Respuesta :
Answer:
The final pressure inside the balloon will be [tex]P_2=0.0359\ atm[/tex].
Explanation:
Given initial temperature of the balloon is [tex]T_1=12.8\°C[/tex]
And the final temperature of the balloon is [tex]T_2=-47.7\°C[/tex]
Initially, the balloon is at atmospheric pressure [tex]P_1=1\ atm[/tex] and volume be [tex]V_1[/tex]
And the final pressure in the balloon is [tex]P_2[/tex] and the volume be [tex]V_2[/tex]
Also, it is given in the question that the final volume became twenty-two times the original volume.
We can write [tex]V_2=22V_1[/tex]
Now, using ideal gas equation.
[tex]\frac{P_2V_2}{P_1V_1}=\frac{nRT_2}{nRT_2}[/tex]
Where, [tex]R[/tex] is the gas constant. And [tex]n[/tex] is moles of substance inside the balloon.
In our problem the value of [tex]n[/tex] will be the same for both cases. Also, [tex]R[/tex] is the gas constant.
[tex]\frac{P_2V_2}{P_1V_1}=\frac{T_2}{T_1}[/tex]
Also, we have [tex]V_2=22V_1[/tex] from the question.
[tex]\frac{P_2\times 22V_1}{P_1V_1}=\frac{T_2}{T_1}\\\\\frac{P_2\times 22}{P_1}=\frac{T_2}{T_1}[/tex]
We need to convert the temperature from [tex]\°C[/tex] to Kelvin.
[tex]T_1=12.8\°C\\T_1=12.8+273.15=285.95\\T_2=-47.7\°C\\T_2=-47.7+273.15=225.45[/tex]
Plug these values we get,
[tex]\frac{P_2\times 22}{1}=\frac{225.45}{285.95}\\\\22\times P_2=0.789\\P_2=\frac{0.789}{22}\\P_2=0.0359\ atm.[/tex]
So, the final pressure inside the balloon will be [tex]P_2=0.0359\ atm[/tex].
