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2 years ago

What happens to the average kinetic energy of a gas when the particles of the gas collide against each other at a constant

Chemistry

2 answers:

aksik [14]2 years ago

8 0

Answer:

The average kinetic energy remains constant.

Explanation:

took the test

Flura [38]2 years ago

6 0

The average kinetic energy of a gas when the particles of the gas collide against each other at a constant temperature and volume will not changes.

<h3>What is average kinetic energy?</h3>

The product of half the mass of each gas molecule with the square of root mean square speed is the average kinetic energy of the molecules.

And this average kinetic energy is deirectly proportional to the absolute temperature of the system but in the given question temperature and volume of the gas is constant. So that there is no change in average kinetic energy of molecules when they collide.

Hence no change in the average kinetic energy of a gas takes place.

To know more about average kinetic energy, visit the below link:

brainly.com/question/13868723

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IrinaK [193]

Answer:

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3 years ago

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Describe the difference between pure substance and mixtures ( give an example each)

kondaur [170]

Answer:

a pure substance consists only of one element or one compound

a mixture consists of two or more different substances, not chemically joined together.

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pure substance : Hydrogen gas - Diamond - Gold metal.

mixture : water and oil - mixtures of sand and water - trail mix

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2 years ago

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Th e molar absorption coeffi cient of a substance dissolved in water is known to be 855 dm3 mol−1 cm−1 at 270 nm. To determine t

Olegator [25]

Answer : The percentage reduction in intensity is 79.80 %

Explanation :

Using Beer-Lambert's law :

$A=\epsilon \times C\times l$

$A=\log \frac{I_o}{I}$

$\log \frac{I_o}{I}=\epsilon \times C\times l$

where,

A = absorbance of solution

C = concentration of solution = $3.25mmol.dm^{3-}=3.25\times 10^{-3}mol.dm^{-3}$

l = path length = 2.5 mm = 0.25 cm

$I_o$ = incident light

$I$ = transmitted light

$\epsilon$ = molar absorptivity coefficient = $855dm^3mol^{-1}cm^{-1}$

Now put all the given values in the above formula, we get:

$\log \frac{I_o}{I}=(855dm^3mol^{-1}cm^{-1})\times (3.25\times 10^{-3}mol.dm^{-3})\times (0.25cm)$

$\log \frac{I_o}{I}=0.6947$

$\frac{I_o}{I}=10^{0.6947}=4.951$

If we consider $I_o$ = 100

then, $I=\frac{100}{4.951}=20.198$

Here 'I' intensity of transmitted light = 20.198

Thus, the intensity of absorbed light $I_A$ = 100 - 20.198 = 79.80

Now we have to calculate the percentage reduction in intensity.

$\% \text{reduction in intensity}=\frac{I_A}{I_o}\times 100$

$\% \text{reduction in intensity}=\frac{79.80}{100}\times 100=79.80\%$

Therefore, the percentage reduction in intensity is 79.80 %

3 0

3 years ago

In the 4th paragraph, it says disturbances like "accelerated erosion, or change in vegetation cover" tend to persist and grow ov

deff fn [24]

Answer:

Erosion is the washing away of the uppermost part of the earth crust. Once erosion starts it only gets worse because, erosion wash away further soil, causing the eroding channel to futher expand, increasing the erosion.

Lack of vegetative cover allows erosion which washes away soil nutrients, further worsening the soil against soil vegetative cover.

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3 years ago

A) Calculate the osmotic pressure difference between seawater and fresh water. For simplicity, assume thatall the dissolved salt

never [62]

Answer:

a) Δπ = 1.264 atm

b) W = 128 joules

c) ΔH >> W ( a factor greater than 17,000 )

Explanation:

a) The osmotic pressure, π , is determined by :

π = nRT/V, where n= moles of solute

R= 0.0821 Latm/kmol

T = 300 K

calling π(sw) osmotic pressure for for sea water and π (fw) for fresh water,

salinity of sea water = 3.5 g / 1L water (assuming only NaCl for the salts)

salinity of fresh water = 0.5 parts per thousand (range: 0- 0.5 ppt)

πsw = (3.5 g/58.44 g/mol) (0.0821 Latm/Kmol) (300 K ) /1 L = 1.475 atm

πfw = (0.5 g/58.44 g/mol) (0.0821 Latm/Kmol) (300 K ) /1 L = 0.211 atm

d water = 1 g/cm³

Δ π = (1.475 - 0.211) = 1.264 atm

b) W = Δπ V = 1.426 atm x 1L = 1.43 L-atm

1 L-atm = 101.33 j

W = 101.33 j/ Latm x 1.43 Latm = 128 joules

c) ΔH = Q₁ + nΔH vap, where

Q₁ = heat required to bring the solution from 300 K to boiling, 373 K

ΔH vap = heat of vaporization

Q = mCΔT = 1000 g x 4.186 j x 73 K = 305.6 j = 0.3056 kj

ΔH vap = (1000 g/ 18 g/mol ) 40.7 kj/mol = 2,261 kj

ΔH = 0.3056 kj + 2,261 kj = 2,261.3 kj

Note = Q << ΔH vap and we could have neglected it.

This result shows why nobody talks about evaporation of sea water to produce fresh water ΔH >> W

6 0

3 years ago