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

[Helly123]

Homework Statement

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 ?

 

[DrClaude]

Note: always report numerical values with units.

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 $n=1\ \mathrm{mol}$?

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.

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.

 

[Helly123]

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

 

[DrClaude]

yes, it wasn't stated on the text

Is it compatible with the conditions at point A?

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.

Q = ΔU + W

Would that be the heat released?

 

[Helly123]

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

 

[DrClaude]

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.

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

 

[Helly123]

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)

 

[DrClaude]

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.

 

[Helly123]

You're skipping some steps. First write P as a function V.

$$ P = 5 - V $$
... isn't it..

 

[DrClaude]

$$ P = 5 - V $$
... isn't it..

Absolutely not. Does this even make sense in terms of units?

 

[Helly123]

Absolutely not. Does this even make sense in terms of units?

P(V) = 5 - V

 

[DrClaude]

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?

 

[Helly123]

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

 

[DrClaude]

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.

 

[Helly123]

This is not correct. Look carefully at the diagram. In any case, you should report values that have units.

$$ W = \int_{1}^{4} P*dV $$
$$ W = \int_{1}^{4} \frac{nRT}{V}* dV $$
$$ W = nRT \int_{1}^{4} \frac{1}{V}* dV $$

 

[DrClaude]

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

$$ W = \int_{1}^{4} P*dV $$
$$ W = \int_{1}^{4} \frac{nRT}{V}* dV $$
$$ W = nRT \int_{1}^{4} \frac{1}{V}* dV $$

You are taking $T$ out of the integral. Is it constant?

 

[Helly123]

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..

 

[DrClaude]

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
$$
W = \int_{V_0}^{4 V_0} P dV
$$

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:
$$
P = a +b V
$$

 

[Helly123]

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?

 

[DrClaude]

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

In SI units, V is never 8 or 2.

 

[Helly123]

In SI units, V is never 8 or 2.

So.. what it supposed to be...?
Why can't we use the $$\int \frac{1}{V} dV $$

 

[DrClaude]

So.. what it supposed to be...?

Take point B, for example: $V = V_0 = 2 \times 10^{-3} \ \mathrm{m}^3$

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.

 

[Helly123]

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

 

[DrClaude]

How do i know? According to Law $$ \frac{P1V1}{T1} = \frac{P2V2}{T2} $$ T at B and C are the same

Yes, but that doesn't mean that T is the same at every point along the curve. You can check what an isotherm looks like on a PV diagram at http://www.aplusphysics.com/courses/honors/thermo/thermodynamics.html

 

[Helly123]

Yes, but that doesn't mean that T is the same at every point along the curve. You can check what an isotherm looks like on a PV diagram at http://www.aplusphysics.com/courses/honors/thermo/thermodynamics.html View attachment 218633

I'll try to do it again. Thanks for the help

 

[Chestermiller]

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?

 

[Chestermiller]

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

 

[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?

I'll try it again, thanks

 

[Helly123]

Yes, but that doesn't mean that T is the same at every point along the curve. You can check what an isotherm looks like on a PV diagram at http://www.aplusphysics.com/courses/honors/thermo/thermodynamics.html
View attachment 218632 View attachment 218633

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

 

[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?

$$\Delta H$$ 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