Question

A balloon filled with helium gas has a volume of 500 mL at a pressure of 1 atm. The balloon is released and reaches an altitude of 6.5 km, where the pressure is 0.5 atm. If the temperature has remained the same, what volume does the gas occupy at this height?

Answer 1

Answer:Volume of the gas occupy at this height will be 1 L.

Explanation:

Volume of the balloon when pressure of helium gas was 1 atm :

= 500 mL = 0.5 L =$V_1$ (1000 mL= 1 L)

Volume of the balloon when the pressure of helium gas is 0.5 atm :

V=$V_2$

According to Boyle's Law:

$Pressure\propto \frac{1}{Volume}$ (At constant temperature)

$P_1V_1=P_2V_2$

$P_1=1 atm, P_2=0.5 atm$

$1 atm\times 0.5 L=0.5 atm\times V$

$V=\frac{1 atm\tiomes 0.5 L}{0.5 atm}= 1 L$

Volume of the gas occupy at this height will be 1 L.

Answer 2

The volume of the gas at the height (i.e 6.5 Km) is 1000 mL

Data obtained from the question

•Initial volume (V₁) = 500 mL

•Initial pressure (P₁) = 1 atm

•Temperature = Constant

•Final pressure (P₂) = 0.5 atm

•Final volume (V₂) =?

The final volume of the gas can be obtained by using the Boyle's law equation as shown below:

P₁V₁ = P₂V₂

1 × 500 = 0.5 × V₂

500 = 0.5 × V₂

Divide both side by 0.5

V₂ = 500 / 0.5

V₂ = 1000 mL

Learn more about gas laws:

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