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A gaseous mixture contains \(56 \mathrm{~g}\) of \(\mathrm{N}_{2}, 44 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(16 \mathrm{~g}\) of \(\mathrm{CH}_{4}\). The total pressure of mixture is \(720 \mathrm{~mm}\) of \(\mathrm{Hg}\). The partial pressure of methane is (a) \(75 \mathrm{~mm}\) (b) \(160 \mathrm{~mm}\) (c) \(180 \mathrm{~mm}\) (d) \(215 \mathrm{~mm}\)

Short Answer

Expert verified
The partial pressure of methane is 180 mm Hg (option c).

Step by step solution

01

Calculate Moles of Each Gas

First, we calculate the number of moles of each gas using the formula: \( ext{moles} = \frac{ ext{mass}}{ ext{molar mass}} \).- For \( \mathrm{N}_2 \): Molar mass of \( \mathrm{N}_2 \) is \( 28 \mathrm{~g/mol} \). Thus, the moles are \( \frac{56}{28} = 2 \) moles.- For \( \mathrm{CO}_2 \): Molar mass of \( \mathrm{CO}_2 \) is \( 44 \mathrm{~g/mol} \). Thus, the moles are \( \frac{44}{44} = 1 \) mole.- For \( \mathrm{CH}_4 \): Molar mass of \( \mathrm{CH}_4 \) is \( 16 \mathrm{~g/mol} \). Thus, the moles are \( \frac{16}{16} = 1 \) mole.
02

Determine Total Moles of Gas

Add the moles of all the gases together to get the total moles of gas in the mixture:- Total moles = \( 2 + 1 + 1 = 4 \text{ moles} \).
03

Find Mole Fraction of Methane

The mole fraction of a gas is given by the formula:\[ \text{Mole fraction of } \mathrm{CH_4} = \frac{\text{moles of } \mathrm{CH_4}}{\text{total moles}} = \frac{1}{4} = 0.25 \].
04

Calculate Partial Pressure of Methane

Use the formula for partial pressure: \( P_i = x_i \times P_{\text{total}} \), where \( P_i \) is the partial pressure and \( x_i \) is the mole fraction.Plug in the values: \[ P_{\mathrm{CH_4}} = 0.25 \times 720 = 180 \text{ mm Hg} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mole Calculation
When dealing with chemical reactions and mixtures, calculating moles is a fundamental skill. Moles provide a way to count and manage chemical substances effectively. In essence, one mole equals Avogadro's number (\(6.022 \times 10^{23}\) particles) of any given molecule or atom. This concept is crucial, as chemical reactions often occur at the molecular level. For any substance, the number of moles is calculated using the formula:

\[ \text{Moles} = \frac{\text{Mass of substance in grams}}{\text{Molar mass of the substance (g/mol)}} \]

For example, if you have 56 grams of \(\mathrm{N}_2\), given that the molar mass of \(\mathrm{N}_2\) is 28 g/mol, the number of moles is calculated as:

  • \(\mathrm{N}_2\): \(\frac{56}{28} = 2\) moles
  • \(\mathrm{CO}_2\): \(\frac{44}{44} = 1\) mole
  • \(\mathrm{CH}_4\): \(\frac{16}{16} = 1\) mole
This calculation allows us to understand the proportion of each component involved.
Gaseous Mixtures
Gaseous mixtures consist of two or more different gases mixed together but not chemically combined. Each gas in the mixture behaves independently and contributes to the total pressure of the system.

In our example, a gaseous mixture contains \(\mathrm{N}_2\), \(\mathrm{CO}_2\), and \(\mathrm{CH}_4\). Determining the behavior or properties of this mixture involves understanding the individual properties of each gas and how they relate to each other. The total pressure of a gaseous mixture is determined by the sum of the partial pressures of the constituent gases. The Dalton's Law of Partial Pressures helps us in determining the pressure exerted by one individual gas.

The steps involved in analyzing gaseous mixtures often start from calculating the number of moles, the total pressure, to ultimately finding the partial pressures. Let's emphasize that in a mixture, pressures add up independently:

  • The pressure of a single gas is considered its partial pressure.
  • Using the mole fraction, we can predict each gas's contribution to the total pressure.
This concept forms the basis for various calculations and predictions regarding the behavior of gaseous mixtures.
Mole Fraction
The mole fraction is a way of expressing the concentration of a component in a mixture. It's particularly useful when dealing with gases, as it's related directly to their partial pressures and thus to their individual behaviors in a mixture.

**Understanding Mole Fraction:**
  • It indicates the ratio of the number of moles of a particular gas to the total number of moles in the mixture.
  • The sum of all mole fractions in a mixture equals 1.
  • For the gas \(\mathrm{CH}_4\) in our exercise, the mole fraction is calculated as:
    \[\text{Mole fraction of } \mathrm{CH_4} = \frac{\text{Moles of } \mathrm{CH_4}}{\text{Total moles in the mixture}} = \frac{1}{4} = 0.25 \]
The specific mole fraction not only tells us about the proportion of \(\mathrm{CH}_4\) but helps calculate its partial pressure using the formula:
\[ P_i = x_i \times P_{\text{total}} \]
This principle helps in many real-world applications, from understanding environmental gas emissions to designing industrial chemical processes.

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Most popular questions from this chapter

Which of the following statement(s) is/are incorrect? (a) A gas can be liquefied at a temperature ' \(\mathrm{T}\) ' such that \(\mathrm{T}<\mathrm{T}_{c}\) and \(\mathrm{p}=\mathrm{P}_{\mathrm{C}}-\mathrm{T}_{\mathrm{c}}\) and \(\mathrm{P}_{\mathrm{c}}\) are critical tem- perature and pressure. (b) Rise in the compressibility factor with increasing pressure is due to equal contribution of both a and b (Van der Waal's parameter). (c) The fraction of molecules having speeds in the range of \(\mathrm{u}\) to \(\mathrm{u}+\) du of a gas of molar mass ' \(\mathrm{M}\) ' at temperature ' \(\mathrm{T}\) ' is the same as that of gas of molar mass ' \(2 \mathrm{M}^{\prime}\) at temperature ' \(\mathrm{T} / 2^{\prime}\) (d) The product of pressure and volume of a fixed amount of a gas is independent of temperature.

The values of van der Waals constant ' \(\alpha\) ' for the gases \(\mathrm{O}_{2}, \mathrm{~N}_{2}, \mathrm{NH}_{3}\) and \(\mathrm{CH}_{4}\) are \(1.360,1.390,4.170\) and \(2.253\) L atm. mol \(^{2}\) respectively. The gas which can most easily be liquefied is (a) \(\mathrm{O}_{2}\) (b) \(\mathrm{N}_{2}\) (c) \(\mathrm{NH}_{3}\) (d) \(\mathrm{CH}_{4}\)

If two moles of ideal gas at \(540 \mathrm{~K}\) has volume \(44.8 \mathrm{~L}\), then its pressure will be (a) \(1 \mathrm{~atm}\) (b) \(2 \mathrm{~atm}\) (c) \(3 \mathrm{~atm}\) (d) \(4 \mathrm{~atm}\)

The term that accounts for intermolecular force in van der Waals equation for non ideal gas is (a) \(\mathrm{RT}\) (b) \(\mathrm{V}-\mathrm{b}\) (c) \(\left(\mathrm{P}+\alpha \mathrm{V}^{2}\right)\) (d) \((\mathrm{RT})^{-1}\)

Which of the following expressions correctly represents the relationship between the average molar kinetic energy, K.E. of \(\mathrm{CO}\) and \(\mathrm{N}_{2}\) molecules at the same temperature? (a) \(\overline{\mathrm{K} . \mathrm{E}_{\mathrm{c}} \mathrm{CO}}<\overline{\mathrm{K} \cdot \mathrm{E}}_{\mathrm{N}_{2}}\) (b) \(\overline{\mathrm{K} . \mathrm{E}}_{\mathrm{CO}}>\overline{\mathrm{K}} \cdot \mathrm{E}_{\mathrm{N}_{2}}\) (c) \(\mathrm{K} \cdot \mathrm{E}_{\mathrm{C} O}=\mathrm{K} \cdot \mathrm{E}_{\mathrm{N}_{2}}\) (d) cannot be predicted unless volumes of the gases are given.

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