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Chapter 16: Problem 9

Determine the volume occupied by \(4.0 \mathrm{~g}\) of oxygen \((M=32 \mathrm{~kg} / \mathrm{kmol})\) at S.T.P. Method 1 Use the Ideal Gas Law directly: $$ \begin{array}{c} P V=\left(\frac{m}{M}\right) R T \\ V=\left(\frac{1}{P}\right)\left(\frac{m}{M}\right) R T=\frac{\left(4.0 \times 10^{-3} \mathrm{~kg}\right)(8314 \mathrm{~J} / \mathrm{kmol} \cdot \mathrm{K})(273 \mathrm{~K})}{\left(1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)(32 \mathrm{~kg} / \mathrm{kmol})}=2.8 \times 10^{-3} \mathrm{~m}^{3} \end{array} $$ Method 2 Under S.T.P., 1 kmol occupies \(22.4 \mathrm{~m}^{3}\). Therefore, 32 kg occupies \(22.4 \mathrm{~m}^{3}\), and so \(4 \mathrm{~g}\) occupies $$ \left(\frac{4.0 \mathrm{~g}}{32000 \mathrm{~g}}\right)\left(22.4 \mathrm{~m}^{3}\right)=2.8 \times 10^{-3} \mathrm{~m}^{3} $$

Short Answer

Expert verified
The volume is \(2.8 \times 10^{-3} \text{ m}^3\).

Step by step solution

01

Apply the Ideal Gas Law

The Ideal Gas Law is given by the equation: \[ PV = \left(\frac{m}{M}\right) RT \]Where \( P \) is the pressure, \( V \) is the volume, \( m \) is the mass of the gas, \( M \) is the molar mass, \( R \) is the universal gas constant, and \( T \) is the temperature.
02

Substitute Known Values into the Ideal Gas Law

Given: mass \( m = 4.0 \text{ g} = 4.0 \times 10^{-3} \text{ kg} \), molar mass \( M = 32 \text{ kg/kmol} \), \( R = 8314 \text{ J/(kmol} \cdot \text{K)} \), \( T = 273 \text{ K} \), and pressure \( P = 1.01 \times 10^5 \text{ N/m}^2 \) at STP.Substitute these values into the Ideal Gas Law to solve for \( V \):\[ V = \frac{\left(4.0 \times 10^{-3} \right)(8314)(273)}{(1.01 \times 10^{5})(32)} \]
03

Calculate Volume Using Ideal Gas Law

Perform the calculation:\[ V = \frac{(4.0 \times 10^{-3})(8314)(273)}{(1.01 \times 10^{5})(32)} \approx 2.8 \times 10^{-3} \text{ m}^3 \] The volume of oxygen gas using the Ideal Gas Law is \( V = 2.8 \times 10^{-3} \text{ m}^3 \).
04

Use Standard Conditions at STP

Under standard conditions, 1 kmol of a gas occupies 22.4 \( \text{m}^3 \). Thus, 32 kg (1 kmol) of oxygen occupies \( 22.4 \text{ m}^3 \).
05

Calculate Volume Using STP Conversion

Since the given mass is 4 g, convert it to kmol first: \( \frac{4 \text{ g}}{32000 \text{ g/kmol}} = \frac{1}{8000} \text{ kmol} \).Multiply the equivalent volume occupied by 1 kmol (22.4 m鲁) to find the volume occupied by 4 g:\[ V = \left(\frac{4}{32000}\right) \times 22.4 \approx 2.8 \times 10^{-3} \text{ m}^3 \]
06

Confirm Volume Under Both Methods

Both calculation methods, using the Ideal Gas Law and using STP conversion ratios, give the same result:The volume of 4.0 g of oxygen at STP is \( 2.8 \times 10^{-3} \text{ m}^3 \).

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Key Concepts

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

Standard Temperature and Pressure (STP)
Standard Temperature and Pressure, abbreviated as STP, is a reference point used in chemistry to denote a set of conditions for experiments. It assumes a temperature of 273 K or 0掳C and a pressure of 1 atm (101325 Pa). These standard conditions are essential as they allow scientists to compare different experimental results or calculate gas behaviors under a consistent set of parameters. At STP, gases have predictable behaviors that make calculations straightforward, such as the fact that 1 mole of an ideal gas occupies 22.4 liters. This helps in simplifying the calculation of gas volumes and provides a baseline for energy content or reaction rates. Knowing STP conditions can help you predict how gases will behave in real-world situations.
molar mass
Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It relates to the molecular weight of a compound, reflecting the mass of the constituent atoms. In the context of gases, knowing the molar mass is crucial for volume calculations and applying the Ideal Gas Law. For example, oxygen (",\,\mathrm{O}_2") has a molar mass of 32 g/mol, which means one mole of oxygen weighs 32 grams. When given a mass of a gas, you can convert it to moles using the formula:
  • Moles = \( \frac{\text{mass}}{\text{molar mass}} \)
This conversion is critical as it allows you to determine the number of moles for use in gas law equations like PV=nRT, to find pressure, volume, or temperature of the gas.
universal gas constant
The universal gas constant, denoted as R, is a vital component in the Ideal Gas Law equation, PV = nRT. It connects the macroscopic volume, pressure, and temperature of a gas to its microscopic properties and is expressed as 8.314 J/(mol路K) in SI units. This value allows the conversion of a gas's energy content between physical and chemical processes. Essentially, R provides a bridge between the energy associated with a mole of substance and the gas's physical parameters. When using the Ideal Gas Law to calculate gas properties under various conditions, R helps dictate how temperature shifts might cause changes in pressure or volume across different contexts.
volume calculation
In chemistry, volume calculation at STP is simplified with foundational principles like the Ideal Gas Law and specific conditions. Using both methods from the exercise, you can see how this works.
  • **Method 1**: Use the Ideal Gas Law for precise calculations:
Volume (V) can be computed by rearranging the Ideal Gas Law as: \[ V = \left(\frac{nRT}{P}\right) \]where "n" is the number of moles \(=\frac{\text{mass}} {\text{molar mass}}\). This approach takes into account the exact conditions of temperature and pressure.
  • **Method 2**: Use known gas properties at STP:
At STP, the volume of a gas can be deduced from the molar volume; 1 mole of gas = 22.4 L. Hence, for a given mass of gas, the volume can be quickly calculated if the conditions are standard. This method offers a quick solution when conditions align perfectly with STP.

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Most popular questions from this chapter

A 3.0-liter tank contains oxygen gas at \(20^{\circ} \mathrm{C}\) and a gauge pressure of \(25 \times 10^{5} \mathrm{~Pa}\). What mass of oxygen is in the tank? The molecular mass of oxygen gas is \(32 \mathrm{~kg} / \mathrm{kmol}\). Assume atmospheric pressure to be \(1 \times 10^{5} \mathrm{~Pa}\). The absolute pressure of the gas is $$ P=\text { (Gauge pressure) }+\left(\text { Atmospheric pressure) }=(25+1) \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}=26 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right. $$ From the Ideal Gas Law, with \(M=32 \mathrm{~kg} / \mathrm{kmol}\), $$ \begin{array}{c} P V=\left(\frac{m}{M}\right) R T \\ \left(26 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(3.0 \times 10^{-3} \mathrm{~m}^{3}\right)=\left(\frac{m}{32 \mathrm{~kg} / \mathrm{kmol}}\right)\left(8314 \frac{\mathrm{J}}{\mathrm{kmol} \cdot \mathrm{K}}\right)(293 \mathrm{~K}) \end{array} $$ Solving this equation gives \(m\), the mass of gas in the tank, as \(0.10 \mathrm{~kg}\).

A \(15-\mathrm{cm}\) -long test tube of uniform bore is lowered, open-end down, into a fresh-water lake. How far below the surface of the lake must the water level be in the tube if one-third of the tube is to be filled with water? Let \(h\) be the depth of the water in the tube below the lake's surface. The air pressure \(P_{2}\) in the tube at depth \(h\) must equal atmospheric pressure \(P_{\mathrm{a}}\) plus the pressure of water at that depth: $$ P_{2}=P_{\mathrm{A}}+\rho g h $$ The Ideal Gas Law gives us the value of \(P_{2}\) as $$ P_{2}=\left(P_{1}\right)\left(\frac{V_{1}}{V_{2}}\right)\left(\frac{T_{2}}{T_{1}}\right)=\left(1.01 \times 10^{5} \mathrm{~Pa}\right)\left(\frac{3}{2}\right)(1.00)=1.50 \times 10^{5} \mathrm{~Pa} $$ Then, from the relation between \(P_{2}\) and \(h\). $$ h=\frac{P_{2}-P_{a}}{\rho g}=\frac{0.50 \times 10^{5} \mathrm{~Pa}}{\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)}=5.1 \mathrm{~m} $$ where atmospheric pressure has been taken as \(100 \mathrm{kPa}\).

Find the density of methane \((M=16 \mathrm{~kg} / \mathrm{kmol})\) at \(20^{\circ} \mathrm{C}\) and \(5.0 \mathrm{~atm}\). Use \(P V=(m / M) R T\) and \(\rho=m / V\) to get $$ \rho=\frac{P M}{R T}=\frac{\left(5.0 \times 1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)(16 \mathrm{~kg} / \mathrm{kmol})}{(8314 \mathrm{~J} / \mathrm{kmol} \cdot \mathrm{K})(293 \mathrm{~K})}=3.3 \mathrm{~kg} / \mathrm{m}^{3} $$

On a day when the barometer reads \(75.23 \mathrm{~cm}\), a reaction vessel holds \(250 \mathrm{~mL}\) of ideal gas at \(20.0^{\circ} \mathrm{C}\). An oil manometer \(\left(\rho=810 \mathrm{~kg} / \mathrm{m}^{3}\right)\) reads the pressure in the vessel to be \(41.0 \mathrm{~cm}\) of oil and below atmospheric pressure. What volume will the gas occupy under S.T.P.?

A mass of oxygen occupies \(0.0200 \mathrm{~m}^{3}\) at atmospheric pressure, \(101 \mathrm{kPa}\), and \(5.0^{\circ} \mathrm{C}\). Determine its volume if its pressure is increased to \(108 \mathrm{kPa}\) while its temperature is changed to \(30^{\circ} \mathrm{C}\). From $$ \frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}} \quad \text { we have } \quad V_{2}=V_{1}\left(\frac{P_{1}}{P_{2}}\right)\left(\frac{T_{2}}{T_{1}}\right) $$ But \(T_{1}=5+273=278 \mathrm{~K}\) and \(T_{2}=30+273=303 \mathrm{~K} ;\) consequently, $$ V_{2}=\left(0.0200 \mathrm{~m}^{3}\right)\left(\frac{101}{108}\right)\left(\frac{303}{278}\right)=0.0204 \mathrm{~m}^{3} $$
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