Question

A 140.0 l weather balloon filled with 5.50 moles of helium has a small leak. if the helium leaks at a rate of 10.0 mmol/hr, what is the volume of the balloon after 45.0 hours?

Answer 1

Answer : The volume of the balloon after 45.0 hour is 128.5 L

Explanation :

First we have to calculate the moles of helium leaked.

In 1 hour, the number of moles of helium leaked = 10.0 mmol

In 45.0 hour, the number of moles of helium leaked = $\frac{45.0hr]{1hr}\times 10.0=450mmol=0.450mole$

conversion used : 1 mmol = 0.001 mole

Now we have to calculate the number of moles of helium remaining in the balloon.

Moles of helium remaining in the balloon = Total moles of helium - Moles of helium leaked

Moles of helium remaining in the balloon = 5.50 - 0.450 = 5.05 mole

Now we have to calculate the volume of the balloon after 45.0 hour.

According to the Avogadro's law, the volume of gas is directly proportional to the number of moles of gas at same pressure and temperature. That means,

$V\propto n$

or,

$\frac{V_1}{V_2}=\frac{n_1}{n_2}$

where,

$V_1$ = initial volume of gas  = 140.0 L

$V_2$ = final volume of gas = ?

$n_1$ = initial moles of gas  = 5.50 mole

$n_2$ = final moles of gas  = 5.05 mole

Now we put all the given values in this formula, we get

$\frac{140.0L}{V_2}=\frac{5.50mole}{5.05mole}$

$V_2=128.5L$

Therefore, the volume of the balloon after 45.0 hour is 128.5 L