Question

What is the pressure inside such a balloon (in atm) if it starts out at sea level with a temperature of 12.8°C and rises to an altitude where its volume is twenty-two times the original volume and its temperature is −47.7°C? (Enter your answer to at least 3 decimal places)

Answer 1

Answer:

The final pressure inside the balloon will be $P_2=0.0359\ atm$.

Explanation:

Given initial temperature of the balloon is $T_1=12.8\°C$

And the final temperature of the balloon is $T_2=-47.7\°C$

Initially, the balloon is at atmospheric pressure $P_1=1\ atm$ and volume be $V_1$

And the final pressure in the balloon is $P_2$ and the volume be $V_2$

Also, it is given in the question that the final volume became twenty-two times the original volume.

We can write $V_2=22V_1$

Now, using ideal gas equation.

$\frac{P_2V_2}{P_1V_1}=\frac{nRT_2}{nRT_2}$

Where, $R$ is the gas constant. And $n$ is moles of substance inside the balloon.

In our problem the value of $n$ will be the same for both cases. Also, $R$ is the gas constant.

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

Also, we have $V_2=22V_1$ from the question.

$\frac{P_2\times 22V_1}{P_1V_1}=\frac{T_2}{T_1}\\\\\frac{P_2\times 22}{P_1}=\frac{T_2}{T_1}$

We need to convert the temperature from $\°C$ to Kelvin.

$T_1=12.8\°C\\T_1=12.8+273.15=285.95\\T_2=-47.7\°C\\T_2=-47.7+273.15=225.45$

Plug these values we get,

$\frac{P_2\times 22}{1}=\frac{225.45}{285.95}\\\\22\times P_2=0.789\\P_2=\frac{0.789}{22}\\P_2=0.0359\ atm.$

So, the final pressure inside the balloon will be $P_2=0.0359\ atm$.