Effect of relative motion on pressure of gas in a container

[Aayush Verma]

Homework Statement

I am a high school student and I tried to solve a problem on Kinetic Theory of Gases and ended up being confused at a question given below.

as far as my understanding goes, the pressure exerted by an ideal gas inside walls of a cubic container according to assumptions and postulates of kinetic theory is given by the equation given below. as a gas molecule inside the container (at a particular instant when the rocket is flying upwards) is not in contact of the vessel, it shall hit with the container's bottom wall and there should be a very large impulse imparted to the wall by collision between gas molecule and the wall (assume the collision is elastic)

and so the pressure of whole mixture should increase on all walls, and gravity will not affect the speed of the gas along vertical very much wether the gas is light or heavy because ultimately, the impulse is so large, the change in pressure will be significant and uniform enough in the container.

this is what i thought and then applied conservation of momentum to verify the factor given in option C but the expression came out something close, not exact. the correct option is B.

now I looked at the formula of pressure exerted by the gas on a wall of the container and what I found was that this formula was derived in the frame of reference of the container in which the gas is kept and also we notice this phenomena during the time where the container is already moving with the rocket so if it was initially at rest and then it started to move along with the rocket so basically the increase in pressure is not observed.

what we observed is the changed pressure which remains constant in our observation from ground. I want to know if my thinking is correct or not.

Relevant Equations

The equation link is not working so i posted the image

 

[Mister T]

Does the fact that Earth is in motion affect the pressure of the gas inside the container?

 

[Aayush Verma]

Does the fact that Earth is in motion affect the pressure of the gas inside the container?

No

 

[Mister T]

Why not? Shouldn't Earth's motion have the same effect as the rocket's?

 

[Aayush Verma]

But how can it affect it ? I mean if the rocket is hypothetically not very far from earth's surface, i.e., within the range of kilometers where effect of altitude on gravitational pull on rocket by earth is not significant,
and the rotation of our earth from which we observe is also not significant up until that instant such that it might make the trajectory appear curved which can make us think that the velocity is constant with changing direction.
So i guess according to that, the motion is rectillinear from ground approximately and then it is possible for pressure to remain constant.

 

[Mister T]

It can't. That's the point.

 

[haruspex]

the pressure of whole mixture should increase on all walls,

Increase as a result of what? The velocity of the spacecraft?
You need to distinguish between bulk motion and random motion. In the ground frame, if the rocket's velocity is $\vec v$ then the momentum of a gas molecule is $m_r(\vec v+\vec v_r)$, where $\Sigma m_r\vec v_r=0$. So the pressure exerted on the walls is the same fore and aft, and unaffected by the rocket's motion.
Or more simply, work in the frame of the rocket. The velocity is given as constant, so it's inertial.

Of course, if the rocket is accelerating it is a different story.

 

[berkeman]

the correct option is B.

I think you mean (2).

Actually, I think (4) is correct too, depending on how "different" they mean. The box is still in Earth's gravitational field, after all...

 

[kuruman]

Not so fast guys. The equation for the kinetic interpretation of pressure is $$pV=\frac{2}{3}\bar K$$ where $\bar K$ is the average kinetic energy (monatomic gas). The standard textbook derivation of this considers atoms bouncing elastically and transferring momentum per unit time per unit area to the walls which, macroscopically is the pressure. In the textbook derivation, the gas is in a container at rest in the lab frame.

What if the container's speed increases? Would the internal energy of the gas increase? It looks like it will because the rms speed will increase. I did a back-of-the-envelope (BOTE) calculation of an atom that starts with speed $v_0$ from the left wall of a 1-D container of length $L$. Simultaneously with the atom, the container starts accelerating to the right with constant acceleration $a$.

After a collision with a wall of infinite mass, the final velocity of the atom $v_{\!f}$ is related to the initial velocity $v_i$ by $$~v_{\!f}=−v_i+2V_{\text{wall}}.$$ The velocity of the container is not affected by the collision. One then has to use the SUVAT equations to find the collision time and atom velocity after each collision.

It is convenient to define a dimensionless quantity $\beta\equiv \sqrt{1−2aL/v_0^2.}$
The velocity of the atom after the first collision is $v_1=v_0(1−2\beta).$
The velocity of the atom after the second collision is $v_2=v_0(3−2\beta).$

Now $\beta <1$ which means that $|v_1|<|v_0|.$ This is expected because both the atom and the wall are moving in the same direction when the first collision occurs. However, note that after the first round trip, the speed $v_2$ is greater than $v_0$. Thus, the speed of the atom increases after each round trip. This means that $v_{\text{rms}}$ increases and so does the temperature for as long as the acceleration is non-zero. If the temperature rises and the volume stays constant, the pressure must rise.

Since $v_{\text{rms}}$ depends on $\beta$, the pressure rise will also depend on $\beta$. None of the four choices offered fits this. There must be acceleration because the container must be at rest on the Earth in order to be filled first and then acquire a velocity of 500 m/s. There is no magic wand to wave that will cause the container to move from zero to 500 m/s without acceleration.

 

[haruspex]

Not so fast guys. The equation for the kinetic interpretation of pressure is $$pV=\frac{2}{3}\bar K$$ where $\bar K$ is the average kinetic energy (monatomic gas). The standard textbook derivation of this considers atoms bouncing elastically and transferring momentum per unit time per unit area to the walls which, macroscopically is the pressure. In the textbook derivation, the gas is in a container at rest in the lab frame.

What if the container's speed increases? Would the internal energy of the gas increase? It looks like it will because the rms speed will increase. I did a back-of-the-envelope (BOTE) calculation of an atom that starts with speed $v_0$ from the left wall of a 1-D container of length $L$. Simultaneously with the atom, the container starts accelerating to the right with constant acceleration $a$.

After a collision with a wall of infinite mass, the final velocity of the atom $v_{\!f}$ is related to the initial velocity $v_i$ by $$~v_{\!f}=−v_i+2V_{\text{wall}}.$$ The velocity of the container is not affected by the collision. One then has to use the SUVAT equations to find the collision time and atom velocity after each collision.

It is convenient to define a dimensionless quantity $\beta\equiv \sqrt{1−2aL/v_0^2.}$
The velocity of the atom after the first collision is $v_1=v_0(1−2\beta).$
The velocity of the atom after the second collision is $v_2=v_0(3−2\beta).$

Now $\beta <1$ which means that $|v_1|<|v_0|.$ This is expected because both the atom and the wall are moving in the same direction when the first collision occurs. However, note that after the first round trip, the speed $v_2$ is greater than $v_0$. Thus, the speed of the atom increases after each round trip. This means that $v_{\text{rms}}$ increases and so does the temperature for as long as the acceleration is non-zero. If the temperature rises and the volume stays constant, the pressure must rise.

Since $v_{\text{rms}}$ depends on $\beta$, the pressure rise will also depend on $\beta$. None of the four choices offered fits this. There must be acceleration because the container must be at rest on the Earth in order to be filled first and then acquire a velocity of 500 m/s. There is no magic wand to wave that will cause the container to move from zero to 500 m/s without acceleration.

Yes, I confess I misinterpreted the question, thinking it was about variation in pressure within the container while at the given velocity.

 

[Aayush Verma]

Not so fast guys. The equation for the kinetic interpretation of pressure is $$pV=\frac{2}{3}\bar K$$ where $\bar K$ is the average kinetic energy (monatomic gas). The standard textbook derivation of this considers atoms bouncing elastically and transferring momentum per unit time per unit area to the walls which, macroscopically is the pressure. In the textbook derivation, the gas is in a container at rest in the lab frame.

What if the container's speed increases? Would the internal energy of the gas increase? It looks like it will because the rms speed will increase. I did a back-of-the-envelope (BOTE) calculation of an atom that starts with speed $v_0$ from the left wall of a 1-D container of length $L$. Simultaneously with the atom, the container starts accelerating to the right with constant acceleration $a$.

After a collision with a wall of infinite mass, the final velocity of the atom $v_{\!f}$ is related to the initial velocity $v_i$ by $$~v_{\!f}=−v_i+2V_{\text{wall}}.$$ The velocity of the container is not affected by the collision. One then has to use the SUVAT equations to find the collision time and atom velocity after each collision.

It is convenient to define a dimensionless quantity $\beta\equiv \sqrt{1−2aL/v_0^2.}$
The velocity of the atom after the first collision is $v_1=v_0(1−2\beta).$
The velocity of the atom after the second collision is $v_2=v_0(3−2\beta).$

Now $\beta <1$ which means that $|v_1|<|v_0|.$ This is expected because both the atom and the wall are moving in the same direction when the first collision occurs. However, note that after the first round trip, the speed $v_2$ is greater than $v_0$. Thus, the speed of the atom increases after each round trip. This means that $v_{\text{rms}}$ increases and so does the temperature for as long as the acceleration is non-zero. If the temperature rises and the volume stays constant, the pressure must rise.

Since $v_{\text{rms}}$ depends on $\beta$, the pressure rise will also depend on $\beta$. None of the four choices offered fits this. There must be acceleration because the container must be at rest on the Earth in order to be filled first and then acquire a velocity of 500 m/s. There is no magic wand to wave that will cause the container to move from zero to 500 m/s without acceleration.

I agree with you but our duration of observation is in the state where rocket has achieved the required speed and is moving with zero acceleration, so that makes the 2nd option correct. although you are also correct to think that practically it must have accelerated to be where it is.

 

[kuruman]

I agree with you but our duration of observation is in the state where rocket has achieved the required speed and is moving with zero acceleration, so that makes the 2nd option correct. although you are also correct to think that practically it must have accelerated to be where it is.

That may be so. However, option 2 says that the "pressure remains the same" which is a bit unclear because one wonders "the same as what?" It would have been better to have said "stays constant while the rocket is moving at 500 m/s."