Respuesta :
Answer:
The volume would become 2.34 L if the temperature dropped to 11.0 °C
Explanation:
From the question, A balloon was filled to a volume of 2.50 L when the temperature was 30.0 °C.
To determine what the volume will become if the temperature dropped to 11.0 °C, we will use one of the Gas laws that relates Volume and Temperature. The Gas law that relates Volume and Temperature is the Charlse' law which states that, "the Volume of a fixed mass of gas is directly proportional to its Temperature ( in Kelvin) provided that the Pressure remains constant"
That is
V ∝ T ( at constant pressure)
Where V is the Volume
and T is the Temperature in Kelvin
Then, we can write that
V = kT
Where k is the constant of proportionality
Then,
[tex]\frac{V}{T} = k[/tex]
Hence, we can write that
[tex]\frac{V_{1} }{T_{1} } = \frac{V_{2} }{T_{2} } = \frac{V_{3} }{T_{3} } ...[/tex]
∴ [tex]\frac{V_{1} }{T_{1} } = \frac{V_{2} }{T_{2} }[/tex]
Where [tex]{V_{1} }[/tex] is the initial volume
[tex]{T_{1} }[/tex] is the initial temperature
[tex]{V_{2} }[/tex] is the final volume
and [tex]{T_{2} }[/tex] is the final temperature
From the question,
[tex]{V_{1} }[/tex] = 2.50 L
[tex]{T_{1} }[/tex] = 30.0 °C = 30.0 + 273.15 K = 303.15 K
[tex]{V_{2} }[/tex] = ??
[tex]{T_{2} }[/tex] = 11.0 °C = 11.0 + 273.15 K = 284.15 K
Putting the values into the equation, we get
[tex]\frac{2.50}{303.15} = \frac{V_{2} }{284.15}[/tex]
∴ [tex]V_{2} = \frac{2.50 \times 284.15}{303.15}[/tex]
[tex]V_{2} = 2.34 L[/tex]
Hence, the volume would become 2.34 L if the temperature dropped to 11.0 °C.
The final volume of the balloon is equal to 0.92 Liter.
Given the following data:
- Initial volume = 2.50 L
- Initial temperature = 30°C
- Final temperature = 11°C
To determine the final volume of the balloon, we would apply Charles's law:
Mathematically, Charles law is given by the formula;
[tex]\frac{V}{T} =k\\\\\frac{V_1}{T_1} = \frac{V_2}{T_2}[/tex]
Where;
- T is the temperature of an ideal gas.
- V is the volume of an ideal gas.
Substituting the given parameters into the formula, we have;
[tex]\frac{2.5}{30} = \frac{V_2}{11} \\\\2.5 \times 11 = 30V_2\\\\27.5 = 30V_2\\\\V_2 = \frac{27.5}{30}[/tex]
Final volume = 0.92 Liter.
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