Respuesta :
A) The temperature increases by a factor 6
We can use the ideal gas equation:
[tex]pV=nRT[/tex]
where
p is the pressure
V is the volume
n is the number of moles
R is the gas constant
T is the temperature
We can also rewrite it as
[tex]\frac{pV}{T}=nR[/tex]
The gas is in a sealed container - this means the amount of gas is fixed, so n is constant. Since R is constant too, the term on the right in the equation is constant. So we can rewrite the equation as:
[tex]\frac{p_1V_1}{T_1}=\frac{p_2 V_2}{T_2}[/tex]
where in this problem we have:
[tex]V_2 = 2V_1[/tex] (the volume is doubled)
[tex]p_2 = 3 p_1[/tex] (the pressure is tripled)
re-arranging the equation, we find the change in temperature:
[tex]\frac{T_2}{T_1}=\frac{p_2 V_2}{p_1 V_1}=\frac{(3 p_1)(2 V_1)}{p_1 V_1}=6[/tex]
so, the temperature increases by a factor 6.
B) The temperature increases by a factor 1.5.
We can use again the same equation:
[tex]\frac{p_1V_1}{T_1}=\frac{p_2 V_2}{T_2}[/tex]
Where in this case:
[tex]V_2 = \frac{V_1}{2}[/tex] (the volume is halved)
[tex]p_2 = 3 p_1[/tex] (the pressure is tripled)
So, we can find the change in temperature:
[tex]\frac{T_2}{T_1}=\frac{p_2 V_2}{p_1 V_1}=\frac{(3 p_1)(\frac{V_1}{2})}{p_1 V_1}=\frac{3}{2}=1.5[/tex]
So, the temperature increases by 1.5 times.
Considering the combined law equation:
- Part A: the temperature increases by a factor of 6.
- Part B: the temperature increases by a factor of 1.5.
Gay-Lussac's Law
Gay-Lussac's Law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move faster. Then the number of shocks against the walls increases, that is, the pressure increases. That is, the gas pressure is directly proportional to its temperature.
In summary, when there is a constant volume, as the temperature increases, the gas pressure increases. And when the temperature decreases, gas pressure decreases.
Gay-Lussac's law can be expressed mathematically as follows:
[tex]\frac{P}{T}=k[/tex]
where:
- P= pressure
- T= temperature
- k= Constant
Boyle's Law
Pressure and volume are related by Boyle's law, which says that the volume occupied by a given mass of gas at constant temperature is inversely proportional to pressure.
Boyle's law is expressed mathematically as:
P× V=k
where:
- P= pressure
- V= volume
- k= Constant
Charles's Law
Finally, Charles's Law consists of the relationship that exists between the volume and the temperature of a certain amount of ideal gas, which is maintained at a constant pressure.
This law says that 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. That is, the volume is directly proportional to the temperature of the gas.
In summary, Charles' law is a law that says that when the amount of gas and pressure are kept constant, the ratio between volume and temperature will always have the same value:
[tex]\frac{V}{T}=k[/tex]
where:
- V= volume
- T= temperature
- k= Constant
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 an final state 2, the following will be true:
[tex]\frac{P1xV1}{T1}=\frac{P2xV2}{T2}[/tex]
Part A
In this case, you know that:
- Volume is double: V2= 2×V1
- Pressure is tripled: P2= 3×P1
So, replacing in the combined law equation:
[tex]\frac{P1xV1}{T1}=\frac{3xP1x2xV1}{T2}[/tex]
Solving:
[tex]\frac{P1xV1}{T1}=\frac{6xP1xV1}{T2}[/tex]
[tex]\frac{T2}{T1}=\frac{6xP1xV1}{P1xV1}[/tex]
[tex]\frac{T2}{T1}=6[/tex]
T2= 6×T1
Finally, the temperature increases by a factor of 6.
Part B
In this case, you know that:
- Volume is halved: V2= V1÷ 2= [tex]\frac{V1}{2}[/tex]
- Pressure is tripled: P2= 3×P1
So, replacing in the combined law equation:
[tex]\frac{P1xV1}{T1}=\frac{3xP1x\frac{V1}{2} }{T2}[/tex]
Solving:
[tex]\frac{P1xV1}{T1}=\frac{3xP1xV1}{2T2}[/tex]
[tex]\frac{T2}{T1}=\frac{3xP1xV1}{2P1xV1}[/tex]
[tex]\frac{T2}{T1}=\frac{3}{2} =1.5[/tex]
T2= 1.5×T1
Finally, the temperature increases by a factor of 1.5.
Learn more about the ideal gas laws:
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