Question

A gas-filled balloon with a volume of 2.90 L at 1.20 atm and 20°C is allowed to rise to the stratosphere (about 30 km above the surface of the Earth), where the temperature and pressure are −23°C and 3.00 × 10−3 atm, respectively. Calculate the final volume of the balloon.

Answer 1

Answer: The final volume of the balloon is 990 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 = 1.20 atm

`P_2` = final pressure of gas =
`3.00* 10^(-3)atm`

`V_1` = initial volume of gas = 2.90 L

`V_2` = final volume of gas = ?

`T_1` = initial temperature of gas =
`20^oC=273+20=293K`

`T_2` = final temperature of gas =
`-23^oC=273-23=250K`

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

`(1.20* 2.90)/(293)=(3.00* 10^(-3)* V_2)/(250K)`

`V_2=990L`

Answer 2

The final volume is 990.8 L

Step-by-step explanation:

Let calculate the moles of gas in the first situation:

P . V = n . R . T

1.20 atm . 2.90 L = n . 0.082 . 293K

(1.20 atm . 2.90 L) / (0.082 . 293K) = 0.145 moles

This are the same moles in the second situation:

P . V = n . R . T

0.003atm . V = 0.145 moles . 0.082 . 250K

V = (0.145 moles . 0.082 . 250K) / 0.003atm

V = 990.8 L