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

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 = 1.20 atm

$P_2$ = final pressure of gas = $3.00\times 10^{-3}atm$

$V_1$ = 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:

$\frac{1.20\times 2.90}{293}=\frac{3.00\times 10^{-3}\times V_2}{250K}$

$V_2=990L$

Answer 2

Answer:

The final volume is 990.8 L

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