All categories

3 years ago

Describe a situation in your life when you might see the ideal gas law exhibited.

Chemistry

1 answer:

statuscvo [17]3 years ago

8 0

Answer:

The ideal gas law is expressed mathematically by the ideal gas equation as follows;

P·V = n·R·T

Where;

P = The gas pressure

V = The volume of the gas

n = The number of moles of the gas present

R = The universal gas constant

T = The temperature of the gas

A situation where the ideal gas law is exhibited is in the atmosphere just before rainfall

The atmospheric temperature of the area expecting rainfall drops, (when there is appreciable blockage of the Sun's rays by cloud covering) followed by increased wind towards the area, which indicates that the area was in a state of a low pressure, 'P', and or volume, 'V', or a combination of both low pressure and volume P·V

When the entry flow of air into the area is observed to have reduced, the temperature of the air in the area is simultaneously sensed to have risen slightly, therefore, the combination of P·V is seen to be proportional to the temperature, 'T', and the number of moles of air particles, 'n' in the area

Explanation:

You might be interested in

how many grams of solid lithium must be added to liquid after in order to obtain 15.0 L of hydrogen gas at STP?

alexdok [17]

Answer:

9.29g

Explanation:

your welcome

3 0

3 years ago

A mole of oxygen and a mole of hydrogen (at STP) have all of the following in common EXCEPT

Drupady [299]

Answer:

Root mean squared velocity is different.

Explanation:

Hello!

In this case, since we have a mixture of oxygen and nitrogen at STP, which is defined as a condition whereas T = 298 K and P = 1 atm, we can infer that these gases have the same temperature, pressure, volume and moles but a different root mean squared velocity according to the following formula:

$v_{rms}=\sqrt{\frac{3RT}{MM} }$

Since they both have a different molar mass (MM), nitrogen (28.02 g/mol) and oxygen (32.02 g/mol), thus we infer that nitrogen would have a higher root mean squared velocity as its molar mass is less than that of oxygen.

Best regards!

8 0

2 years ago

A sample of argon gas collected at a pressure of 0.918 atm and a temperature of 279 K is found to occupy a volume of 719 millili

babymother [125]

Yeah i have 0 idea i’m sorry.

3 0

3 years ago

A measured sample of argon gas has a volume of 20.0 L at a pressure of 660 mm Hg. What is the final volume

valkas [14]

Answer:

C . 24 L

Explanation:

Given data:

Initial volume of gas = 20.0 L

Initial pressure of gas = 660 mmHg

Final volume = ?

Final pressure = 550 mmHg

Solution:

The given problem will be solved through the Boly's law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume

Now we will put the values in formula,

P₁V₁ = P₂V₂

660 mmHg × 20.0 L = 550 mmHg × V₂

V₂ = 13200 mmHg. L/ 550 mmHg

V₂ = 24 L

3 0

3 years ago

Substitute natural gas (SNG) is a gaseous mixture containing CH4(g) that can be used as a fuel. One reaction for the production

Lynna [10]

Answer:

ΔH° of the reaction is -747.54kJ

Explanation:

Based on gas law, it is possible to find the ΔH of a reaction using ΔH of half reactions.

Using the reactions:

(1) C(graphite) + 1/2O₂(g) → CO(g) ΔH° = -110.5 kJ

(2) CO(g) + 1/2O₂(g) → CO₂(g) ΔH° = -283.0 kJ

(3) H₂(g) + 1/2O₂(g) → H₂O(l) ΔH° = -285.8 kJ

(4) C(graphite) + 2H₂(g) → CH₄(g) ΔH° = -74.81 kJ

(5) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH° = -890.3 kJ

The sum of 4×(4) + (5) gives:

4C(graphite) + 8H₂(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) + 3CH₄(g)

ΔH° = -74.81 kJ ×4 - 890.3 kJ = -1189.54kJ

Now, this reaction - 4×(1) gives:

4CO(g) + 8H₂(g) → CO₂(g) + 2H₂O(l) + 3CH₄(g)

ΔH° = -1189.54kJ - 4×-110.5 = -747.54kJ

Thus ΔH° of the reaction is -747.54kJ

3 0

3 years ago