A helium balloon with an internal pressure of 1.0 atm and a volume of 4.50 L at 20.0⁰C is released. What volume will the balloon occupy at an altitude where the pressure is 60.78 kPa and the temperature is -20.0⁰C?

Answer :

Answer: Volume occupied by volume is 6.48 L at an altitude where the pressure is 60.78 kPa and the temperature is -20.0⁰C

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 = 1 atm = 101.3 kPa

[tex]P_2[/tex] = final pressure of gas = 60.78 kPa

[tex]V_1[/tex] = initial volume of gas =  4.50 L

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

[tex]T_1[/tex] = initial temperature of gas = [tex]20.0^oC=273+20.0=293K[/tex]

[tex]T_2[/tex] = final temperature of gas = [tex]-23^oC=273-23=253K[/tex]

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

[tex]\frac{101.3kPa\times 4.50L}{293K}=\frac{60.78kPa\times V_2}{253K}[/tex]

[tex]V_2=6.48L[/tex]

The volume occupied by volume is 6.48 L at an altitude where the pressure is 60.78 kPa and the temperature is -20.0⁰C

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