Question

A weather balloon with a volume of 3.40774 L

is released from Earth’s surface at sea level.

What volume will the balloon occupy at an

altitude of 20.0 km, where the air pressure is

10 kPa?

Answer in units of L.

Answer 1

Answer: The volume at an altitude of 20.0 km is 34.5289 L

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto \frac{1}{V}$     (At constant temperature and number of moles)

$P_1V_1=P_2V_2$

where,

$P_1$ = initial pressure of gas = 101.325 kPa ( at sea level)

$P_2$ = final pressure of gas = 10 kPa

$V_1$ = initial volume of gas = 3.40774 L

$V_2$ = final volume of gas = ?

Now put all the given values in the above equation, we get the final volume of gas.

$101.325\times 3.40774=10\times V_2$

$V_2=34.5289L$

Thus the volume at an altitude of 20.0 km is 34.5289 L

Therefore, the final pressure of the gas will be 3.75 atm