Question

A very flexible helium-filled balloon is released from the ground into the air at 20. ∘C. The initial volume of the balloon is 5.00 L, and the pressure is 760. mmHg. The balloon ascends to an altitude of 20 km, where the pressure is 76.0 mmHg and the temperature is −50. ∘C. What is the new volume, V2, of the balloon in liters?

Answer 1

Answer:

V = 38.0 L

Explanation:

As we know that number of moles will remains conserved inside the balloon

so we will have

$moles = \frac{PV}{RT}$

here we have

$\frac{P_1V_1}{RT_1} = \frac{P_2V_2}{RT_2}$

now we have

$P_1 = 760 mm Hg$

$P_2 = 76 mm Hg$

$V_1 = 5.00 L$

$T_1 = 20^o C = 293 K$

$T_2 = -50^o C = 223 K$

$\frac{(760mm Hg)(5L)}{R(293)} = \frac{(76mm Hg)(V)}{R(223)}$

$V = 38.0 L$

Answer 2

Answer : The new volume of of the balloon in liters is

Solution :

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 = 760 mm Hg

$P_2$ = final pressure of gas = 76.0 mm Hg

$V_1$ = initial volume of gas = 5.00 L

$V_2$ = final volume of gas = ?

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

$T_2$ = final temperature of gas = $-50^oC=273+(-50)=223K$

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

$\frac{760\times 5.00L}{293K}=\frac{76\times V_2}{223K}$

$V_2=38.0L$

Therefore, the new volume of of the balloon in liters is 38.0