Find the width increases when a monoatomic gas is heated

[Helly123]

1. The problem statement, all variables and given/known

Homework Equations

$\frac {PV1}{T1} = \frac {PV2}{T2}$

The Attempt at a Solution

I find that v2 = 330/273 V1

But how can this related to the distance of P travel?
Also, as the A gas pressed to B, i think the B gas try to resist. But couldn't express it in physics term.. can you help me?

 

[TSny]

How can you express the volume of one of the gases in terms of the area S?

 

[Helly123]

How can you express the volume of one of the gases in terms of the area S?

V = pi r^2 . h for cylindrical
The radius changes.
A gas new radius = 10 + x
Initial radius was 10
I try it but wrong answer.

 

[TSny]

The radius of the cylinder does not change.

$S$ is the cross sectional area which is a constant.

 

[Helly123]

V2 = 330/273 v1
(10 + x) = $\frac{330}{273 }$ 10
X = (3300-2730)/273
X = 570/273
X = 2 cm

 

[TSny]

V2 = 330/273 v1
(10 + x) = $\frac{330}{273 }$ 10

Both A and B will change their lengths.

 

[Helly123]

We do not compare the initial V for A and the final V for A ?
Is the V/T = V/T is for volume between A and B?

 

[TSny]

We need to back up for a minute. When you wrote $\frac {PV_1}{T_1} = \frac{PV_2}{T_2}$, what do the subscripts 1 and 2 refer to?

 

[Helly123]

(10 + x ) 273 = 330 (10 - x)
273x + 330x = 570
X = 0.945 cm

 

[Helly123]

We need to back up for a minute. When you wrote $\frac {PV_1}{T_1} = \frac{PV_2}{T_2}$, what do the subscripts 1 and 2 refer to?

V1 for inital volume of A
V2 is final one

T1 is initial temperature of A
T2 is final one

 

[TSny]

Are the initial and final pressures of gas A the same?

 

[TSny]

(10 + x ) 273 = 330 (10 - x)

This might be correct. Can you explain the reasoning behind it?

 

[Helly123]

Are the initial and final pressures of gas A the same?

the area of contact between piston and and the container is tightly sealed..
What does it mean? No internal change?
The volume and pressure and temperature still able to change?

 

[Helly123]

This might be correct. Can you explain the reasoning behind it?

The h for B area decrease because of the gas A give pressure to it. So the initial h 10 cm decreased by x. Become 10 - x
While the h of area A increase by x. Become 10 + x

 

[Helly123]

This might be correct. Can you explain the reasoning behind it?

Pressure remain constant.

 

[TSny]

The h for B area decrease because of the gas A give pressure to it. So the initial h 10 cm decreased by x. Become 10 - x
While the h of area A increase by x. Become 10 + x

OK

Pressure remain constant.

I'm not sure what you mean here. Are you saying that the final pressure of A equals the final pressure of B?

It would be nice if you would summarize your entire argument for arriving at (10 + x ) 273 = 330 (10 - x).

 

[TSny]

the area of contact between piston and and the container is tightly sealed..
What does it mean?

It means no gas from chamber A can leak into chamber B, and vice versa.

The volume and pressure and temperature still able to change?

Yes, the pressure, volume, and temperature of A can change. The pressure and volume of B can change, but the temperature of B is kept fixed by the "thermostatic bath".

 

[Helly123]

OK

I'm not sure what you mean here. Are you saying that the final pressure of A equals the final pressure of B?

It would be nice if you would summarize your entire argument for arriving at (10 + x ) 273 = 330 (10 - x).

Area A the pressure constant
Volume increase
Temperature increase

Area B
Temperature constant
Pressure volume decrease

For A, v1/T1 = v2/T2 applied

For B, P1V1 = P2V2 applied

So for A
V2 = 330/273 V1

I didnt get why the V1 is x - 10
V2 is x + 10

 

[TSny]

Area A the pressure constant
Volume increase
Temperature increase

Why would the pressure of A remain constant?

Area B
Temperature constant
Pressure volume decrease

Why would the pressure of B decrease?

I didnt get why the V1 is x - 10
V2 is x + 10

The volume of a cylinder is $V = Sh$ where $S$ is the cross-sectional area and $h$ is the length as shown in post #4. As you noted, in the initial states $h_A = h_B=10$ cm and in the final states $h_A = 10+x$ , $h_B = 10 - x$.

From the ideal gas law, show that $\frac{PV}{nT}$ has the same value for any ideal gas in any state. $n$ is the number of moles.

This means $\frac{P_AV_A}{n_AT_A} = \frac{P_BV_B}{n_BT_B}$. On the left you can use any state of gas A. On the right you can use any state of gas B. Decide which states of the gases to use. When you use $V = Sh$ in this equation, what happens to $S$?

 

[Helly123]

P of B decrease because the T kept constant while V is decrease

P of A constant because the T increase cause the V increase

 

[Helly123]

S remains the same. What change is h. Number of moles remain the same

 

[TSny]

P of B decrease because the T kept constant while V is decrease

Rearrange the gas law as P = nRT/V. For gas B, n and T remain constant while V decreases. So, what happens to P?

P of A constant because the T increase cause the V increase

Repeat the above for gas A.

 

[TSny]

S remains the same. What change is h. Number of moles remain the same

Something to think about: How do the pressures of the two gases compare to one another in the initial state of the system? How do the pressures of the two gases compare to one another in the final state of the system?

 

[Helly123]

Rearrange the gas law as P = nRT/V. For gas B, n and T remain constant while V decreases. So, what happens to P?

Repeat the above for gas A.

For Gas B
P = 1/V
V decrease
So P increase

For gas A
P = T/V
T increase
V increase
So P remain the same

P for A initial = P A final

P for B initial less than P for B final

P initial A and B the same
P final A less than P final B

PV/nT for A at final = PV/nT for B at final
P(10+x)/n*330K = P2(10-x)/n*273K ...(1)

Find P final for B (P2) :
PV for B initial = PV for B final
P(10) = P2(10-x)
P2 = P.10/(10-x)

Sub P2 to ...(1)
P(10+x)/n*330K = P.10/n*273K
10+x = 330.10/273
X = (330.10/273 )- 10
X = (3300 - 2730)/273

That is what i get

 

[TSny]

For Gas B
P = 1/V
V decrease
So P increase

OK. For gas B it's not really P = 1/V, but rather P = C/V for some constant C. But, you are right that the pressure for gas B must increase.

For gas A
P = T/V
T increase
V increase
So P remain the same

More precisely, P = C⋅T/V for some constant C. Although T and V both increase, that doesn't imply that P must remain constant. If T increases a lot while V increases a little, P would increase. If T increases a little while V increases a lot, P would decrease.

P initial A and B the same

Yes.

P final A less than P final B

No. In the final state, the piston remains at rest. What does this tell you about the final pressures of A and B?

PV/nT for A at final = PV/nT for B at final

Good. You are almost there. You just need to think about the pressures in the final state and to also think about how the number of moles of A compares to the number of moles of B.

 

[Helly123]

Ok. The piston remain at rest. Means, the pressure between A and B the same.
And the answer is what stated before 0.95 cm

The number of initial molecules for A and B are the same.
Number of molecules can affect the pressure. But will the number of molecules change because of pressure volume or temperature?

If the final pressure the same for A and B, how to prove that by the formula? How to know that T and V increase in A gas is proportional or not?

Could it be that P for A increase, and P for B increase, and during the final rest condition the pressure reached the same pressure?
Which means there is change in amount of molecule for gas B and A, but the amount of molecules A and B not the same

 

[TSny]

The number of initial molecules for A and B are the same.
Number of molecules can affect the pressure. But will the number of molecules change because of pressure volume or temperature?

The number of molecules of each gas will not change. The problem states "the area of contact between the piston and the container is tightly sealed". So, the molecules are trapped on each side of the piston. Changing the pressure or temperature of one of the gases will not change the number of molecules of that gas.

If the final pressure the same for A and B, how to prove that by the formula?

You can't deduce it from the formula. You have to make that conclusion based on the fact that there is no friction between the piston and the container and the fact that the piston remains at rest in the final state.

Could it be that P for A increase, and P for B increase, and during the final rest condition the pressure reached the same pressure?

Yes.

Which means there is change in amount of molecule for gas B and A, but the amount of molecules A and B not the same

There will not be any change in the number of molecules of gas on either side of the piston. To show that the two gases have the same number of molecules, you can set up $\frac{P_AV_A}{n_AT_A} = \frac{P_BV_B}{n_BT_B}$ for the initial state of the system.

 

[Helly123]

Thanks ^^

 

[Helly123]

Is the answer round off to 1?

 

[TSny]

Is the answer round off to 1?

Yes. It appears they are giving choices that are only accurate to one significant figure where the significant figure is either a 1 or a 5. They ask for the "best answer". So, 1 cm is the best choice.