P-V Diagram Question - Can Q1<Q2<Q3?

[Amith2006]

Sir,
An ideal gas of mass m in a state A goes to another state B via 3 different processes as shown in the figure. If Q1, Q2 and Q3 denote the heat absorbed by the gas along the 3 paths, can we say that Q1<Q2<Q3?
I solved it in the following way:
From the figure it is clear that T1<T2<T3. Now, smaller the temperature smaller will be the heat absorbed by the gas. Therefore, Q1<Q2<Q3. Is my argument right?

 

[Amith2006]

Sir,
An ideal gas of mass m in a state A goes to another state B via 3 different processes as shown in the figure. If Q1, Q2 and Q3 denote the heat absorbed by the gas along the 3 paths, can we say that Q1<Q2<Q3?
I solved it in the following way:
From the figure it is clear that T1<T2<T3. Now, smaller the temperature smaller will be the heat absorbed by the gas. Therefore, Q1<Q2<Q3. Is my argument right?

I am herewith attaching the P-V diagram.

 

[Andrew Mason]

Sir,
An ideal gas of mass m in a state A goes to another state B via 3
different processes as shown in the figure. If Q1, Q2 and Q3 denote the
heat absorbed by the gas along the 3 paths, can we say that Q1<Q2<Q3?
I solved it in the following way:
From the figure it is clear that T1<T2<T3. Now, smaller the temperature
smaller will be the heat absorbed by the gas. Therefore, Q1<Q2<Q3. Is my
argument right?

You can do it a couple of different ways. You can
see that if PV is greater for all points on on path 3 than on paths 1
and 2, the temperature must be greater, ie. more heat is present. But
you can also do it by looking at the work done by the gas.

Since: dQ = dU + dW where dW is the work done by the gas, and since dU
is the same for all, the greater heat occurs on the path where the
greatest amount of work is done by the gas.

I can't see your diagram, but assuming it is a PV diagram, the area
under the graph will give you the work done. The path with more area
under it has the greater heat added.

AM