The discussion focuses on calculating the work done on a monoatomic gas using a P-V diagram. Key equations include the internal energy change formula, $$\Delta U = \frac{3}{2}nR(T_2 - T_1)$$, and the work done, $$W = P\Delta V$$. Participants clarify that the total work done in a cycle is represented by the area under the curve, specifically for the BC leg, which is calculated as a trapezoid. The correct values for pressure and volume at points A, B, and C are critical for accurate calculations.
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$$p=4p_0-\frac{p_0}{V_0}(V-V_0)$$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 $$W=\frac{(4p_0+p_0)}{2}(3V_0)=7.5p_0V_0$$Helly123 said:What's the equation for P as f(V) ?
$$\Delta H=n\frac{5}{2}R(T_A-T_C)$$Helly123 said:$$\Delta H$$ 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
Where does it come from?Chestermiller said:$$p=4p_0-\frac{p_0}{V_0}(V-V_0)$$where ##p_0=1\times 10^5\ Pa##, and ##V_0=2\times 10^{-3}\ m^3##.
If i want to use the integral.. can you explain how to do it?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 $$W=\frac{(4p_0+p_0)}{2}(3V_0)=7.5p_0V_0$$
Why 5/2 not 3/2 since it's monoatomic?$$\Delta H=n\frac{5}{2}R(T_A-T_C)$$
Helly123 said:Where does it come from?
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?If i want to use the integral.. can you explain how to do it?
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, $$Q=n\frac{3}{2}R\Delta T+nR\Delta T=n\frac{5}{2}R\Delta T=\Delta H$$Why 5/2 not 3/2 since it's monoatomic?
If I asked you to integrate the equation ##y = mx +b## between ##x=x_1## and ##x=x_2## said:$$\frac {mx^2}{2} + bx |(x1,x2)$$
$$\frac {mx1^2}{2} + bx1 - \frac {mx2^2}{2} - bx2 $$If i want to find W for ABC using integral, W for BC using integral,
how can i do it?
Chestermiller said:For the heat removed in step CA, this is a constant pressure process, so $$Q=\Delta H$$. What is ##\Delta H## for this change?
Gives me 7.5PoVo. Wow, the same as trapezoid area..Chestermiller said: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?
$$\Delta H=n\frac{5}{2}R(T_A-T_C)$$
No way. The two parallel sides are vertical, and the altitude is horizontal.Helly123 said: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
Sure, provided you include the correct work for that step.Helly123 said:If i just sum W + ##\Delta##U
Is it ok?
Thanks :)Chestermiller said:Sure, provided you include the correct work for that step.