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WINSTONCH [101]
4 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]4 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.
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A rod of very small diameter with a mass 2m and length 3L is placed along the xaxis with one end at the origin. An identical rod
rewona [7]

Answer:

coordinates of the center of mass for these two rods

(x_{cm}, y_{cm})= (\frac{3L}{4},  \frac{3L}{4})cm

Explanation:

given

mass of a rod = 2m

length of the rod = 3L

mass of two rods = 2(2m) = 4m

radius = diameter/2 = \frac{3L}{2}

attached is the diagram and solution to the question

5 0
3 years ago
Which layer of the atmosphere has the highest density?
fomenos

Answer:

The Troposphere is the lowermost portion of Earth’s atmosphere. It is the densest layer of the atmosphere and contains approximately 75 percent of the mass of the atmosphere and almost all the water vapor and aerosol. The troposphere extends from the Earth’s surface up to the tropopause where the stratosphere begins.

Explanation:

4 0
3 years ago
A car is traveling at 120 km/h (75 mph). When applied the braking system can stop the car with a deceleration rate of 9.0 m/s2.
Bumek [7]

Answer:

the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15

Explanation:

Given that;

speed of car V  = 120 km/h = 33.3333 m/s

Reaction time of an alert driver = 0.8 sec

Reaction time of an alert driver = 3 sec

extra time taken by sleepy driver over an alert driver = 3 - 0.8 = 2.2 sec

now, extra distance that car will travel in case of sleepy driver  will be'

S_d = V × 2.2 sec

S_d = 33.3333 m/s × 2.2 sec

S_d = 73.3333 m

hence, number of car of additional car length  n will be;

n = S_n / car length

n = 73.3333 m / 5m

n = 14.666 ≈ 15

Therefore, the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15

8 0
3 years ago
Air temperature in a desert can reach 58.0°C (about 136°F). What is the speed of sound (in m/s) in air at that temperature?
Veronika [31]

The approximate speed of sound in dry (0% humidity) air, in meters per second, at temperatures near 0 °C, can be calculated from

c_{air} = (331.3+0.606 \upsilon)

Here

\upsilon = Temperature in Celsius

Replacing with our values we have that

\upsilon=58\° C

c_{air} = (331.3+0.606*58)

c_{air} = 366.1m/s

Therefore the speed of sound in air at that temperature is 366.1m/s

3 0
3 years ago
A fan cart with the fan set to high rolled across the floor. The cart's speeds with the fan on high are shown below. If the fan
Vadim26 [7]

Answer:

Speed of cart's might be less than the high speed after 5 seconds.

Explanation:

Given that,

A fan cart with the fan set to high rolled across the floor.

Let the speed of fan cart with set to high is x\ cm per second.

The fan supplies a force to the cart. If a lower fan speed were used, less force would be applied. This would cause a slower change in the cart's speed. So, the cart would be rolling more slowly than x\ cm per second after 5 seconds. The speed of cart's might be less than x\ cm per second.

Force is needed

A. for a moving object to keep moving at the same speed and direction

B. for a moving object to change its speed

C. for a motionless object to remain still

D. to prevent a moving object from turning

Hence,

Speed of cart's might be less than the high speed after 5 seconds.

3 0
3 years ago
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