Respuesta :
Answer: a) 900 K
b) 1200 K
Explanation:
According to the ideal gas equation:
[tex]PV=nRT[/tex]
P = Pressure of the gas = 6.00 atm
V= Volume of the gas = ?
T= Temperature of the gas = 27°C = 300 K [tex]0^00C=273K[/tex]
R= Gas constant = 0.0821 atmL/K mol
n= moles of gas = 1
[tex]V=\frac{nRT}{P}=\frac{1\times 0.0821\times 300}{6.00}=4.10L[/tex]
a) To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.
Mathematically,
[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]
where,
[tex]P_1\text{ and }T_1[/tex] are the initial pressure and temperature of the gas.
[tex]P_2\text{ and }T_2[/tex] are the final pressure and temperature of the gas.
We are given:
[tex]P_1=6.00atm\\T_1=300K\\P_2=3\times 6.00=18.0atm\\T_2=?[/tex]
Putting values in above equation, we get:
[tex]\frac{6.00}{300K}=\frac{18.0}{T_2}\\\\T_2=900K[/tex]
The final temperature is 900 K
b) The combined gas equation is,
[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]
where,
[tex]P_1[/tex] = initial pressure of gas = 6.00 atm
[tex]P_2[/tex] = final pressure of gas = [tex]2\times 6.00atm=12.0atm[/tex]
[tex]V_1[/tex] = initial volume of gas = 4.10 L
[tex]V_2[/tex] = final volume of gas = [tex]2\times 4.10 L=8.20L[/tex]
[tex]T_1[/tex] = initial temperature of gas = [tex]300K[/tex]
[tex]T_2[/tex] = final temperature of gas =?
Now put all the given values in the above equation, we get:
[tex]\frac{6.00\times 4.10}{300}=\frac{12.0\times 8.20}{T_2}[/tex]
[tex]T_2=1200K[/tex]
The final temperature is 1200 K
Taking into account the Gay- Lussac's law, combined law equation and ideal gas law, you obtain:
a) The final temperature is 900 K.
b) The final temperature is 1200 K.
a)
- Gay Lussac's law
Gay Lussac's law states that the relationship between the temperature and the pressure of a gas when the volume is constant is directly proportional.
In other words, when there is a constant volume, as the temperature increases, the pressure of the gas increases. And when the temperature is decreased, the pressure of the gas decreases.
Mathematically, Gay Lussac's law is expressed as:
[tex]\frac{P}{T} =k[/tex]
Studying two different states, an initial state 1 and a final state 2, it is true that:
[tex]\frac{P1}{T1} =\frac{P2}{T2}[/tex]
In this case, you know:
- P1= 6 atm
- T1= 27 C= 300 K
- P2= 3× 6 atm= 18 atm because the pressure triples
- T2= ?
Replacing:
[tex]\frac{6 atm}{300 K} =\frac{18 atm}{T2}[/tex]
Solving:
[tex]T2x\frac{6 atm}{300 K} =18 atm[/tex]
[tex]T2=\frac{18 atm}{\frac{6 atm}{300 K} }[/tex]
T2= 900 K
In summary, the final temperature is 900 K.
b)
- Charles's Law
Charles's Law is a gas law that relates the volume and temperature of a certain quantity of gas at constant pressure by direct proportionality.
In other words, for a given sum of gas at constant pressure, as the temperature increases, the volume of the gas increases, and as the temperature decreases, the volume of the gas decreases.
In summary, Charles's law is a law that says that when the amount of gas and pressure are kept constant, the quotient that exists between the volume and the temperature will always have the same value:
[tex]\frac{V}{T}=k[/tex]
- Boyle's law
Finally, Boyle's law expresses that the volume occupied by a certain gaseous mass at constant temperature is inversely proportional to the pressure.
Boyle's law is expressed mathematically as:
P×V=k
- Combined law equation
Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law:
[tex]\frac{PxV}{T}=k[/tex]
Studying two different states, an initial state 1 and a final state 2, it is true that:
[tex]\frac{P1xV1}{T1} =\frac{P2xV2}{T2}[/tex]
- Ideal gas law
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, and n is the number of moles of the gas. The universal constant of ideal gases R has the same value for all gaseous substances.
- This case
One mole of oxygen gas is at a pressure of 6.00 atm and a temperature of 27.0°C (300 K). So, you can applied the ideal gas law to determine the initial volume of the gas:
6 atm× V1 = 1 mole×0.082 [tex]\frac{atmL}{molK}[/tex]× 300 K
Solving:
V1 = (1 mole×0.082 [tex]\frac{atmL}{molK}[/tex]× 300 K)÷ 6 atm
V1= 4.1 L
The gas is heated so that both the pressure and volume are doubled. So, you know:
- P1= 6 atm
- V1= 4.1 L
- T1= 300 K
- P2= 2× 6 atm= 12 atm
- V2= 2× 4.1 L= 8.2 L
- T2= ?
Replacing in the combined law equation:
[tex]\frac{6 atmx4.1 L}{300 K} =\frac{12 atmx8.2 L}{T2}[/tex]
Solving:
[tex]T2x\frac{6 atmx4.1 L}{300 K} =12 atmx8.2 L[/tex]
[tex]T2 =\frac{12 atmx8.2 L}{\frac{6 atmx4.1 L}{300 K}}[/tex]
T2= 1200 K
In summary, the final temperature is 1200 K.
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