Question

The partial pressure of ${\mathrm{CH}}_4(g)$ is 0.175 atm and that of ${\mathrm{O}}_2(g)$ is 0.250 $\mathrm{atm}$ in a mixture of the two gases. a. What is the mole fraction of each gas in the mixture? b. If the mixture occupies a volume of 10.5 $\mathrm{L}$ at ${65}^{\circ }\mathrm{C}$ , calculate the total number of moles of gas in the mixture. c. Calculate the number of grams of each gas in the mixture.

Short answer

a. Mole fraction of CH4: $x_{CH4}=\frac{0.175}{0.175+0.250}=0.412$ Mole fraction of O2: $x_{O2}=\frac{0.250}{0.175+0.250}=0.588$ b. Total number of moles (n) = $\frac{(0.175+0.250)(10.5)}{(0.0821)(65+273.15)}$ = 0.406 mol c. Mass of CH4: $(0.412)(0.406)(16.04)$ = 2.68 g Mass of O2: $(0.588)(0.406)(32.00)$ = 7.59 g

Step-by-step solution

a. Mole fraction of each gas

To calculate the mole fractions of both gases, we can use the relation between partial pressure and mole fractions. For methane (CH4): Mole fraction (x) = Partial pressure of CH4 (P_CH4) / Total pressure (P_total) For oxygen (O2): Mole fraction (x) = Partial pressure of O2 (P_O2) / Total pressure (P_total) First, we need to find the total pressure (P_total) which is the sum of partial pressures of CH4 and O2: P_total = P_CH4 + P_O2 Now, calculate the mole fraction of each gas: Mole fraction of CH4 (x_CH4) = P_CH4 / P_total Mole fraction of O2 (x_O2) = P_O2 / P_total

b. Total number of moles

To find the total number of moles of gas in the mixture, we will use the Ideal Gas Law equation: PV = nRT where P is the total pressure of the mixture, V is the volume of the mixture, n is the total number of moles, R is the gas constant ($0.0821\frac{L\cdot atm}{mol\cdot K}$), and T is the temperature in Kelvin. First, we need to convert the temperature from Celsius to Kelvin: T(K) = 65 + 273.15 Now, solve for total moles (n): n = PV / RT

c. Number of grams of each gas

To find the number of grams of CH4 and O2, we need to multiply the mole fraction of each gas by the total number of moles, and then multiply it by the molar mass of each gas. For CH4, the molar mass is 16.04 g/mol. For O2, the molar mass is 32.00 g/mol. Mass of CH4 (m_CH4) = (mole fraction of CH4 x total moles) x molar mass of CH4 Mass of O2 (m_O2) = (mole fraction of O2 x total moles) x molar mass of O2

Key concepts

Partial Pressure

In a gas mixture, each gas exerts its own pressure, known as partial pressure. This is an important concept in understanding how different gases behave when combined. The total pressure of the mixture is the sum of these individual partial pressures.
To calculate the partial pressure, one can use Dalton's Law of Partial Pressures, which states that the total pressure of a gas mixture is equal to the sum of the partial pressures of each component gas.
For example, in our exercise, we find that the partial pressure for methane (${\text{CH}}_4$) is 0.175 atm and that for oxygen (${\text{O}}_2$) is 0.250 atm.
By simply adding these, we get the total pressure of the mixture:

• $P_{\text{total}}=P_{{\text{CH}}_4}+P_{{\text{O}}_2}$
• $P_{\text{total}}=0.175+0.250=0.425\text{atm}$

Partial pressures are essential in applications such as breathing mixtures for divers and atmospheric studies.

Mole Fraction

The mole fraction is a way of expressing the concentration of a particular component in a mixture. It is calculated by dividing the partial pressure of each gas by the total pressure.
In our example, the mole fraction of methane (${\text{CH}}_4$) can be determined by:

• $x_{{\text{CH}}_4}=\frac{P_{{\text{CH}}_4}}{P_{\text{total}}}$
• $x_{{\text{CH}}_4}=\frac{0.175}{0.425}\approx 0.412$

Similarly, the mole fraction for oxygen (${\text{O}}_2$) is calculated by:

• $x_{{\text{O}}_2}=\frac{P_{{\text{O}}_2}}{P_{\text{total}}}$
• $x_{{\text{O}}_2}=\frac{0.250}{0.425}\approx 0.588$

These values indicate the proportion of each gas in the mixture and are fundamental in quantifying how each gas contributes to the total behavior of the mixture.

Molar Mass

The molar mass is crucial for converting between the number of moles and the mass of a substance. It is the mass of one mole of a substance, expressed in grams per mole.
Using the molar mass, we can convert moles of a substance to grams and vice versa. In our example, the molar mass of methane (${\text{CH}}_4$) is 16.04 g/mol, and for oxygen (${\text{O}}_2$), it's 32.00 g/mol.

To find the mass of each gas in the mixture, we use the formula:

• $\text{Mass of gas}=(\text{Mole fraction of gas}\times \text{Total moles})\times \text{Molar mass of gas}$

This calculation allows us to determine the actual mass of each gas present in a given volume or system. The molar mass is foundational in stoichiometry and helps relate chemical reactions to real-world applications.

Temperature Conversion

Temperature conversion is essential when dealing with gas laws. Temperatures in the gas laws must be in Kelvin because Kelvin is the absolute temperature scale, which allows the equations to work properly.
To convert from Celsius to Kelvin, you add 273.15 to the Celsius temperature:

• $T(K)=T(°C)+273.15$

In our exercise, the temperature given is 65°C, so converting to Kelvin gives:

• $T(K)=65+273.15=338.15K$

Using the Kelvin scale ensures that we have a positive temperature value to use in the Ideal Gas Law, $PV=nRT$.
Converting temperature accurately is critical for calculations involving gases, as it directly affects the pressure and volume relationships described by various gas laws.