Q. 8.77 - An Introduction to General, Organic, and Biological Chemistry

Question

A weather balloon is partially filled with helium to allow for expansion at high altitudes. At STP, a weather balloon is filled with enough helium to give a volume of $25.0L$. At an altitude of $30.0km$ and $-35°C$, it has expanded to $2460L$. The increase in volume cause it to burst and a small parachute returns the instruments to Earth.

a. How many grams of helium are added to the balloon?

b. What is the final pressure, in millimeters of mercury, of the helium inside the balloon when it bursts?

Step-by-Step Solution

Answer

a. The mass of Helium added is $4.47g$.

b. The final pressure of the helium inside the balloon is $6.73mmHg$.

Part (a) Part 1: Given Information

We need to find the grams of helium which added to the balloon.

Part (a) Step 2: Simplify

For calculating the mass of the helium, we will use the ideal gas law equation

$PV=nRT ... \left(1\right)$

Since, $n=\frac{mass}{molar mass}$

Substituting the value of $n$ in equation $\left(1\right)$

$PV=\frac{mass}{molar mass}RT ... \left(2\right)$

Considering the given conditions at STP

The pressure $P=1atm$,

the temperature $T=273K$,

the volume of gas $V=25L$

the universal gas constant $R= 0.0821\frac{atm-L}{mol-K}$

the molar mass of helium $=4.003\frac{g}{mol}$

Now, for the mass of the Helium using the equation $\left(2\right)$

$mass=\frac{PV\times molar mass}{RT}$

Substituting the given values

$$\begin{gathered} mass=\frac{1atm\times 25L\times 4.003{\displaystyle \frac{g}{mol}}}{0.0821{\displaystyle \frac{atm-L}{mol-K}}\times 273K}=\frac{100.075}{22.41}g \\ mass=4.47g \end{gathered}$$

Part (b) Step 1: Given Information

We need to find the final pressure.

Part (b) Step 2: Simplify

For calculating the final pressure of Helium inside the balloon, we will use the combined gas law equation

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2} ... \left(1\right)$

Considering the given values

initial pressure $P_1=1atm=760mmHg$,

initial volume $V_1=25L$

initial temperature $T_1=273K$, and

the final volume $V_2=2460L$,

the final temperature $T_2=-35°C=\left(-35+273\right)K=238K$

Now, calculating the final pressure using the equation $\left(1\right)$

$P_2=\frac{P_1{V}_1{T}_2}{V_2{T}_1}$

Substituting the given values

$$\begin{gathered} P_2=\frac{760mmHg\times 25L\times 238K}{2460L\times 273K}=\frac{4,522,000}{371,580}mmHg \\ {P}_2=6.73mmHg \end{gathered}$$