/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 (a) Calculate the mass of nitrog... [FREE SOLUTION] | 91Ó°ÊÓ

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(a) Calculate the mass of nitrogen present in a volume of \(3000 \mathrm{~cm}^{3}\) if the gas is at \(22.0^{\circ} \mathrm{C}\) and the absolute pressure of \(2.00 \times 10^{-13}\) atm is a partial vacuum easily obtained in laboratories. (b) What is the density (in \(\mathrm{kg} / \mathrm{m}^{3}\) ) of the \(\mathrm{N}_{2}\) ?

Short Answer

Expert verified
The mass of Nitrogen is calculated using n * molar mass, and the density of \( \mathrm{N}_2 \) is calculated using \( \rho = \frac{Mass}{V} \) .

Step by step solution

01

Applying the Ideal Gas Law

Using the Ideal Gas Law, \( PV = nRT \), where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. Firstly, convert the volume from cm^3 to m^3, temperature from Celsius to Kelvin and pressure from atm to pascal: \( V = 3000 \times 10^{-6} \, m^3 \), \( T = 22.0 + 273.15 \, K = 295.15 \, K \), and \( P = 2.00 \times 10^{-13} \times 1.013 \times 10^{5} \, Pa \). Solve the equation for n: \( n = PV/RT \) \(
02

Calculate the mass of Nitrogen

Knowing that one mole of any gas contains Avogadro's number (\(6.022 \times 10^{23}\)) of molecules, and the molar mass of nitrogen (\(\mathrm{N}_2\)) is 28.0134 g / mole, convert the number of moles to mass: Mass = n * molar mass = \( n \times 28.0134 \times 10^{-3} \, kg \).
03

Calculate the density of Nitrogen

Finally, use the mass and the volume to calculate the density, \( \rho = \frac{Mass}{V} \) .

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Key Concepts

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

Mass of Nitrogen
The mass of nitrogen in a given volume of gas can be calculated using the Ideal Gas Law and some basic conversions. The Ideal Gas Law equation is given by \( PV = nRT \), where:
  • \( P \) is the pressure,
  • \( V \) is the volume,
  • \( n \) is the number of moles,
  • \( R \) is the gas constant,
  • \( T \) is the temperature in Kelvin.
Let's break down how to calculate the mass of nitrogen.First, it's important to convert the provided units to the standard units used in the Ideal Gas Law:
  • Convert the volume from \( \mathrm{cm}^3 \) to \( \mathrm{m}^3 \) by multiplying by \(10^{-6}\). So, \( 3000 \, \mathrm{cm}^3 = 3000 \times 10^{-6} \, \mathrm{m}^3 \).
  • Convert the temperature from degrees Celsius to Kelvin by adding 273.15. Thus, \( 22.0^\circ \mathrm{C} = 295.15 \, \mathrm{K} \).
  • Convert the pressure from atm to pascal by multiplying by \( 1.013 \times 10^5 \). Therefore, \(2.00 \times 10^{-13} \, \mathrm{atm} = 2.00 \times 10^{-13} \times 1.013 \times 10^5 \, \mathrm{Pa} \).
Now, solve for \( n \) using the rearranged Ideal Gas Law: \( n = \frac{PV}{RT} \). Once \( n \) is known, multiply it by the molar mass of nitrogen \( (28.0134 \, \mathrm{g/mol} = 28.0134 \times 10^{-3} \, \mathrm{kg/mol}) \) to find the mass.This approach will yield the mass of nitrogen in the given conditions.
Density Calculation
Once we determine the mass of a gas, calculating its density is straightforward. Density (\( \rho \)) is a measure of how much mass is contained within a specific volume and is defined by the equation:\[ \rho = \frac{\text{Mass}}{\text{Volume}} \]For our specific example with nitrogen, now that we have the mass calculated using the Ideal Gas Law, we can plug it into our density formula.Considerations for calculating density:
  • Make sure that the mass is in kilograms \((\mathrm{kg})\), which is derived from the number of moles and the molar mass.
  • Ensure the volume is in cubic meters \((\mathrm{m}^3)\), which was previously converted from cubic centimeters in our earlier calculation.
Simply substitute the values into the density formula to obtain the density of nitrogen in \( \mathrm{kg/m}^3 \). This measurement gives a clear idea of how concentrated the nitrogen molecules are within the given volume.
Conversion of Units
Converting units is a crucial step when working with physics equations like the Ideal Gas Law. Using consistent units ensure accurate calculations. Here's how unit conversion applies in our scenario:**Volume Conversion**The volume was initially given in cubic centimeters \((\mathrm{cm}^3)\). To convert to cubic meters \((\mathrm{m}^3)\), multiply by \(10^{-6}\). This is because there are \(100\) centimeters in a meter, so \((\mathrm{cm})^3\) conversion to \((\mathrm{m})^3\) involves \((10^{-2})^3 = 10^{-6}\).**Temperature Conversion**Temperature provided in Celsius \((^\circ \mathrm{C})\) is converted to Kelvin (\(\mathrm{K}\)) by adding 273.15. This is because the Kelvin scale is an absolute temperature scale and reduces temperature-based discrepancies in calculations.**Pressure Conversion**Pressure in atmospheres \((\mathrm{atm})\) is converted to Pascals \((\mathrm{Pa})\) by multiplying by \(1.013 \times 10^5\). The pascal is the SI unit for pressure, ensuring we maintain uniformity with other SI units like meters and kilograms.Unit conversions are not merely procedural steps; they are vital for ensuring the correct application of formulas and achieving reliable results. Always ensure to cross-check your units for consistency across all measurement parameters.

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Most popular questions from this chapter

A parcel of air over a campfire feels an upward buoyant force because the heated air is less dense than the surrounding air. By estimating the acceleration of the air immediately above a fire, one can estimate the fire's temperature. The mass of a volume \(V\) of air is \(n M_{\text {air }},\) where \(n\) is the number of moles of air molecules in the volume and \(M_{\text {air }}\) is the molar mass of air. The net upward force on a parcel of air above a fire is roughly given by \(\left(m_{\text {out }}-m_{\text {in }}\right) g,\) where \(m_{\text {out }}\) is the mass of a volume of ambient air and \(m_{\text {in }}\) is the mass of a similar volume of air in the hot zone. (a) Use the ideal-gas law, along with the knowledge that the pressure of the air above the fire is the same as that of the ambient air, to derive an expression for the acceleration \(a\) of an air parcel as a function of \(\left(T_{\text {out }} / T_{\text {in }}\right),\) where \(T_{\text {in }}\) is the absolute temperature of the air above the fire and \(T_{\text {out }}\) is the absolute temperature of the ambient air. (b) Rearrange your formula from part (a) to obtain an expression for \(T_{\text {in }}\) as a function of \(T_{\text {out }}\) and \(a\). (c) Based on your experience with campfires, estimate the acceleration of the air above the fire by comparing in your mind the upward trajectory of sparks with the acceleration of falling objects. Thus you can estimate \(a\) as a multiple of \(g .\) (d) Assuming an ambient temperature of \(15^{\circ} \mathrm{C}\), use your formula and your estimate of \(a\) to estimate the temperature of the fire.

A large tank of water has a hose connected to it (Fig. P18.61). The tank is sealed at the top and has compressed air between the water surface and the top. When the water height \(h\) has the value \(3.50 \mathrm{~m}\), the absolute pressure \(p\) of the compressed air is \(4.20 \times 10^{5} \mathrm{~Pa}\). Assume that the air above the water expands at constant temperature, and take the atmospheric pressure to be \(1.00 \times 10^{5} \mathrm{~Pa}\). (a) What is the speed with which water flows out of the hose when \(h=3.50 \mathrm{~m} ?\) (b) As water flows out of the tank, \(h\) decreases. Calculate the speed of flow for \(h=3.00 \mathrm{~m}\) and for \(h=2.00 \mathrm{~m} .\) (c) At what value of \(h\) does the flow stop?

A person at rest inhales \(0.50 \mathrm{~L}\) of air with each breath at a pressure of 1.00 atm and a temperature of \(20.0^{\circ} \mathrm{C}\). The inhaled air is \(21.0 \%\) oxygen. (a) How many oxygen molecules does this person inhale with each breath? (b) Suppose this person is now resting at an elevation of \(2000 \mathrm{~m}\) but the temperature is still \(20.0^{\circ} \mathrm{C}\). Assuming that the oxygen percentage and volume per inhalation are the same as stated above, how many oxygen molecules does this person now inhale with each breath? (c) Given that the body still requires the same number of oxygen molecules per second as at sea level to maintain its functions, explain why some people report "shortness of breath" at high elevations.

A hot-air balloon stays aloft because hot air at atmospheric pressure is less dense than cooler air at the same pressure. If the volume of the balloon is \(500.0 \mathrm{~m}^{3}\) and the surrounding air is at \(15.0^{\circ} \mathrm{C}\) what must the temperature of the air in the balloon be for it to lift a total load of \(290 \mathrm{~kg}\) (in addition to the mass of the hot air)? The density of air at \(15.0^{\circ} \mathrm{C}\) and atmospheric pressure is \(1.23 \mathrm{~kg} / \mathrm{m}^{3}\).

A Jaguar XK8 convertible has an eight-cylinder engine. At the beginning of its compression stroke, one of the cylinders contains \(499 \mathrm{~cm}^{3}\) of air at atmospheric pressure \(\left(1.01 \times 10^{5} \mathrm{~Pa}\right)\) and a temperature of \(27.0^{\circ} \mathrm{C}\). At the end of the stroke, the air has been compressed to a volume of \(46.2 \mathrm{~cm}^{3}\) and the gauge pressure has increased to \(2.72 \times 10^{6} \mathrm{~Pa}\). Compute the final temperature.

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