The discussion revolves around calculating the work done on a monoatomic gas as represented in a P-V diagram. Participants are exploring the relationships between pressure, volume, and temperature in the context of thermodynamics, particularly focusing on the work done during a thermodynamic cycle.
The discussion is ongoing, with participants providing guidance on how to approach the integration and questioning the validity of certain assumptions. There is a lack of consensus on specific values and methods, indicating that multiple interpretations are being explored.
Participants note that the values of pressure and volume must be reported with units, and there is an emphasis on ensuring that the equations used are dimensionally consistent. Some constraints arise from the original problem statement, which does not specify certain parameters, leading to varied interpretations.
$$p=4p_0-\frac{p_0}{V_0}(V-V_0)$$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.