Find the work done on a monoatomic gas in this P-V diagram

[Helly123]

For the heat removed in step CA, this is a constant pressure process, so $$Q=\Delta H$$. What is $\Delta H$ for this change?

The x1's and the x2's should be switched in your final result. If I factor the correct final result, I get:
$$\int_{x_1}^{x_2}{ydx}=(x_2-x_1)\frac{(mx_2+b)+(mx_1+b)}{2}$$Now, what if $$p=4p_0-\frac{p_0}{V_0}(V-V_0)=5p_o-\frac{p_0}{V_0}V$$so that $$m=-\frac{p_0}{V_0}$$ and $$b=5p_0$$and $$x=V$$ and $$y = p$$ and $$x_1=V_0$$ and $$x_2=4V_0$$What does your equation give?

Gives me 7.5PoVo. Wow, the same as trapezoid area..

But, why in trapezoid formula the heigh is 3Vo

Isn't it $$ W = \frac{4Vo + Vo}{2}*3Po$$
The height is 3Po

 

[Helly123]

$$\Delta H=n\frac{5}{2}R(T_A-T_C)$$

If i just sum W + $\Delta$U
Is it ok?

 

[Chestermiller]

Gives me 7.5PoVo. Wow, the same as trapezoid area..

But, why in trapezoid formula the heigh is 3Vo

Isn't it $$ W = \frac{4Vo + Vo}{2}*3Po$$
The height is 3Po

No way. The two parallel sides are vertical, and the altitude is horizontal.

 

[Chestermiller]

If i just sum W + $\Delta$U
Is it ok?

Sure, provided you include the correct work for that step.

 

[Helly123]

Sure, provided you include the correct work for that step.

Thanks :)