Question

A hot air balloon has a volume of 2,510,000L at a pressure of 105 kPa and 122 degrees Celcius. If the balloon increased to 2,600,000L and the temperature increased to 150 degrees Celcius, what would be the pressure of the balloon?

Answer 1

Answer: The new pressure of the balloon is 108 kPa

Explanation:

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 105 kPa

$P_2$ = final pressure of gas = ?

$V_1$ = initial volume of gas = 2510000 L

$V_2$ = final volume of gas = 2600000 L

$T_1$ = initial temperature of gas = $122^0C=(122+273)K=395K$

$T_2$ = final temperature of gas = $150^0C=(150+273)K=423K$

Now put all the given values in the above equation, we get:

$\frac{105\times 2510000}{395}=\frac{P_2\times 2600000}{423}$

$P_2=108kPa$

The new pressure of the balloon is 108 kPa