Respuesta :
Answer:
V = 38.0 L
Explanation:
As we know that number of moles will remains conserved inside the balloon
so we will have
[tex]moles = \frac{PV}{RT}[/tex]
here we have
[tex]\frac{P_1V_1}{RT_1} = \frac{P_2V_2}{RT_2}[/tex]
now we have
[tex]P_1 = 760 mm Hg[/tex]
[tex]P_2 = 76 mm Hg[/tex]
[tex]V_1 = 5.00 L[/tex]
[tex]T_1 = 20^o C = 293 K[/tex]
[tex]T_2 = -50^o C = 223 K[/tex]
[tex]\frac{(760mm Hg)(5L)}{R(293)} = \frac{(76mm Hg)(V)}{R(223)}[/tex]
[tex]V = 38.0 L[/tex]
Answer : The new volume of of the balloon in liters is
Solution :
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
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 = 760 mm Hg
[tex]P_2[/tex] = final pressure of gas = 76.0 mm Hg
[tex]V_1[/tex] = initial volume of gas = 5.00 L
[tex]V_2[/tex] = final volume of gas = ?
[tex]T_1[/tex] = initial temperature of gas = [tex]20^oC=273+20=293K[/tex]
[tex]T_2[/tex] = final temperature of gas = [tex]-50^oC=273+(-50)=223K[/tex]
Now put all the given values in the above equation, we get the final volume of gas.
[tex]\frac{760\times 5.00L}{293K}=\frac{76\times V_2}{223K}[/tex]
[tex]V_2=38.0L[/tex]
Therefore, the new volume of of the balloon in liters is 38.0
