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Nitrogen is compressed to a density of \(4 \mathrm{kg} / \mathrm{m}^{3}\) under an absolute pressure of \(400 \mathrm{kPa}\). Determine the temperature in degrees Celsius.

Short Answer

Expert verified
The temperature is approximately 9167.65 °C.

Step by step solution

01

Identify the Given Variables

We are given the density \( \rho = 4 \, \text{kg/m}^3 \), the absolute pressure \( P = 400 \, \text{kPa} = 400,000 \, \text{Pa} \). We need to find the temperature in degrees Celsius.
02

Use the Ideal Gas Law

The relationship between pressure, density, and temperature for an ideal gas is given by the ideal gas law equation, \( PV = nRT \). We can rearrange this for \( T \), and use density to express it in terms of the molar mass \( M \): \( P = \rho RT / M \), so \( T = \frac{PM}{\rho R} \).
03

Input Known Values for Calculations

For nitrogen (\( N_2 \)), the molar mass \( M \) is approximately \( 28.02 \, \text{g/mol} = 28.02 \, \text{kg/kmol} \). The specific gas constant \( R \) for nitrogen is approximately \( 296.8 \, \text{J/(kg K)} \). Substitute these values and given variables into the temperature equation: \[T = \frac{400,000 \, \text{Pa} \times 28.02 \, \text{kg/kmol}}{4 \, \text{kg/m}^3 \times 296.8 \, \text{J/(kg K)}}\].
04

Solve for Temperature in Kelvin

Perform the calculations: \[T = \frac{400,000 \times 28.02}{4 \times 296.8} \approx \frac{11,208,000}{1,187.2} \approx 9440.8 \, \text{K}\].
05

Convert Temperature to Celsius

Since the temperature was initially in Kelvin, convert it to degrees Celsius using the formula \( T_{\text{Celsius}} = T_{\text{Kelvin}} - 273.15 \). Substitute in the calculated temperature: \[T_{\text{Celsius}} = 9440.8 - 273.15 = 9167.65 \, \text{°C}\].

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Key Concepts

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

Nitrogen Properties
Nitrogen, a colorless, odorless, and tasteless gas, is a crucial component of the Earth's atmosphere, comprising about 78% of the air. It is a diatomic molecule, meaning each molecule consists of two nitrogen atoms, often denoted as \( N_2 \). Due to its chemical stability, nitrogen is inert under standard conditions, which means it does not easily react with other substances. This makes it an excellent choice for applications that require a non-reactive atmosphere, such as food packaging and chemical storage.

Its molecular weight, crucial for calculations involving gases, is approximately \( 28.02 \, \text{g/mol} \). This is important in applications like the Ideal Gas Law to determine properties like temperature, pressure, and volume under varying conditions.
Density Calculations
Density, a measure of how much mass is contained in a given volume, is a fundamental concept in gas calculations. For gases, density is typically expressed in \( \text{kg/m}^3 \), which means kilograms per cubic meter. Knowing the density is essential when applying the Ideal Gas Law, as it directly relates to the pressure, temperature, and volume of a gas.

Density can be calculated using the formula \( \rho = \frac{P \cdot M}{R \cdot T} \), derived by manipulating the Ideal Gas Law equation. This formula involves the pressure \( P \), molar mass \( M \), specific gas constant \( R \), and temperature \( T \). In our example, nitrogen's density informs us how closely its molecules are packed together under a specific pressure, making it possible to calculate other properties when dealing with real-world systems.
Temperature Conversion
Temperature is a vital measurement in the study of gases. To convert between Kelvin and Celsius, which are two common temperature units used in scientific contexts, a simple formula is applied:
  • \( T_{\text{Celsius}} = T_{\text{Kelvin}} - 273.15 \)
  • Conversely, \( T_{\text{Kelvin}} = T_{\text{Celsius}} + 273.15 \)
This conversion is crucial in scientific calculations because Kelvin starts at absolute zero, meaning it directly corresponds to the energy of the particles in a substance. Celsius, on the other hand, is more familiar in everyday life. Accurately converting between these units ensures correct calculation outcomes when employing gas laws or interpreting experimental results.
Molar Mass
Molar mass is a measure of the mass of one mole of a substance, expressed in \( \text{g/mol} \). For gases, it plays a critical role in determining how a gas behaves under various conditions when using the Ideal Gas Law. Each molecule of nitrogen has a molar mass of approximately \( 28.02 \, \text{g/mol} \).

In Ideal Gas Law calculations, the molar mass allows us to connect the microscopic properties of a substance (like the number of molecules) with macroscopic properties (like mass and volume). For example, when calculating the temperature of a compressed gas, knowing the molar mass enables the determination of how much space the gas's molecules occupy, thus affecting the resulting calculations.

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Most popular questions from this chapter

In a test to determine the bulk modulus of a liquid it was found that as the absolute pressure was changed from 15 to 3000 psi the volume decreased from 10.240 to 10.138 in. \(^{3}\) Determine the bulk modulus for this liquid.

A rigid-walled cubical container is completely filled with water at \(40^{\circ} \mathrm{F}\) and sealed. The water is then heated to \(100^{\circ} \mathrm{F}\). Determine the pressure that develops in the container when the water reaches this higher temperature. Assume that the volume of the container remains constant and the value of the bulk modulus of the water remains constant and equal to 300,000 psi.

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There are many fluids that exhibit non-Newtonian behavior (see, for example, Video \(V 1.6\) ). For a given fluid the distinction between Newtonian and non- Newtonian behavior is usually based on measurements of shear stress and rate of shearing strain. Assume that the viscosity of blood is to be determined by measurements of shear stress, \(\tau,\) and rate of shearing strain, duldy, obtained from a small blood sample tested in a suitable viscometer. Based on the data given below, determine if the blood is a Newtonian or non-Newtonian fluid. Explain how you arrived at your answer. $$\begin{array}{l|l|l|l|l|l|l|l|l} \tau\left(\mathrm{N} / \mathrm{m}^{2}\right) & 0.04 & 0.06 & 0.12 & 0.18 & 0.30 & 0.52 & 1.12 & 2.10 \\ \hline d u / d y\left(\mathrm{s}^{-1}\right) & 2.25 & 4.50 & 11.25 & 22.5 & 45.0 & 90.0 & 225 & 450 \end{array}$$

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