Chapter 18, Problem 1DE

Interpretation Introduction

To determine: The mass of helium in grams inside the blimp.

Answer to Problem 1DE

Solution: The mass of helium inside the blimp is $9.39\times {10}^5\text{g}$ and it corresponds to option (e).

Explanation of Solution

Given

The volume of helium inside the blimp is $5.74\times {10}^6\text{L}$ .
The temperature of helium is $25°\text{C}$ .
The absolute pressure of helium is $1.00\text{atm}$ .
The conversion of $°\text{C}$ to Kelvin is done as,

$T°\text{C}=T+273\text{K}$

Hence, the conversion of $25°\text{C}$ into Kelvin is ,

$\begin{array}{c}25°\text{C}=25+273\text{K}\\ =298\text{K}\end{array}$

The relation between number of moles of a gas at the given values of $T,P\text{and}V$ for an ideal gas is given by the formula,

$\begin{array}{c}PV=nRT\\ n=\frac{PV}{RT}\end{array}$

Where,

• $n$ is the number of moles of the gas.
• $P$ is the absolute pressure of the gas.
• $T$ is the absolute temperature of the gas.
• $R$ is the gas constant $\left(0.08206\text{L-atm/mol-K}\right)$ .
• $V$ is the volume of the gas.

Now, substitute the values of $T,P,R\text{and}V$ in above equation.

$\begin{array}{c}n=\frac{1.00\text{atm}\times 5.74\times {10}^6\text{L}}{0.08206\text{L-atm/mol-K}\times 298\text{K}}\\ =2.347\times {10}^5\text{mol}\end{array}$

The atomic mass of helium is $4.0\text{g/mol}$ .

The mass of the helium is calculated using the formula,

$\text{Mass}\text{of}\text{He}=\text{Moles}\text{of}\text{He}\times \text{Atomic}\text{mass}\text{of}\text{He}$

Substitute the value of moles and atomic mass of $\text{He}$ in the above equation.

$\begin{array}{c}\text{Mass}\text{of}\text{He}=2.347\times {10}^5\text{mol}\times 4.00\text{g/mol}\\ =\text{9}\text{.39}\times {\text{10}}^5\text{g}\end{array}$

Hence, the mass of helium in grams inside the blimp is $9.39\times {10}^5\text{g}$ .

Conclusion

The mass of helium inside the blimp is $9.39\times {10}^5\text{g}$ .