Question

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?

Answer 1

Answer: The volume is 70872 L

Step-by-step explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

`(P_1V_1)/(T_1)=(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:

`(1* 67000)/(273)=(0.98* V_2)/(283)`

`V_2=70872L`

Thus the volume is 70872 L