How Does Process B Work If PV Is Constant and Internal Energy Changes?

[bartersnarter]

Homework Statement

A gas contained in a piston-cylinder assembly undergoes two processes, A and B, between the same end states, 1 and 2.
State 1: P₁ = 1 bar, V₁ = 1 m³, U₁ = 400 kJ
State 2: P₂ = 10 bar, V₁ = 0.1 m³, U₁ = 450 kJ
Process A: Constant-volume process from state 1 to a pressure of 10 bar, followed by a onstant-pressure process to state 2.
Process B: Process from 1 to 2 during which the pressure volume relation is PV = constant.
Kinetic and potential effects can be ignored.

Homework Equations

U = Q - W
PV = constant

The Attempt at a Solution

For this problem, I have trouble grasping the concepts. In a prior physics class I remember learning that PV = constant is representative of an isothermal process. This means that there is no change in temperature. No change in temperature also means that there is no change in internal energy. How then is process B possible? I've done all the math and I've gotten the right values, but I don't understand the logic. I also don't understand how to draw a PV diagram for such a process. My first guess was just a curve similar to y = ln(x), but apparently the solutions show a curve that looks like a quarter ellipse. I imagine this has something to do with the first predicament.
Here is the textbook's answer for the PV diagrams. Process A is simple, I understand that, but why does process B look like a quarter ellipse instead of a y = ln(x) curve?

 

[haruspex]

Why do you think it should be y=ln(x)? The chart is linear-linear, not log-linear.
But I disagree with the curve shown also. Look at the point (accidentally) marked B. The PV there is about 6.

 

[bartersnarter]

Ah, I'm sorry, I meant y = 1/x, not y = ln(x).

 

[haruspex]

Ah, I'm sorry, I meant y = 1/x, not y = ln(x).

Yes, I'd agree with that. So the graph is wrong, and the 450kJ is a puzzle. Maybe it's not to be considered an ideal gas, though I've no idea whether such a discrepancy is realistic.