All categories

3 years ago

Use the kinetic-molecular theory to explain the compression and expansion of gases.

Physics

1 answer:

erma4kov [3.2K]3 years ago

8 0

Answer:

According to the kinetic-molecular theory of gases, gases are composed of atoms and molecules (particles), where the distance among these elements is very large compared to their own size, therefore the total volume occupied by these particles is only a small part of the total volume occupied by the whole gas.

In other words: a gas has enough empty spaces in its total volume, therefore low density, which gives it the property of being highly compressible.

In this same sense, the kinetic-molecular theory of gases starts from the first principle of thermodynamics, establishing a relationship between heat and movement, since all material is composed of particles that are in motion (to a certain extent , depending on the state of matter), and in the case of gases the movement is greater, which is a strong indication of heat (thermal energy).

Now, gases can change their volume in two ways:

-By a change in temperature (heat transfer).

-By a change of pressure

So, this change or variation in volume is related to the work the gas does to change from an initial volume to a final volume.

Then, if in this process the volume decreases, it is said that the gas has been compressed (compression work); but if, on the contrary, the volume increases, the gas will have expanded (expansion work).

In this way, with the increase of the temperature of the gas, it is expanded by the increase of the kinetic energy of its molecules, but if an external pressure is applied it is compressed.

You might be interested in

When a falling body is at distance above the surface which is 4 times the Earth radius what is the acceleration due to Earth gra

Hoochie [10]

Answer:

Explanation:

g = GM/R²

as gravity decreases by the inverse of the square of the distance, increasing the distance by 4 times will reduce gravity to 1/16 that at the surface

9.8 / 16 = 0.61 m/s²

8 0

3 years ago

How much kinetic energy does a proton gain if it is accelerated, with no friction, through a potential difference of 1.00 V? The

ra1l [238]

Answer:

If energy is conserved, then the sum of the potential energy and the kinetic energy is a constant.

Assuming the proton starts from rest, so it's kineitc energy is zero, but it has a potential energy, PE equal to:

PE = qV

where q =1.6 x 10^-19 C

and V = 1.00 V

Assuming the proton no longer experiences the potential energy and it is all converted to kinetic energy then:

PE* = 0,

KE* = 1/(2mv^2)

Now since

PE + KE = Total energy =PE* + KE*

Therefore,

qV + 0 = 0 + 1/2mv^2

KE = qV = 1.6 10^-19 J

4 0

3 years ago

As shown in the diagram, two forces act on an object. The forces have magnitudes F1 = 5.7 N and F2 = 1.9 N. What third force wil

galina1969 [7]

Answer:

Second option 6.3 N at 162° counterclockwise from

F1->

Explanation:

Observe the attached image. We must calculate the sum of all the forces in the direction x and in the direction y and equal the sum of the forces to 0.

For the address x we have:

$-F_3sin(b) + F_1 = 0$

For the address and we have:

$-F_3cos(b) + F_2 = 0$

The forces $F_1$ and $F_2$ are known

$F_1 = 5.7\ N\\\\F_2 = 1.9\ N$

We have 2 unknowns ( $F_3$ and b) and we have 2 equations.

Now we clear $F_3$ from the second equation and introduce it into the first equation.

$F_3 = \frac{F_2}{cos (b)}$

Then

$-\frac{F_2}{cos (b)}sin(b)+F_1 = 0\\\\F_1 = \frac{F_2}{cos (b)}sin(b)\\\\F_1 = F_2tan(b)\\\\tan(b) = \frac{F_1}{F_2}\\\\tan(b) = \frac{5.7}{1.9}\\\\tan^{-1}(\frac{5.7}{1.9}) = b\\\\b= 72\°\\\\m = b +90\\\\\m= 162\°$

Then we find the value of $F_3$

$F_3 = \frac{F_1}{sin(b)}\\\\F_3 =\frac{5.7}{sin(72\°)}\\\\F_3 = 6.01 N$

Finally the answer is 6.3 N at 162° counterclockwise from

F1->

7 0

3 years ago

f your risk-aversion coefficient is A = 4 and you believe that the entire 1926–2015 period is representative of future expected

siniylev [52]

Answer:

The portfolio should invest 48.94% in equity while 51.05% in the T-bills.

Explanation:

As the complete question is not given here ,the table of data is missing which is as attached herewith.

From the maximized equation of the utility function it is evident that

$Weight=\frac{E_M-r_f}{A\sigma_M^2}$

For the equity, here as

$Weight$ is percentage of the equity which is to be calculated
${E_M-r_f}$ is the Risk premium whose value as seen from the attached data for the period 1926-2015 is 8.30%
$A$ is the risk aversion factor which is given as 4.
$\sigma_M$ is the standard deviation of the portfolio which from the data for the period 1926-2015 is 20.59