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Chapter 16: Problem 12

At \(18^{\circ} \mathrm{C}\) and \(765 \mathrm{mmHg}, 1.29\) liters of an ideal gas has a mass of \(2.71\) g. Compute the molecular mass of the gas. Use \(P V=(m / M) R T\) and the fact that \(760 \mathrm{mmHg}=1.00\) atm to obtain $$ M=\frac{m R T}{P V}=\frac{(0.00271 \mathrm{~kg})(8314 \mathrm{~J} / \mathrm{kmol} \cdot \mathrm{K})(291 \mathrm{~K})}{\left[(765 / 760)\left(1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\right]\left(0.00129 \mathrm{~m}^{3}\right)}=50.0 \mathrm{~kg} / \mathrm{kmol} $$

Short Answer

Expert verified
The molecular mass of the gas is 50.0 kg/kmol.

Step by step solution

01

Convert Units

First, we convert the volume from liters to cubic meters and the pressure from mmHg to atm. Since \(1.29 \text{ liters} = 0.00129 \text{ m}^3\) and from the problem, \(765 \text{ mmHg} = \frac{765}{760} \text{ atm}\).
02

Convert Temperature

Convert the temperature from Celsius to Kelvin using the formula \(T(K) = T(\degree C) + 273.15\). Hence, the temperature is \(18 + 273.15 = 291.15 \text{ K}\).
03

Rewrite the Ideal Gas Equation

The given equation is \(PV = \frac{m}{M}RT\). Rearrange this equation to solve for the molecular mass \(M\): \(M = \frac{mRT}{PV}\).
04

Substitute Known Values

Substitute into the equation \(M = \frac{mRT}{PV}\):\[m = 2.71 \text{ g} = 0.00271 \text{ kg},\, R = 8314 \text{ J/kmol} \cdot \text{K},\, T = 291.15 \text{ K},\, P = \frac{765}{760} \times 1.01 \times 10^5 \text{ N/m}^2,\, V = 0.00129 \text{ m}^3.\]
05

Calculate the Molecular Mass

After substituting the values, calculate:\[M = \frac{(0.00271)(8314)(291.15)}{\left(\frac{765}{760}\right)(1.01 \times 10^5)(0.00129)}\]Perform operations step by step to find \(M\).
06

Simplify and Solve

Carry out the multiplication and division operations:\[M = \frac{(0.00271)(8314)(291.15)}{(1.0066)(1.01 \times 10^5)(0.00129)}\]This simplifies and calculates to \(M \approx 50.0 \) kg/kmol.

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Key Concepts

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

Molecular Mass Calculation
Calculating the molecular mass of a gas can seem complicated at first glance, but it's quite straightforward when broken down into simple steps. In this exercise, the molecular mass is calculated using the ideal gas law equation: \[ PV = \frac{m}{M} RT \]This equation allows us to solve for the molecular mass \(M\) through rearrangement, resulting in:\[ M = \frac{mRT}{PV} \]Where:- \(m\) is the mass of the gas,- \(R\) is the universal gas constant,- \(T\) is the temperature in Kelvin,- \(P\) is the pressure in Pascals,- \(V\) is the volume in cubic meters.By substituting the known values into this equation, we can easily compute the molecular mass.
Unit Conversion
Handling unit conversion is essential in any physics or chemistry problem, especially when using formulas like the ideal gas law. In our scenario, several conversions are required. - **Volume Conversion**: Convert from liters to cubic meters. - Given: \(1.29 \text{ liters}\) needs conversion to m鲁. - Conversion factor: \(1 \text{ liter} = 0.001 \text{ m}^3\). - Therefore, \(1.29 \text{ liters} = 1.29 \times 0.001 = 0.00129 \text{ m}^3\).- **Pressure Conversion**: Convert pressure from mmHg to atm. - Given: \(765 \text{ mmHg}\). - Conversion factor: \(760 \text{ mmHg} = 1 \text{ atm}\). - Therefore, \(765 \text{ mmHg} = \frac{765}{760} \text{ atm}\).These conversions allow us to express all quantities in the units necessary for the ideal gas law.
Temperature Conversion
Temperature must be in Kelvin for many scientific equations, including the ideal gas law. In this exercise, you convert Celsius to Kelvin. The process is simple:The formula to convert Celsius to Kelvin is:\[ T(K) = T(\degree C) + 273.15 \]Here, the temperature is given as \(18^{\circ} C\). Thus:\[ T(K) = 18 + 273.15 = 291.15 \text{ K} \]This straightforward conversion ensures that the temperature is compatible with the other units in the gas law equation.
Pressure Conversion
Pressure is often measured in mmHg (millimeters of mercury), but for gas law calculations, atm (atmospheres) or Pa (Pascals) is preferable. Here, you need to convert mmHg to atm and ultimately to Pascals, as these units align with the ideal gas equation.- **Convert mmHg to atm**: - Given: \(765 \text{ mmHg}\). - Conversion factor: \(760 \text{ mmHg} = 1 \text{ atm}\). - Conversion: \(\text{Pressure} = \frac{765}{760} \text{ atm}\).- **Convert atm to Pascals**: - 1 atm is equivalent to \(1.01 \times 10^5 \text{ N/m}^2\). - Therefore: - \[765 \text{ mmHg} = \left(\frac{765}{760}\right) \times 1.01 \times 10^5 \text{ N/m}^2\] These conversions ensure our pressure measurements align perfectly with the other units in the formulas we use.

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Most popular questions from this chapter

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

A fish emits a \(2.0-\mathrm{mm}^{3}\) bubble at a depth of \(15 \mathrm{~m}\) in a lake. Find the volume of the bubble as it reaches the surface. Assume its omnoraturo dooc not chonore The absolute pressure in the bubble at a depth \(h\) is $$ P=\rho g h+\text { Atmospheric pressure } $$ where \(\rho=1000 \mathrm{~kg} / \mathrm{m}^{3}\) and atmospheric pressure is about \(100 \mathrm{kPa}\). At \(15 \mathrm{~m}\), $$ P_{1}=\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.8 \mathrm{~m} / \mathrm{s}^{3}\right)(15 \mathrm{~m})+100 \mathrm{kPa}=247 \mathrm{kPa} $$ and at the surface, \(P_{2}=100 \mathrm{kPa}\). Following the usual procedure, $$ V_{2}=V_{1}\left(\frac{P_{1}}{P_{2}}\right)\left(\frac{T_{2}}{T_{1}}\right)=\left(2.0 \mathrm{~mm}^{3}\right)\left(\frac{247}{100}\right)(1.0)=4.9 \mathrm{~mm}^{3} $$

A 2.0-mg droplet of liquid nitrogen is present in a \(30 \mathrm{~mL}\) tube as it is sealed off at very low temperature. What will be the nitrogen pressure in the tube when it is warmed to \(20{ }^{\circ} \mathrm{C}\) ? Express your answer in atmospheres. (M for nitrogen is \(28 \mathrm{~kg} / \mathrm{kmol}\).) Use \(P V=(m / M) R T\) to find $$ \begin{aligned} P &=\frac{m R T}{M V}=\frac{\left(2.0 \times 10^{-6} \mathrm{~kg}\right)(8314 \mathrm{~J} / \mathrm{kmol} \cdot \mathrm{K})(293 \mathrm{~K})}{(28 \mathrm{~kg} / \mathrm{kmol})\left(30 \times 10^{-6} \mathrm{~m}^{3}\right)}=5800 \mathrm{~N} / \mathrm{m}^{2} \\ &=\left(5800 \mathrm{~N} / \mathrm{m}^{2}\right)\left(\frac{1.0 \mathrm{~atm}}{1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}}\right)=0.057 \mathrm{~atm} \end{aligned} $$

On a day when atmospheric pressure is \(76 \mathrm{cmHg}\), the pressure gauge on a tank reads the pressure inside to be \(400 \mathrm{cmHg}\). The gas in the tank has a temperature of \(9^{\circ} \mathrm{C}\). If the tank is heated to \(31^{\circ} \mathrm{C}\) by the Sun, and if no gas exits from it, what will the pressure gauge read? $$ \frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}} \quad \text { and } \quad P_{2}=P_{1}\left(\frac{T_{2}}{T_{1}}\right)\left(\frac{V_{1}}{V_{2}}\right) $$ But gauges on tanks usually read the difference in pressure between inside and outside; this is called the gauge pressure. Therefore, $$ P_{1}=76 \mathrm{cmHg}+400 \mathrm{cmHg}=476 \mathrm{cmHg} $$ Also, \(V_{1}=V_{2}\). We then have $$ P_{2}=(476 \mathrm{~cm} \mathrm{Hg})\left(\frac{273+31}{273+9}\right)(1.00)=513 \mathrm{cmHg} $$ The gauge will read \(513 \mathrm{cmHg}-76 \mathrm{cmHg}=437 \mathrm{cmHg}\).

A gas is sealed into a closed container. The gas is heated so that its temperature rises from \(100^{\circ} \mathrm{C}\) to \(200{ }^{\circ} \mathrm{C}\). If the initial absolute pressure in the container was \(2.00\) atm, what will its new value be?
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