/*! 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 10 A 2.0-mg droplet of liquid nitro... [FREE SOLUTION] | 91Ó°ÊÓ

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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 $$ \begin{aligned} P &=\frac{m R T}{M V}=\frac{\left(2.0 \times 10^{-6} \mathrm{~kg}\right)(8314 \mathrm{~J} / \mathrm{kmol} \cdot \mathrm{K})(293 \mathrm{~K})}{(28 \mathrm{~kg} / \mathrm{kmol})\left(30 \times 10^{-6} \mathrm{~m}^{3}\right)}=5800 \mathrm{~N} / \mathrm{m}^{2} \\ &=\left(5800 \mathrm{~N} / \mathrm{m}^{2}\right)\left(\frac{1.0 \mathrm{~atm}}{1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}}\right)=0.057 \mathrm{~atm} \end{aligned} $$

Short Answer

Expert verified
The nitrogen pressure in the tube is 0.057 atm.

Step by step solution

01

Identify the Given Values

We have a droplet of liquid nitrogen with a mass of 2.0 mg (or \(2.0 \times 10^{-6}\) kg). The volume of the tube is 30 mL (which is \(30 \times 10^{-6}\) m\(^3\)). The temperature is given as 20\(^{\circ}\)C, which needs to be converted to Kelvin by adding 273, resulting in 293 K.
02

Use the Ideal Gas Law Formula

The problem states to use the formula \(PV = \frac{m}{M} RT\), which is derived from the ideal gas law \(PV = nRT\). Here, \(m\) is the mass, \(M\) is the molar mass of nitrogen \(28\; \mathrm{kg/kmol}\), \(R\) is the universal gas constant \(8314\; \mathrm{J/(kmol \cdot K)}\), and \(T\) is the temperature in Kelvin.
03

Substitute the Values into the Equation

Plug in the known values into the equation: \(P = \frac{(2.0 \times 10^{-6} \; \mathrm{kg}) \times 8314 \; \mathrm{J/(kmol \cdot K)} \times 293 \; \mathrm{K}}{28 \; \mathrm{kg/kmol} \times 30 \times 10^{-6} \; \mathrm{m^3}}\). This simplifies to \(P = 5800 \; \mathrm{N/m^2}\).
04

Convert Pressure to Atmospheres

To convert the pressure from Newtons per square meter to atmospheres, use the conversion factor \(1.0\; \mathrm{atm} = 1.01 \times 10^5\; \mathrm{N/m^2}\). Therefore, \(P = \frac{5800 \; \mathrm{N/m^2}}{1.01 \times 10^5 \; \mathrm{N/m^2/atm}} = 0.057\; \mathrm{atm}\).

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Key Concepts

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

Pressure Calculation
When working with gases, calculating the pressure is fundamental to understanding how gases behave in different conditions. The pressure of a gas can be found using the ideal gas law equation: \[ P = \frac{mRT}{MV} \]Where:
  • \(P\) is the pressure of the gas.
  • \(m\) is the mass of the gas.
  • \(R\) is the universal gas constant with a value of 8314 J/(kmol·K).
  • \(T\) is the temperature in Kelvin.
  • \(M\) is the molar mass of the gas.
  • \(V\) is the volume of the container.
To obtain the pressure in the desired units, it’s important to use consistent units throughout the formula. After determining the pressure in Newtons per square meter (N/m²), you might need to further convert it to atmospheres (atm) for convenience, especially in scientific and academic settings. This involves using the conversion factor: \(1.0\; \, \mathrm{atm} = 1.01 \times 10^5 \; \, \mathrm{N/m^2}\). Hence, the pressure in atmospheres is calculated by dividing the pressure in N/m² by this conversion factor.
Temperature Conversion
Temperature conversion is a necessary step when dealing with gas laws because the equations used typically require temperature in Kelvin. The Kelvin scale is used in scientific calculations since it starts from absolute zero, providing a true reflection of thermal energy. If the temperature is given in Celsius, as it often is in problem statements, converting to Kelvin is simple.To convert Celsius to Kelvin, the equation is:\[ K = ^{\circ}C + 273 \]For example, if you have a temperature of \(20{ }^{\circ} \, \mathrm{C}\), converting to Kelvin involves adding 273, which gives 293 K.This conversion is crucial because gas laws like the ideal gas law rely on temperature in Kelvin. Without this conversion, calculations might lead to inaccurate or erroneous results, impacting the conclusions drawn from the analysis.
Unit Conversion
Unit conversion is fundamental when applying formulas like the ideal gas law as measurements often come in a variety of units. Using consistent units prevents errors in computation:- **Mass Conversion:** If mass is provided in milligrams (mg), it must be converted to kilograms (kg) since the molar mass is in kg/kmol. For example, a 2.0 mg mass is equivalent to \(2.0 \times 10^{-6} \, \mathrm{kg}\).
- **Volume Conversion:** Volume measurements might be in milliliters (mL) and need conversion to cubic meters (m³) because the gas constant \(R\) is expressed in J/(kmol·K) and assumes SI units including volume in m³. Converting 30 mL to m³ results in \(30 \times 10^{-6} \, \mathrm{m}^3\).
Ensuring your units are consistent with the constants and formulas utilized in calculations is essential. This attention to detail prevents discrepancies and ensures that the output is reliable and comparable to other scientific data.
Molar Mass
Molar mass is a measure of the mass of a substance’s molecules and is typically reported in kg/kmol or g/mol. It is an integral part of calculating properties of substances through equations like the ideal gas law. In the context of gas laws, the molar mass allows us to relate the macroscopic properties of a gas with its molecular properties by converting mass into moles. For nitrogen in our example, the molar mass is given as 28 kg/kmol. This indicates that one kilomole of nitrogen, which contains Avogadro's number of molecules, has a mass of 28 kg. It’s crucial to accurately use molar mass in calculations for ideal gas law applications, as errors may distort the relationships between pressure, volume, and temperature. Proper understanding and usage of molar mass ensure precise computation, yielding meaningful insights from gas law equations.

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Most popular questions from this chapter

A 500-mL sealed flask contains nitrogen at a pressure of \(76.00\) cmHg. A tiny glass tube lies at the bottom of the flask. Its volume is \(0.50 \mathrm{~mL}\) and it contains hydrogen gas at a pressure of \(4.5\) atm. Suppose the glass tube is now broken so that the hydrogen fills the flask. What is the new pressure in the flask?

Compute the volume of \(8.0\) g of helium \((M=4.0 \mathrm{~kg} / \mathrm{kmol})\) at 15 \({ }^{\circ} \mathrm{C}\) and \(480 \mathrm{mmHg}\). Use \(P V=(m / M) R T\) to obtain $$ V=\frac{m R T}{M P}=\frac{(0.0080 \mathrm{~kg})(8314 \mathrm{~J} / \mathrm{kmol} \cdot \mathrm{K})(288 \mathrm{~K})}{(4.0 \mathrm{~kg} / \mathrm{kmol})\left[(480 / 760)\left(1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\right]}=0.075 \mathrm{~m}^{3}=75 \text { liters } $$

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 certain mass of hydrogen gas occupies \(370 \mathrm{~mL}\) at \(16^{\circ} \mathrm{C}\) and 150 kPa. Find its volume at \(-21^{\circ} \mathrm{C}\) and \(420 \mathrm{kPa}\). $$ \begin{array}{c} \frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}} \quad \text { leads to } \quad V_{2}=V_{1}\left(\frac{P_{1}}{P_{2}}\right)\left(\frac{T_{2}}{T_{1}}\right) \\\ V_{2}=(370 \mathrm{~mL})\left(\frac{150 \mathrm{kPa}}{420 \mathrm{kPa}}\right)\left(\frac{273 \mathrm{~K}-21 \mathrm{~K}}{273 \mathrm{~K}+16 \mathrm{~K}}\right)=115 \mathrm{~mL} \end{array} $$

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 \((\rho=810\) \(\mathrm{kg} / \mathrm{m}^{3}\) ) 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.?

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