ΔU, Q, W of thermodynamic process

[Any Help]

Homework Statement

A monatomic ideal gas undergoes the thermodynamic process shown in the PV diagram. Determine whether each of the values ΔU, Q, and W for the gas is positive, negative, or zero.

Homework Equations

ΔU=Q-W
W=integral under the curve
PV=nRT T=PV/nR

The Attempt at a Solution

direction is negative then the integral negative then work is negative
ΔU depends on change in temp, and here pressure increases and volume decreases so it was Ti=P0x2V0/nR and at final it becomes Tf=2P0xV0/nR Tf=Ti then it stays the same no change in temp then ΔU =0
therefore 0=Q-W Q=W but W negative then Q also negative
Is that correct? I need only to check my answer...
and is there any other way to determine Q from Graph (without looking at W or ΔU) ?

 

[DrClaude]

Is that correct? I need only to check my answer...

Yes.

and is there any other way to determine Q from Graph (without looking at W or ΔU) ?

You could draw adiabats, one passing through point $2V_0,P_0$ and one through point $V_0,2P_0$, and see if you are going closer to the origin ($Q<0$) or farther away from the origin ($Q>0$), but it is easier to calculate it as you did.

 

[Any Help]

draw adiabats

how we draw them? do you mean to draw hyperbolas passing through them?

 

[DrClaude]

I mean that you draw the curves $P = V^{-\gamma}$ that pass through the given points.

 

[rude man]

Careful with how W is defined. Physicists usually take W as the work done BY the gas, which is what you did & that's correct.

But chemists often (or usually?) take W as the work done ON the gas in which case your W would be positive and the 1st law would read ΔU = Q + W.