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Chapter 9: Problem 1233

\(2 \mathrm{~g}\) of \(\mathrm{O}_{2}\) gas is taken at \(27^{\circ} \mathrm{C}\) and pressure \(76 \mathrm{~mm} \mathrm{Hg}\). Find out volume of gas (ln liter) (A) \(3.08\) (B) \(44.2\) (C) \(2.05\) (D) \(2.44\)

Short Answer

Expert verified
The volume of the gas is approximately 2.05 liters. The correct answer is (C).

Step by step solution

01

Convert the given mass of O₂ to moles

We are given the mass of O₂ (2 grams). To find the number of moles, we need to divide the mass by the molar mass of O₂. The molar mass of O₂ is 16 g/mol (for each oxygen atom) multiplied by 2 (since there are two oxygen atoms in a molecule of O₂), which is 32 g/mol: Number of moles (n) = (mass of O₂) / (molar mass of O₂) n = 2 g / 32 g/mol = 1/16 mol
02

Convert temperature to Kelvin

The given temperature (T) is 27°C. We need to convert this to Kelvin, because the ideal gas law requires the temperature to be in Kelvin. To do this, simply add 273.15: T = 27°C + 273.15 = 300.15 K
03

Convert pressure to atmospheres

We are given the pressure of the gas in mm Hg. To use the ideal gas law, we need to convert the pressure to atmospheres. Since there are 760 mm Hg in 1 atmosphere, we have: Pressure (P) = 76 mm Hg * (1 atm / 760 mm Hg) = 0.1 atm
04

Use the ideal gas law to find the volume

We now have all the necessary information to use the ideal gas law (PV=nRT). The only unknown is the volume (V). We can rearrange the equation to solve for V: V = nRT / P Now plug in the necessary values. The universal gas constant (R) is always 0.0821 L⋅atm/(mol⋅K): V = (1/16 mol)(0.0821 L⋅atm/(mol⋅K))(300.15 K) / (0.1 atm)
05

Calculate the volume of the gas

Now perform the computation: V = (1/16)(0.0821)(300.15) / (0.1) V ≈ 2.046 The volume of the gas is approximately 2.05 liters. The correct answer is (C).

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Most popular questions from this chapter

1 mole of gas occupies a volume of \(100 \mathrm{~m} 1\) at \(50 \mathrm{~mm}\) pressure. What is the volume occupied by two moles of gas at $100 \mathrm{~mm}$ pressure and at same temperature (A) \(50 \mathrm{ml}\) (B) \(200 \mathrm{ml}\) (C) \(100 \mathrm{ml}\) (D) \(500 \mathrm{~m} 1\)

If three molecules have velocities \(0.5,1\) and 2 the ratio of rms speed and average speed is (The velocities are in \(\mathrm{km} / \mathrm{s}\) ) (A) \(0.134\) (B) \(1.34\) (C) \(1.134\) (D) \(13.4\)

A cylinder contains \(10 \mathrm{~kg}\) of gas at pressure of $10^{\prime} \mathrm{N} / \mathrm{m}^{2}$. The quantity of gas taken out of the cylinder, if final pressure is \(2.5 \times 10^{6} \mathrm{Nm}^{-2}\). will be (temperature of gas is constant) (A) \(5.2 \mathrm{~kg}\) (B) \(3.7 \mathrm{~kg}\) (C) \(7.5 \mathrm{~kg}\) (D) \(1 \mathrm{~kg}\)

What is the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2 atm and temperature $17^{\circ} \mathrm{C} ?$ Take the radius of nitrogen molecule to be 1A. Molecular mass of nitrogen \(=28\), $\mathrm{k}_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{JK}^{-1}, 1 \mathrm{~atm}=1.013 \times 10^{5} \mathrm{Nm}^{-2}$ (A) \(2.2 \times 10^{-7} \mathrm{~m}, 2.58 \times 10^{9}\) (B) \(1.1 \times 10^{-7} \mathrm{~m}, 4.58 \times 10^{8}\) (C) \(1.1 \times 10^{-7} \mathrm{~m}, 4.58 \times 10^{9}\) (D) \(2.2 \times 10^{-7} \mathrm{~m}, 3.58 \times 10^{9}\)

At room temperature \(\left(27^{\circ} \mathrm{C}\right)\), the rms speed of the molecules of certain diatomic gas is found to be $1930 \mathrm{~m} / \mathrm{s}$. The gas is (A) \(\mathrm{O}_{2}\) (B) \(\mathrm{Cl}_{2}\) (C) \(\mathrm{H}_{2}\) (D) \(\mathrm{F}_{2}\)
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