How do water particles move in a wave? A. They move forward with the wave. B. They move in a circular motion. C. They move up an
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Considering the ideal gas law and STP conditions, the volume of the gas at a pressure of 286.5 kPa and a temperature of 12.9°C is 1.8987 L.
<h3>Definition of STP condition</h3>The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.
<h3>Ideal gas law</h3>Ideal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.
The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:
P×V = n×R×T
where:
- P is the gas pressure.
- V is the volume that occupies.
- T is its temperature.
- R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
- n is the number of moles of the gas.
In first place, you can apply the following rule of three: if by definition of STP conditions 22.4 L are occupied by 1 mole of carbon dioxide, 5.13 L (5.13 dm³= 5.13 L, being 1 dm³= 1 L) are occupied by how many moles of carbon dioxide?
<u><em>amount of moles of carbon dioxide= 0.229 moles</em></u>
Then, you know:
- P= 286.5 kPa= 2.8275352 atm (being 1 kPa= 0.00986923 atm)
- V= ?
- T= 12.9 C= 285.9 K (being 0°C= 273 K)
- R= 0.082
- n= 0.229 moles
Replacing in the ideal gas law:
2.8275352 atm× V = 0.229 moles×0.082 × 285.9 K
Solving:
V= (0.229 moles×0.082 × 285.9 K)÷ 2.8275352 atm
<u><em>V= 1.8987 L</em></u>
Finally, the volume of the gas at a pressure of 286.5 kPa and a temperature of 12.9°C is 1.8987 L.
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STP conditions:
the ideal gas law:
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Answer:
34.28 L ( 1.5*22.4 L)
Explanation:
Calculation of the moles of aluminum as:-
Mass = 55 g
Molar mass of aluminum = 26.981539 g/mol
The formula for the calculation of moles is shown below:
Thus,
According to the reaction:-
4 moles of aluminum react with 3 moles of oxygen gas
1 mole of aluminum react with moles of oxygen gas
2.0384 moles of aluminum react with moles of oxygen gas
Moles of oxygen gas = 1.5288 moles
At STP,
Pressure = 1 atm
Temperature = 273.15 K
Using ideal gas equation as:
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
1 atm × V = 1.5288 mol × 0.0821 L.atm/K.mol × 273.15 K
⇒V = 34.28 L ( 1.5*22.4 L)