Chemistry: 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?

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if the balloon is 67,000 L at STP, what volume would

paralaw61772

29.01.2022

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,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

P_1 = initial pressure of gas (STP) = 1 atm

P_2 = final pressure of gas = 99.2 kPa = 0.98 atm (1kPa=0.0098atm)

V_1 = initial volume of gas = 67000 L

V_2 = final volume of gas = ?

T_1 = initial temperature of gas (STP) = 273K

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

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

$\frac{1\times 67000}{273}=\frac{0.98\times V_2}{283}$

V_2=70872L

Thus the volume is 70872 L

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