Chapter 5, Problem 1QAP

Interpretation Introduction

Interpretation: The volume of the tank in liters, amount of methane gas in grams, and the temperature of the tank in Kelvin should be determined.

Concept introduction: An arithmetical multiplier which is used for converting a quantity expressed in one unit into another equivalent set of units is said to be conversion factor.

The relation between mass, molar mass and number of moles is represented as follows:

  $PV=nRT$

Here, P is pressure, V is volume, n is number of moles, R is Universal gas constant and T is temperature.

Answer to Problem 1QAP

Volume of the tank is 37.9 L

Mass of the gas is 19.89 g

The temperature in Kelvin is 296.48 K.

Explanation of Solution

Volume of a gas in a container is equal to the volume of the container. Conversion factor for gallon to liters is as follows

  $1\text{ gallon}=3.79\text{ L}$

Convert 10 gallons into liters using the conversion factor.

  $\begin{array}{l}10\text{ gallon =10 gallon}\times \frac{\text{3}\text{.79 L}}{\text{1 gallon}}\\ \text{ =37}\text{.9 L}\end{array}$

Therefore, volume of the tank is 37.9 L

Mass of a gas is equal to the product of its number of moles and molar mass.

Molar mass of methane, CH₄ is 16 g/mol.

  $\begin{array}{l}\text{Mass of a gas = moles of the gas }\times \text{ molar mass of the gas}\\ \text{ =1}\text{.243 mol}\times \frac{16g}{1\text{ mol}}\\ \text{ =19}\text{.89 g}\end{array}$

Therefore, mass of the methane gas is 19.89 g.

Expression for the conversion of temperature from $°F$ to K is as below:

  $\text{Kelvin}=\left(°F-32\right)\times \frac{5}{9}+273.15$

Plug in the given details in the expression for conversion of temperature,

  $\begin{array}{l}=(74-32)\times \frac{5}{9}+273.15\\ =296.48K\end{array}$

Hence, the temperature in Kelvin is 296.48 K.

Conclusion

Thus, the volume of the tank is 37.9 L

Mass of the gas is 19.89 g

The temperature in Kelvin is 296.48 K.