Problem 16
Question
Pressure of a mixture of \(4 \mathrm{~g}\) of \(\mathrm{O}_{2}\) and \(2 \mathrm{~g}\) of \(\mathrm{H}_{2}\) confined in a bulb of \(1.0 \mathrm{~L}\) capacity at \(0^{\circ} \mathrm{C}\) is (a) \(25.18 \mathrm{~atm}\) (b) \(31.205 \mathrm{~atm}\) (c) \(40.215 \mathrm{~atm}\) (d) \(15.210 \mathrm{~atm}\)
Step-by-Step Solution
Verified Answer
The pressure of the mixture is 25.18 atm, so the answer is (a).
1Step 1: Calculate the Number of Moles of O2
First, calculate the moles of \( \mathrm{O}_2 \). The molar mass of \( \mathrm{O}_2 \) is 32 g/mol. Therefore, the number of moles of \( \mathrm{O}_2 \) can be calculated as:\[\text{{moles of }} \mathrm{O}_2 = \frac{\text{{mass of }} \mathrm{O}_2}{\text{{molar mass of }} \mathrm{O}_2} = \frac{4\, \text{g}}{32\, \text{g/mol}} = 0.125\, \text{mol}\]
2Step 2: Calculate the Number of Moles of H2
Next, calculate the moles of \( \mathrm{H}_2 \). The molar mass of \( \mathrm{H}_2 \) is 2 g/mol. Calculate the number of moles as follows:\[\text{{moles of }} \mathrm{H}_2 = \frac{\text{{mass of }} \mathrm{H}_2}{\text{{molar mass of }} \mathrm{H}_2} = \frac{2\, \text{g}}{2\, \text{g/mol}} = 1\, \text{mol}\]
3Step 3: Calculate the Total Number of Moles
Add the number of moles of \( \mathrm{O}_2 \) and \( \mathrm{H}_2 \) to find the total number of moles:\[\text{Total moles} = 0.125 + 1 = 1.125\, \text{mol}\]
4Step 4: Use the Ideal Gas Law to Find Pressure
Use the ideal gas law to calculate the pressure. The ideal gas law is \( PV = nRT \).- \(P\) is the pressure in atmospheres, which we need to find.- \(V\) is the volume, given as 1.0 L.- \(n\) is the total moles, which is 1.125 mol.- \(R\) is the gas constant, \(0.0821 \frac{\text{L} \cdot \text{atm}}{\text{mol} \cdot \text{K}}\).- \(T\) is the temperature in Kelvin. At \(0^{\circ}\text{C}\), \(T = 273.15\, \text{K}\).Rearranging the ideal gas law gives:\[P = \frac{nRT}{V} = \frac{1.125 \cdot 0.0821 \cdot 273.15}{1.0} = 25.18\, \text{atm}\]
5Step 5: Match the Answer
Compare the calculated pressure with the provided options. The calculated pressure is \(25.18\, \text{atm}\), which matches option (a).
Key Concepts
Moles CalculationMolar MassGas ConstantTemperature Conversion
Moles Calculation
To find the number of moles in a given substance, you'll need to use its mass and its molar mass. The formula to determine the number of moles is:
- \( ext{Moles} = \frac{\text{Mass (in grams)}}{\text{Molar mass (g/mol)}} \)
Molar Mass
The molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It is determined by summing the atomic masses of the elements in a compound, considering their respective quantities. For instance:
- The molar mass of \( \text{O}_2 \) is 32 g/mol. This is because oxygen has an atomic mass of approximately 16, and \( \text{O}_2 \) has two oxygen atoms.
- For \( \text{H}_2 \), the molar mass is 2 g/mol since hydrogen has an atomic mass of about 1, and \( \text{H}_2 \) has two atoms of hydrogen.
Gas Constant
The gas constant \( R \) is crucial in calculations involving gas laws. It bridges the relationship between moles, volume, temperature, and pressure in the ideal gas law. The value of the gas constant \( R \) is:
- \( R = 0.0821 \frac{\text{L} \cdot \text{atm}}{\text{mol} \cdot \text{K}} \)
- \( P \) is the pressure in atmospheres,
- \( V \) is the volume in liters,
- \( n \) is the number of moles,
- \( T \) is the temperature in Kelvin.
Temperature Conversion
To work with gas laws, converting temperature to Kelvin from Celsius is necessary because the Kelvin scale is absolute. The Kelvin scale is directly related to the energy of the particles; it starts at absolute zero, the point where particle motion stops. The conversion from Celsius to Kelvin is straightforward:
- \( T_{\text{Kelvin}} = T_{\text{Celsius}} + 273.15 \)
Other exercises in this chapter
Problem 14
By the ideal gas law the pressure of \(0.60 \mathrm{~mol} \mathrm{NH}_{3}\) gas in a \(3.00\) litre vessel at \(25^{\circ} \mathrm{C}\) is (a) \(48.9 \mathrm{~a
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At what Centigrade temperature will be the volume of a gas at \(0^{\circ} \mathrm{C}\) double of itself, when pressure remains constant? (a) \(0^{\circ} \mathrm
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A closed container contains equal number of oxygen and hydrogen molecules at a total pressure of \(740 \mathrm{~mm}\). If oxygen is removed from the system then
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What are the conditions under which the relation between volume (V) and number of moles (n) of gas is plotted? \((\mathrm{P}=\) pressure; \(\mathrm{T}=\) temper
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