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Alexandra [31]
3 years ago
12

A real gas will behave most like an ideal gas under conditions of ________.

Physics
1 answer:
KengaRu [80]3 years ago
4 0

Answer: high temperature and low pressure

Explanation:

The Ideal Gas equation is:  

P.V=n.R.T  

Where:  

P is the pressure of the gas  

V is the volume of the gas

n the number of moles of gas  

R=0.0821\frac{L.atm}{mol.K} is the gas constant  

T is the absolute temperature of the gas in Kelvin

According to this law, molecules in gaseous state do not exert any force among them (attraction or repulsion) and the volume of these molecules is small, therefore negligible in comparison with the volume of the container that contains them.  

Now, real gases can behave approximately to an ideal gas, under the conditions described above and taking into account the following:  

When <u>temperature is high</u> a real gas approximates to ideal gas, because the molecules move quickly, preventing the repulsion or attraction forces to take effect.  In addition, at <u>low pressures</u>, the volume of molecules is negligible.

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Answer:

Explanation:

At the time of a body achieving terminal velocity, the drag force becomes equal to the weight of the body less the buoyant force by the surrounding medium which can be represented by the following equation

\frac{4\pi\times r^3(d-\rho)}{3} =6\pi\times n\times r\times v

Where r is radius of the body , d is density of the material of the body σ is density of the medium and n is coefficient of viscosity of the medium and v is terminal velocity.

Simplifying

v = \frac{2\times r^2(d-\rho)}{9\times n}

Assuming the value of density of air as 1.225 kg/m³ and putting other given values in the formula we get

v = [tex]\frac{2\times (1.2\times10^{-5})^2(2182-1.225)}{9\times 1.8\times10^{-5}}[/tex]

v = 387 x 10⁻⁵ m/s

Terminal velocity = 387 x 10⁻⁵ m/s

Time taken to fall a distance of 100 m

= \frac{100}{387\times10^{-5}}

= 2.6 x 10⁴ s.

5 0
3 years ago
The atmosphere of Mercury and Mars are very thin. What effect does the thin atmosphere have on the temperature on the surface of
KengaRu [80]

Answer:

Very hot during the day and very cold at night.

Explanation:

Due to the thin atmosphere, they have very hot climate during the day time and very cold climate at night. This happens because they contain very low amounts of greenhouse gases. These gases retain the heat at night. The atmosphere also prevents excessive light and UV rays from entering. The thin  atmosphere leads to many asteroids and comets hitting the surface of the planet. On earth, these asteroids usually, burn up in the mesosphere layer of the atmosphere. These asteroid collisions cause massive fires. This in turn,  causes the temperature to increase during the day. During the night time, massive fires cannot burn due to the low temperature because of the lack of greenhouse gases.

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Which of the following choices describes the type of image formed by a plane mirror?
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True, the law of inertia effects both moving and non-moving objects.
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The initial kinetic energy imparted to a 0.25 kg bullet is 1066 J. The acceleration of gravity is 9.81 m/s 2 . Neglecting air re
lubasha [3.4K]

Answer:

The range of the bullet is 0.435 kilometers.

Explanation:

According to the problem, maximum height is equal to the range of the bullet. That is:

\Delta x = \Delta y

Where:

\Delta x - Range of the bullet, measured in meters.

\Delta y - Maximum height of the bullet, measured in meters.

By the Principle of Energy Conservation, gravitational potential energy reaches its maximum at the expense of the initial kinetic energy. That is to say:

K_{1} = U_{2}

Where:

K_{1} - Kinetic energy at point 1, measured in joules.

U_{1} - Gravitational potential energy at point 2, measured in joules, and:

U_{2} = m\cdot g \cdot \Delta y

Where:

m - Mass of the bullet, measured in kilograms.

g - Gravitational constant, measured in meters per square second.

The maximum height is now cleared:

K_{1} = m\cdot g \cdot \Delta y

\Delta y = \frac{K_{1}}{m\cdot g}

If K_{1} = 1066\,J, m = 0.25\,kg and g = 9.81\,\frac{m}{s^{2}}, the maximum height is now computed:

\Delta y = \frac{1066\,J}{(0.25\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}

\Delta y = 434.791\,m

\Delta y = 0.435\,km

Lastly, the range of the bullet is 0.435 kilometers.

3 0
3 years ago
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