Question

A thin-walled vessel of volume $V$, held at constant temperature, contains a gas which slowly leaks out through a small hole of area $A$. The outside pressure is low enough that leakage back into the vessel is negligible. (a) Show that the rate of decrease of the total number of particles $N$ inside the container is given by $\frac{dN}{dt} = -\frac{1}{4}n\bar{v}A$ where $n = N/V$ is the particle density in the container. (b) Use the equation of state of an ideal gas $\bar{p} = nk_BT$ to show that the rate of change of the pressure inside the container is given by $\frac{d\bar{p}}{dt} = -\frac{\bar{v}A}{4V}\bar{p}$ (c) Solve this equation and find the time required for the pressure to decrease to 1/e of its original value. Express your answer in terms of $A$, $V$ and the mean molecular speed $\bar{v}$.

          A thin-walled vessel of volume $V$, held at constant temperature, contains a gas which slowly leaks out through a small hole of area $A$. The outside pressure is low enough that leakage back into the vessel is negligible.
(a) Show that the rate of decrease of the total number of particles $N$ inside the container is given by
$\frac{dN}{dt} = -\frac{1}{4}n\bar{v}A$
where $n = N/V$ is the particle density in the container.
(b) Use the equation of state of an ideal gas $\bar{p} = nk_BT$ to show that the rate of change of the pressure inside the container is given by
$\frac{d\bar{p}}{dt} = -\frac{\bar{v}A}{4V}\bar{p}$
(c) Solve this equation and find the time required for the pressure to decrease to 1/e of its original value. Express your answer in terms of $A$, $V$ and the mean molecular speed $\bar{v}$.

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A thin-walled vessel of volume V, held at constant temperature, contains a gas which slowly leaks out through a small hole of area A. The outside pressure is low enough that leakage back into the vessel is negligible.
(a) Show that the rate of decrease of the total number of particles N inside the container is given by
(dN)/(dt) = -(1)/(4)nv̅A
where n = N/V is the particle density in the container.
(b) Use the equation of state of an ideal gas p̅ = nkBT to show that the rate of change of the pressure inside the container is given by
(dp̅)/(dt) = -(v̅A)/(4V)p̅
(c) Solve this equation and find the time required for the pressure to decrease to 1/e of its original value. Express your answer in terms of A, V and the mean molecular speed v̅.

Added by Timothy D.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Prachi Joshi

02/14/2022

Step 1

Step 1: Start with the ideal gas law, which relates pressure, volume, number of particles, and temperature: pV = NkT, where k is the Boltzmann constant. Show more…

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A thin-walled vessel of volume V, held at constant temperature, contains a gas which slowly leaks out through a small hole of area A. The outside pressure is low enough that leakage back into the vessel is negligible. (a) Show that the rate of decrease of the total number of particles N inside the container is given by dN/dt = -(1/4)(n)(bar v)A where n = N/V is the particle density in the container. (b) Use the equation of state of an ideal gas p = nkT to show that the rate of change of the pressure inside the container is given by dp/dt = -((bar v)A)/(4V)p (c) Solve this equation and find the time required for the pressure to decrease to 1/e of its original value. Express your answer in terms of A, V, and the mean molecular speed (bar v).

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Transcript

00:01 Hello everyone.

00:02 So here in this question, it is given that there is a container and two liquids are filled here.

00:07 Liquid of first is having density d.

00:11 Second liquid is having density 2b and both are filled up to the height, h divided by 2.

00:18 A block is immersed into this liquid and the above liquid is having the length of 3l by 4 of the block and lower liquid is having length l by 4 of the block.

00:29 The area of this container is a and the area of this block is a by 5 and density we have to find out the liquid.

00:37 So the above liquid will apply the buoyed force b1 on this block and the lower liquid will apply the buoyed force b2 on this block.

00:48 So we will write b1 plus b2 that is equals to mg because that will be balanced by mg.

01:01 Mg can also be written as volume of block, density of block and gravity.

01:09 So, simplifying this b1 and b2, we can also write v1 -g plus v2 -r -2g, that is equal to v -r -g.

01:26 G will cancel out from all the size, that is from left -hand side and right -hand side.

01:32 So we can write it as v1, row 1 plus v2, row 2, that is equal to v row.

01:41 V1, we can write as a1, l1, row 1 plus a2, l2, row 2, that is equal to alr.

01:55 Now, if we write a1, that is the area of this liquid, that will be a, l1, this is 3 ,000.

02:05 L by 4 because this much part of the block is immersed into this liquid.

02:11 So it will be 3l by 4.

02:13 D.

02:17 Here, a2 will be a.

02:19 Length that is immersed in the second liquid is l by 4.

02:23 D...

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