Question

A weather balloon is filled to a volume of 2 liters with helium gas at a pressure of 101 kPa, at a temperature of 22oC. The balloon is released and floats to a height where the pressure is 90 kPa and the air temperature is now 16oC. What is the new volume of the weather balloon?

Answer 1

Answer: The new volume of  weather balloon will be 2.2 Liters

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 =  101 kPa

$P_2$ = final pressure of gas = 90 kpa

$V_1$ = initial volume of gas = 2 L

$V_2$ = final volume of gas = ? L

$T_1$ = initial temperature of gas = $22^oC=273+22=295K$

$T_2$ = final temperature of gas = $16^oC=273+16=289K$

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

$\frac{101\times 2L}{295K}=\frac{90\times V_2}{289K}$

$V_2=2.2L$

Therefore, the volume will be 2.2 Liters.

Answer 2

Answer:

B. 2.20L

Explanation: