icehawks18
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if the balloon is 67,000 L at STP, what volume would it be if the air temperature cool to10.0°C in the pressure dropped to 99.2kPa as the balloon went higher?

Respuesta :

Answer: The volume is 70872 L

Explanation:

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 (STP) = 1 atm

[tex]P_2[/tex] = final pressure of gas = 99.2 kPa = 0.98 atm  (1kPa=0.0098atm)

[tex]V_1[/tex] = initial volume of gas  = 67000 L

[tex]V_2[/tex] = final volume of gas = ?

[tex]T_1[/tex] = initial temperature of gas (STP)  = 273K

[tex]T_2[/tex] = final temperature of gas = [tex]10.0^0C=(273+10)K=283K[/tex]

Now put all the given values in the above equation, we get:

[tex]\frac{1\times 67000}{273}=\frac{0.98\times V_2}{283}[/tex]

[tex]V_2=70872L[/tex]

Thus the volume is 70872 L

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