1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
WINSTONCH [101]
3 years ago
6

At which of the following temperature and pressure levels would a gas be most likely to follow the ideal gas law? A. 0 K and 100

Pa B. 273 K and 1 Pa C. 100 K and 273 Pa D. 273 K and 100 kPa
Physics
1 answer:
bulgar [2K]3 years ago
6 0
The Ideal Gas Law makes a few assumptions from the Kinetic-Molecular Theory. These assumptions make our work much easier but aren't true under all conditions. The assumptions are,

1) Particles of a gas have virtually no volume and are like single points.
2) Particles exhibit no attractions or repulsions between them.
3) Particles are in continuous, random motion.
4) Collisions between particles are elastic, meaning basically that when they collide, they don't lose any energy.
5) The average kinetic energy is the same for all gasses at a given temperature, regardless of the identity of the gas.

It's generally true that gasses are mostly empty space and their particles occupy very little volume. Gasses are usually far enough apart that they exhibit very little attractive or repulsive forces. When energetic, the gas particles are also in fairly continuous motion, and without other forces, the motion is basically random. Collisions absorb very little energy, and the average KE is pretty close.

Most of these assumptions are dependent on having gas particles very spread apart. When is that true? Think about the other gas laws to remember what properties are related to volume.

A gas with a low pressure and a high temperature will be spread out and therefore exhibit ideal properties.

So, in analyzing the four choices given, we look for low P and high T.

A is at absolute zero, which is pretty much impossible, and definitely does not describe a gas. We rule this out immediately.

B and D are at the same temperature (273 K, or 0 °C), but C is at 100 K, or -173 K. This is very cold, so we rule that out.

We move on to comparing the pressures of B and D. Remember, a low pressure means the particles are more spread out. B has P = 1 Pa, but D has 100 kPa. We need the same units to confirm. Based on our metric prefixes, we know that kPa is kilopascals, and is thus 1000 pascals. So, the pressure of D is five orders of magnitude greater! Thus, the answer is B.
You might be interested in
A worker pushes horizontally on a large crate with a force of 265.0 N and the crate moved 4.3 m. How much work was done?
trapecia [35]

Answer:

<h2>1139.5 J</h2>

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question we have

work done = 265 × 4.3 = 1139.5

We have the final answer as

<h3>1139.5 J</h3>

Hope this helps you

8 0
2 years ago
At time t=0, a particle is located at the point (3,6,9). It travels in a straight line to the point (5,2,7), has speed 8 at (3,6
Elis [28]

The particle has constant acceleration according to

\vec a(t)=2\,\vec\imath-4\,\vec\jmath-2\,\vec k

Its velocity at time t is

\displaystyle\vec v(t)=\vec v(0)+\int_0^t\vec a(u)\,\mathrm du

\vec v(t)=\vec v(0)+(2\,\vec\imath-4\,\vec\jmath-2\,\vec k)t

\vec v(t)=(v_{0x}+2t)\,\vec\imath+(v_{0y}-4t)\,\vec\jmath+(v_{0z}-2t)\,\vec k

Then the particle has position at time t according to

\displaystyle\vec r(t)=\vec r(0)+\int_0^t\vec v(u)\,\mathrm du

\vec r(t)=(3+v_{0x}t+t^2)\,\vec\imath+(6+v_{0y}t-2t^2)\,\vec\jmath+(9+v_{0z}t-t^2)\,\vec k

At at the point (3, 6, 9), i.e. when t=0, it has speed 8, so that

\|\vec v(0)\|=8\iff{v_{0x}}^2+{v_{0y}}^2+{v_{0z}}^2=64

We know that at some time t=T, the particle is at the point (5, 2, 7), which tells us

\begin{cases}3+v_{0x}T+T^2=5\\6+v_{0y}T-2T^2=2\\9+v_{0z}T-T^2=7\end{cases}\implies\begin{cases}v_{0x}=\dfrac{2-T^2}T\\\\v_{0y}=\dfrac{2T^2-4}T\\\\v_{0z}=\dfrac{T^2-2}T\end{cases}

and in particular we see that

v_{0y}=-2v_{0x}

and

v_{0z}=-v_{0x}

Then

{v_{0x}}^2+(-2v_{0x})^2+(-v_{0x})^2=6{v_{0x}}^2=64\implies v_{0x}=\pm\dfrac{4\sqrt6}3

\implies v_{0y}=\mp\dfrac{8\sqrt6}3

\implies v_{0z}=\mp\dfrac{4\sqrt6}3

That is, there are two possible initial velocities for which the particle can travel between (3, 6, 9) and (5, 2, 7) with the given acceleration vector and given that it starts with a speed of 8. Then there are two possible solutions for its position vector; one of them is

\vec r(t)=\left(3+\dfrac{4\sqrt6}3t+t^2\right)\,\vec\imath+\left(6-\dfrac{8\sqrt6}3t-2t^2\right)\,\vec\jmath+\left(9-\dfrac{4\sqrt6}3t-t^2\right)\,\vec k

4 0
3 years ago
so I know you can solve this either by using Vox or Voy. I'm getting 3.08s when using Vox and 3.14s for Voy way. For Voy I'm usi
Tatiana [17]

The person's horizontal position is given by

x=v_0\cos40^\circ t

and the time it takes for him to travel 56.6 m is

56.6\,\mathrm m=\left(24.0\,\dfrac{\mathrm m}{\mathrm s}\right)\cos40^\circ t\implies t=3.08\,\mathrm s

so your first computed time is the correct one.

The question requires a bit of careful reading, and I think there may be a mistake in the problem. The person's vertical velocity v_y at time t is

v_y=v_{0y}-gt

which tells us that he would reach the ground at about t=3.15\,\mathrm s. In this time, he would have traveled

x=v_{0x}(3.15\,\mathrm s)=57.9\,\mathrm m

But we're told that he is caught by a net at 56.6 m, which would mean that the net cannot have been placed at the same height from which he was launched. However, it's possible that the moment at which he was launched doesn't refer to the moment the cannon went off, but rather the moment at which the person left the muzzle of the cannon a fraction of a second after the cannon was set off. After this time, the person's initial vertical velocity v_{0y} would have been a bit smaller than \left(24.0\,\frac{\mathrm m}{\mathrm s}\right)\sin40^\circ.

7 0
2 years ago
Who wants to be freinds
brilliants [131]

Hello there, Goodmorning and God Bless...Here is your answer to your following question:

yessss i do!!!

5 0
2 years ago
Read 2 more answers
Starting from a pillar, you run 200 m east (the + x-direction) at an average speed of 5.0 m/s and then run 280 m west at an aver
zalisa [80]

Answer:

Total time taken=110 seconds

Total distance traveled=480m

Explanation:

First of all, we find the total time taken:

For that, we use the formula : Distance/Speed= Time

Time for part 1 : 200/5=40 seconds

Time for part 2 : 280/4=70seconds

Total time taken=110 seconds

Total distance traveled=480m

Average Speed= 480/110=4.36 m/s

Total displacement=200-280=-80m (Since this is displacement, we need to find the distance between the initial and final point. Also, I've taken east direction as positive and west as negative)

Average Velocity=-80/110=-0.72 m/s

OR 0.72m/s towards west.

3 0
2 years ago
Other questions:
  • 1) My 14V car battery could be used to charge my laptop, but I need to use an inverter to first convert it to a standard 120V. T
    13·1 answer
  • Which of the following does not affect the apparent brightness of a star.
    11·2 answers
  • Physics question i appreciate your help please
    15·2 answers
  • A river flows from south to north at 5.4 km/hr. on the west bank of this river, a boat launches and travels perpendicular to the
    5·1 answer
  • The drawing shows a golf ball passing through a windmill at a miniature golf course. The windmill has 8 blades and rotates at an
    11·1 answer
  • When carrying extra weight, the space formed between the top of your head and the two axles of the motorcycle is referred to as
    10·1 answer
  • The three factors that determine the amount of potential energy in an object are ________, ___________, and _________.
    12·2 answers
  • Is the movement of wind kinetic energy ?
    12·1 answer
  • A golf ball is struck with a velocity of 80 ft/s as shown. Determine the speed at which it strikes the ground at b and the time
    12·2 answers
  • A fighter plane is descending at 30 m/s. The pilot ejects, and the ejector seat accelerates him upwards at 120 m/s2 for 2 second
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!