Respuesta :
Answer : The volume of the balloon will be 35.5 L at 80°C
Explanation :
The pressure is kept constant. Therefore any change in temperature would affect the volume. This can be explained with the help of Charles Law
According to this law, at constant pressure, the volume of an ideal gas varies directly with its absolute temperature ( K)
In equation form, the law can be stated as
[tex] \frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}} [/tex]
We have been given
V₁ = 30.0 L
We need temperature in kelvin unit .
T₁ = 25 °C = 25 + 273 = 298 K
V₂ = ?
T₂ = 80 + 273 = 353 K
Let us plug in these values in Charles law equation.
[tex] \frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}} [/tex]
[tex] \frac{30.0 L}{298K} = \frac{V_{2}}{353K } [/tex]
On cross multiplication, we get
[tex] V_{2} \times 298K = 30.0 L \times 353 K [/tex]
[tex] V_{2} = \frac{10590L.K}{298 K} [/tex]
[tex] V_{2} = 35.5 L [/tex]
The volume of the balloon will be 35.5 L
The final volume of the balloon has been 35 L.
The gas in the balloon has been assumed to be an ideal gas. The relationship between volume and temperature at constant pressure has been given by:
[tex]\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2} [/tex]
Computation for the volume of Balloon
The initial volume of balloon has been, [tex]V_1=30\;\rm L[/tex]
The initial temperature of the balloon has been, [tex]T_1=25\;^\circ \text C\\ T_1=298\;\rm K[/tex]
The final temperature of the balloon has been, [tex]T_2=80\;^\circ \text C\\ T_2=353\;\rm K[/tex]
The final volume of the balloon, [tex]V_2[/tex] has been given by substituting the values as:
[tex]\dfrac{30}{298}=\dfrac{V_2}{353}\\\\ V_2= \dfrac{30\;\times\;353}{298}\;\text L\\\\ V_2=35\;\rm L[/tex]
The final volume of the balloon has been 35 L.
Learn more about ideal gas, here:
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