, x = 1.25

So I got:

But I got this problem incorrect. What errors have I made?

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