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Chapter 13: Problem 38

At \(0.0^{\circ} \mathrm{C}\) and \(1.00 \mathrm{atm}, 1.00 \mathrm{mol}\) of a gas occupies a volume of \(0.0224 \mathrm{m}^{3} .\) (a) What is the number density? (b) Estimate the average distance between the molecules. (c) If the gas is nitrogen \(\left(\mathrm{N}_{2}\right),\) the principal component of air, what is the total mass and mass density?

Short Answer

Expert verified
Answer: The number density of nitrogen gas is approximately 44.64 mol/m鲁, the average distance between molecules is about 3.30 脳 10鈦烩伖 m, the total mass of the gas is 28 g or 0.028 kg, and the mass density is approximately 1.25 kg/m鲁.

Step by step solution

01

Finding number density

To find the number density, we can use the formula for number density: Number density = number of moles/volume We are given that we have 1.00 mol of gas occupying 0.0224 m鲁 volume. So, we can plug these values into the formula to find the number density: Number density = 1.00 mol/0.0224 m鲁
02

Calculating number density

Now, we can calculate the number density: Number density 鈮� 44.64 mol/m鲁 In part (a), we found the number density of the gas molecules.
03

Finding average distance between molecules

Now let's find the average distance between the molecules in the gas. We can use the formula for the average distance between molecules which is: Average distance = (volume of one molecule /number density)^(1/3) The volume of one molecule can be found by dividing the total volume by the number of gas molecules, which is equal to the number of moles times Avogadro's number (6.022 脳 10虏鲁 particles/mol): Volume of one molecule = (0.0224 m鲁) / (1.00 mol 脳 6.022 脳 10虏鲁 particles/mol)
04

Calculating average distance between molecules

Now, we can calculate the average distance using the values we have found: Average distance 鈮� ((0.0224 m鲁) / (1.00 mol 脳 6.022 脳 10虏鲁 particles/mol))^(1/3) Average distance 鈮� 3.30 脳 10鈦烩伖 m The estimated average distance between molecules is 3.30 脳 10鈦烩伖 m in part (b).
05

Calculate the mass of one molecule of \(\mathrm{N}_{2}\) gas

To find the total mass of 1.00 mol of nitrogen gas, we need to know the molecular weight of one molecule of N鈧�. Nitrogen has an atomic mass of approximately 14 amu, so a molecule of \(\mathrm{N}_{2}\) will have a weight of: Mass of one molecule of N鈧� = 2 脳 14 amu = 28 amu Now converting this mass to grams per mole: 28 amu 脳 (1 g/mol) / (1 amu) = 28 g/mol So, the mass of 1.00 mol of nitrogen gas will be 28 g.
06

Calculate the mass density of the gas

We found the total mass of the gas in Step 5, so we can now calculate the mass density using the formula: Mass density = mass/volume Mass density = (28 g) / (0.0224 m鲁) Now, we can convert grams to kilograms: 28 g 脳 (1 kg / 1000 g) = 0.028 kg
07

Calculate mass density using the mass in kilograms

Now, we can calculate the mass density in kg/m鲁: Mass density = (0.028 kg) / (0.0224 m鲁) Mass density 鈮� 1.25 kg/m鲁 In part (c), we found the total mass of the nitrogen gas to be 28 g or 0.028 kg, and mass density to be approximately 1.25 kg/m鲁.

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Most popular questions from this chapter

Show that, for an ideal gas, $$ P=\frac{1}{3} \rho v_{\mathrm{rms}}^{2} $$ where \(P\) is the pressure, \(\rho\) is the mass density, and \(v_{\mathrm{rms}}\) is the rms speed of the gas molecules.
As a Boeing 747 gains altitude, the passenger cabin is pressurized. However, the cabin is not pressurized fully to atmospheric $\left(1.01 \times 10^{5} \mathrm{Pa}\right),$ as it would be at sea level, but rather pressurized to \(7.62 \times 10^{4} \mathrm{Pa}\). Suppose a 747 takes off from sea level when the temperature in the airplane is \(25.0^{\circ} \mathrm{C}\) and the pressure is \(1.01 \times 10^{5} \mathrm{Pa} .\) (a) If the cabin temperature remains at \(25.0^{\circ} \mathrm{C},\) what is the percentage change in the number of moles of air in the cabin? (b) If instead, the number of moles of air in the cabin does not change, what would the temperature be?

What is the total internal kinetic energy of 1.0 mol of an ideal gas at \(0.0^{\circ} \mathrm{C}\) and 1.00 atm?
Agnes Pockels \((1862-1935)\) was able to determine Avogadro's number using only a few household chemicals, in particular oleic acid, whose formula is \(\mathrm{C}_{18} \mathrm{H}_{34} \mathrm{O}_{2}\) (a) What is the molar mass of this acid? (b) The mass of one drop of oleic acid is \(2.3 \times 10^{-5} \mathrm{g}\) and the volume is $2.6 \times 10^{-5} \mathrm{cm}^{3} .$ How many moles of oleic acid are there in one drop? (c) Now all Pockels needed was to find the number of molecules of oleic acid. Luckily, when oleic acid is spread out on water, it lines up in a layer one molecule thick. If the base of the molecule of oleic acid is a square of side \(d\), the height of the molecule is known to be \(7 d .\) Pockels spread out one drop of oleic acid on some water, and measured the area to be \(70.0 \mathrm{cm}^{2}\) Using the volume and the area of oleic acid, what is \(d ?\) (d) If we assume that this film is one molecule thick, how many molecules of oleic acid are there in the drop? (e) What value does this give you for Avogadro's number?
A temperature change \(\Delta T\) causes a volume change \(\Delta V\) but has no effect on the mass of an object. (a) Show that the change in density $\Delta \rho\( is given by \)\Delta \rho=-\beta \rho \Delta T$. (b) Find the fractional change in density \((\Delta \rho / \rho)\) of a brass sphere when the temperature changes from \(32^{\circ} \mathrm{C}\) to \(-10.0^{\circ} \mathrm{C}\)
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