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

In summary: In the integral you just wrote, you put the limits of integration...At point A, when P = 1, V = 1while point B, when P = 4, V = 1point C, P = 1, V = 4At point A, when P = 1, V = 1while point B, when P = 4, V = 1point C, P = 1, V = 4At point A, when P = 1, V = 1while point B, when P = 4, V = 1point C, P = 1, V = 4At point A, when P = 1, V = 1while point B, when
  • #1
Helly123
581
20

Homework Statement


297iao.png


Homework Equations



internal change = $$\frac{3}{2}n.R.(T2 - T1)$$
Work = P.ΔV

The Attempt at a Solution


1) T2 = $$\frac {P2. V2 . T1 }{P1 . V1} = 1.2 * 10^3$$

2) Q = Internal change =
$$\frac{3}{2}n.R.(T2 - T1) $$
$$ = \frac{3}{2} * 1 *8.3*10^{-3}*(12*10^2 - 3*10^2) $$
$$ = 112.05 J $$ (the answer is 9*10^2)

3) work BC = the are of triangle ABC = $$9*10^2 $$ (the answer is 1.5*10^3)

4) Heat released is it the same as Heat final - Heat initial ?
 

Attachments

  • 297iao.png
    297iao.png
    19.7 KB · Views: 2,024
Last edited by a moderator:
Physics news on Phys.org
  • #2
Note: always report numerical values with units.

Helly123 said:
2) Q = Internal change =
$$\frac{3}{2}n.R.(T2 - T1) $$
$$ = \frac{3}{2} * 1 *8.3*10^{-3}*(12*10^2 - 3*10^2) $$
$$ = 112.05 J $$ (the answer is 9*10^2)
Why did you set ##n=1\ \mathrm{mol}##?

Helly123 said:
3) work BC = the are of triangle ABC = $$9*10^2 $$ (the answer is 1.5*10^3)
The total work done by the cycle is the area subtended by the cycle. You are asked only for the work for the BC leg of the cycle.

Helly123 said:
4) Heat released is it the same as Heat final - Heat initial ?
No. You get the heat released by considering the change in internal energy and the work done.
 
  • #3
DrClaude said:
Note: always report numerical values with units.Why did you set ##n=1\ \mathrm{mol}##?
yes, it wasn't stated on the text


The total work done by the cycle is the area subtended by the cycle. You are asked only for the work for the BC leg of the cycle.
do you mean the length BC? work = P.ΔV = which P to be used?

No. You get the heat released by considering the change in internal energy and the work done.

Q = ΔU + W
 
  • #4
Helly123 said:
yes, it wasn't stated on the text
Is it compatible with the conditions at point A?

Helly123 said:
do you mean the length BC? work = P.ΔV = which P to be used?
As P is a function of V, you have to integrate.

Helly123 said:
Q = ΔU + W
Would that be the heat released?
 
  • #5
DrClaude said:
Is it compatible with the conditions at point A?
mol is 0.08 mol
so, internal change = $$
\frac{3}{2}* 0.08*8.3*9*10^2$$ = 9*10^2
As P is a function of V, you have to integrate.
is the function P = 5-V
how to find P as function of V ?
Would that be the heat released?
heat released Q < 0 ; W < 0
-Q = ΔU - W
 
  • #6
Helly123 said:
is the function P = 5-V
how to find P as function of V ?
P is a linear function of V, so P = a + bV. Use the values of P and V at point A and B to find the constants a and b.

Helly123 said:
heat released Q < 0 ; W < 0
-Q = ΔU - W
Be careful here. If Q is the energy of the system, then Q = ΔU - W, as you wrote previously. The energy released is -Q = -ΔU + W
 
  • #7
DrClaude said:
P is a linear function of V, so P = a + bV. Use the values of P and V at point A and B to find the constants a and b.
$$\int_{1}^{4} 5v - \frac{v^2}{2} $$ = 7.5 (wrong answer)
 
  • #8
You're skipping some steps. First write P as a function V.

Second, are are using incorrect values. For instance, the volume is never 1 or 4.
 
  • #9
DrClaude said:
You're skipping some steps. First write P as a function V.
$$ P = 5 - V $$
... isn't it..
 
  • #10
Helly123 said:
$$ P = 5 - V $$
... isn't it..
Absolutely not. Does this even make sense in terms of units?
 
  • #11
DrClaude said:
Absolutely not. Does this even make sense in terms of units?
P(V) = 5 - V
 
  • #12
Helly123 said:
P(V) = 5 - V
Are the units of the left-hand side of the equation the same as those on the right-hand side?

Lets take it one step back. What are the values of P and V at points A and B?
 
  • #13
DrClaude said:
Are the units of the left-hand side of the equation the same as those on the right-hand side?

Lets take it one step back. What are the values of P and V at points A and B?
point A, when P = 1, V = 1
while point B, when P = 4, V = 1
point C, P = 1, V = 4
 
  • #14
Helly123 said:
point A, when P = 1, V = 1
while point B, when P = 4, V = 1
point C, P = 1, V = 4
This is not correct. Look carefully at the diagram. In any case, you should report values that have units.
 
  • #15
DrClaude said:
This is not correct. Look carefully at the diagram. In any case, you should report values that have units.
$$ W = \int_{1}^{4} P*dV $$
$$ W = \int_{1}^{4} \frac{nRT}{V}* dV $$
$$ W = nRT \int_{1}^{4} \frac{1}{V}* dV $$
 
  • #16
We're going in circles. You should really take a step back.

In the integral you just wrote, you put the limits of integration of the volume to 1 and 4. It makes no sense to write ##V=1## or ##V=4##. What units would that be? Hint: the volume is never 4 m3 or 4 cm3 or 4 l or 4
Helly123 said:
$$ W = \int_{1}^{4} P*dV $$
$$ W = \int_{1}^{4} \frac{nRT}{V}* dV $$
$$ W = nRT \int_{1}^{4} \frac{1}{V}* dV $$
You are taking ##T## out of the integral. Is it constant?
 
  • #17
DrClaude said:
We're going in circles. You should really take a step back.

In the integral you just wrote, you put the limits of integration of the volume to 1 and 4. It makes no sense to write ##V=1## or ##V=4##. What units would that be? Hint: the volume is never 4 m3 or 4 cm3 or 4 l or 4
volume is dm3
$$ nRT \int_{2}^{8} \frac{1}{V} dV$$
You are taking ##T## out of the integral. Is it constant?
T is constant right..
 
  • #18
Helly123 said:
volume is dm3
$$ nRT \int_{2}^{8} \frac{1}{V} dV$$
Concerning the volume, this is better, but I think you should keep it more symbolic, and introduce numerical values later. It is better to use
$$
W = \int_{V_i}^{V_f} P dV
$$
or at least
$$
W = \int_{V_0}^{4 V_0} P dV
$$

Helly123 said:
T is constant right..
No. An isotherm in a PV diagram is definitely not a straight line. (Simple check: use the ideal gas law to calculate the temperature at points B and C).

You need to find the equation for P:
$$
P = a +b V
$$
 
  • #19
DrClaude said:
Concerning the volume, this is better, but I think you should keep it more symbolic, and introduce numerical values later. It is better to use
$$
W = \int_{V_i}^{V_f} P dV
$$
or at least
$$
W = \int_{V_0}^{4 V_0} P dV
$$No. An isotherm in a PV diagram is definitely not a straight line. (Simple check: use the ideal gas law to calculate the temperature at points B and C).

$$ \frac{4P*V}{T1} = \frac{P*4V}{T2}
T1 = T2 $$

You need to find the equation for P:
$$
P = a +b V
$$

10^5 = a + b(8)
4*10^5 = a + b(2)

10^5 = 4*10^5 - 2b + 8b
-3*10^5 = 6b
b = -5*10^4

a = 5*10^5

P = a +b V
P = 5*10^4 (10 - V)

$$ \int_{Vi}^{Vf} 5*10^4 (10 - V) . dV $$
$$ \int_{2}^{8} 5*10^4 (10 - V) . dV $$
$$ 5*10^4 \int_{2}^{8} (10 - V) . dV $$
$$ 5*10^4 * ( 10V - \frac{V^2}{2} |(2)(8) ) $$
$$ 5*10^4 * ( 80 - 20 - (\frac{64}{2} - \frac{4}{2}) )$$
answer is 1.5 * 10^6
the right answer is 1.5 * 10^3
why?
 
  • #20
Helly123 said:
10^5 = a + b(8)
4*10^5 = a + b(2)
In SI units, V is never 8 or 2.
 
  • #21
DrClaude said:
In SI units, V is never 8 or 2.
So.. what it supposed to be...?
Why can't we use the $$\int \frac{1}{V} dV $$
 
  • #22
Helly123 said:
So.. what it supposed to be...?
Take point B, for example: ##V = V_0 = 2 \times 10^{-3} \ \mathrm{m}^3##

Helly123 said:
Why can't we use the $$\int \frac{1}{V} dV $$
You have
$$
W = \int_{V_B}^{V_C} P(V) dV = n R \int_{V_B}^{V_C} \frac{T(V)}{V} dV
$$
where I have explicitly indicated that both P and T depend on V in the BC segment.
 
  • Like
Likes Helly123
  • #23
DrClaude said:
Take point B, for example: ##V = V_0 = 2 \times 10^{-3} \ \mathrm{m}^3##You have
$$
W = \int_{V_B}^{V_C} P(V) dV = n R \int_{V_B}^{V_C} \frac{T(V)}{V} dV
$$
where I have explicitly indicated that both P and T depend on V in the BC segment.
How do i know? According to Law $$ \frac{P1V1}{T1} = \frac{P2V2}{T2} $$ T at B and C are the same
 
  • #24

Attachments

  • PV-Processes.png
    PV-Processes.png
    7.2 KB · Views: 883
  • PV-HeatRemoved_Soln.png
    PV-HeatRemoved_Soln.png
    11.2 KB · Views: 944
  • Like
Likes Helly123
  • #26
Helly123: What is the formula for the area of a trapezoid? Did you notice from the diagram that the area under the line BC is a trapezoid? What are the actual pressures at B and C, and what are what are the actual volumes at these points?
 
  • #27
For the heat removed in step CA, this is a constant pressure process, so $$Q=\Delta H$$. What is ##\Delta H## for this change?
 
  • #28
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?
I'll try it again, thanks
 
  • #30
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?
$$\Delta H$$ is $$\Delta U ?$$
$$\Delta U$$ is the same in every segments of diagram.. so, 9*10^2 J.
While the $$W = P\Delta V$$
$$ = 10^5*6*10^{-3} $$
Q = 1.5 * 10^3
 
  • #31
Helly123 said:
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 $$W=\frac{(4p_0+p_0)}{2}(3V_0)=7.5p_0V_0$$
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
$$\Delta H=n\frac{5}{2}R(T_A-T_C)$$
 
  • Like
Likes Helly123
  • #32
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##.
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?
 
  • #33
Helly123 said:
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, $$Q=n\frac{3}{2}R\Delta T+nR\Delta T=n\frac{5}{2}R\Delta T=\Delta H$$
 
  • Like
Likes Helly123
  • #34
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?
 
  • #35
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.
 
  • Like
Likes Helly123
<h2>1. What is a monoatomic gas?</h2><p>A monoatomic gas is a type of gas that consists of single atoms of an element. Examples of monoatomic gases include helium, neon, and argon.</p><h2>2. How is work done on a monoatomic gas?</h2><p>Work is done on a monoatomic gas when the gas expands or compresses, causing a change in volume. This change in volume results in a change in pressure, which is what is represented on a P-V diagram.</p><h2>3. How is work calculated on a P-V diagram?</h2><p>The work done on a monoatomic gas can be calculated by finding the area under the curve on a P-V diagram. This area represents the change in volume and pressure, which is equal to the work done on the gas.</p><h2>4. What is the significance of the P-V diagram for a monoatomic gas?</h2><p>The P-V diagram is a graphical representation of the work done on a monoatomic gas. It helps to visualize the change in volume and pressure, and allows for the calculation of work done on the gas.</p><h2>5. How does the P-V diagram relate to the first law of thermodynamics?</h2><p>The first law of thermodynamics states that energy cannot be created or destroyed, only transferred. The P-V diagram shows the transfer of energy in the form of work done on the gas, which relates to the first law of thermodynamics.</p>

1. What is a monoatomic gas?

A monoatomic gas is a type of gas that consists of single atoms of an element. Examples of monoatomic gases include helium, neon, and argon.

2. How is work done on a monoatomic gas?

Work is done on a monoatomic gas when the gas expands or compresses, causing a change in volume. This change in volume results in a change in pressure, which is what is represented on a P-V diagram.

3. How is work calculated on a P-V diagram?

The work done on a monoatomic gas can be calculated by finding the area under the curve on a P-V diagram. This area represents the change in volume and pressure, which is equal to the work done on the gas.

4. What is the significance of the P-V diagram for a monoatomic gas?

The P-V diagram is a graphical representation of the work done on a monoatomic gas. It helps to visualize the change in volume and pressure, and allows for the calculation of work done on the gas.

5. How does the P-V diagram relate to the first law of thermodynamics?

The first law of thermodynamics states that energy cannot be created or destroyed, only transferred. The P-V diagram shows the transfer of energy in the form of work done on the gas, which relates to the first law of thermodynamics.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
740
  • Introductory Physics Homework Help
Replies
4
Views
469
  • Introductory Physics Homework Help
Replies
12
Views
664
  • Introductory Physics Homework Help
Replies
3
Views
879
  • Introductory Physics Homework Help
Replies
19
Views
1K
  • Introductory Physics Homework Help
Replies
16
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
599
  • Introductory Physics Homework Help
Replies
4
Views
819
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
Back
Top