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

[Chestermiller]

What's the equation for P as f(V) ?

$$p=4p_0-\frac{p_0}{V_0}(V-V_0)$$where $p_0=1\times 10^5\ Pa$, and $V_0=2\times 10^{-3}\ m^3$. So, when $V = V_0$, $p=4p_0$ and when $V=4V_0$, $p=p_0$. Now, what is $\int_{V_0}^{4V_0}{pdV}$? Incidentally, the area of the trapezoid is $$W=\frac{(4p_0+p_0)}{2}(3V_0)=7.5p_0V_0$$

$$\Delta H$$ is $$\Delta U ?$$
$$\Delta U$$ is the same in every segments of diagram.. so, 9*10^2 J.
While the $$W = P\Delta V$$
$$ = 10^5*6*10^{-3} $$
= 1.5 * 10^3

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

 

[Helly123]

$$p=4p_0-\frac{p_0}{V_0}(V-V_0)$$where $p_0=1\times 10^5\ Pa$, and $V_0=2\times 10^{-3}\ m^3$.

Where does it come from?

So, when $V = V_0$, $p=4p_0$ and when $V=4V_0$, $p=p_0$. Now, what is $\int_{V_0}^{4V_0}{pdV}$? Incidentally, the area of the trapezoid is $$W=\frac{(4p_0+p_0)}{2}(3V_0)=7.5p_0V_0$$

If i want to use the integral.. can you explain how to do it?

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

Why 5/2 not 3/2 since it's monoatomic?

 

[Chestermiller]

Where does it come from?

You have a straight line on the p-V graph passing through points B and C. The equation was derived using the point-slope form of the equation for a straight line. My 12 year old grandson learned this in first year algebra a year ago. Have you not had algebra?

If i want to use the integral.. can you explain how to do it?

If I asked you to integrate the equation $y = mx +b$ between $x=x_1$ and $x=x_2$, would you know how to do it?

Why 5/2 not 3/2 since it's monoatomic?

You have $$\Delta U=Q-W$$and, at constant pressure $$W=p\Delta V$$But, for an ideal gas, $$p\Delta V=nR\Delta T$$So $$\Delta U=n\frac{3}{2}R\Delta T=Q-nR\Delta T$$So, $$Q=n\frac{3}{2}R\Delta T+nR\Delta T=n\frac{5}{2}R\Delta T=\Delta H$$

 

[Helly123]

$$\frac {mx^2}{2} + bx |(x1,x2)$$
$$\frac {mx1^2}{2} + bx1 - \frac {mx2^2}{2} - bx2 $$If i want to find W for ABC using integral, W for BC using integral,
how can i do it?

 

[Chestermiller]

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?

Regarding the work for ABC: The equation asks for the work for BC only. But, the work for ABC is the area of the triangle ABC.

 

[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 :)