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We know that for an adiabatic process \(p V^{\gamma}=\) a constant. Evaluate "a constant" for an adiabatic process involving exactly \(2.0 \mathrm{~mol}\) of an ideal gas passing through the state having exactly \(p=1.0 \mathrm{~atm}\) and \(T=300 \mathrm{~K}\). Assume a diatomic gas whose molecules rotate but do not oscillate.

Short Answer

Expert verified
The constant for the adiabatic process is \( pV^{1.4} \) evaluated with the specific volume.

Step by step solution

01

Identify the Value of \( \gamma \) for a Diatomic Gas

For diatomic gases that only rotate, the specific heat ratio \( \gamma \) is typically \( \frac{7}{5} = 1.4 \).
02

Use the Ideal Gas Law to Find Volume

Given \( p = 1.0 \; \text{atm} \) (which we convert to Pa by multiplying by \( 1.01325 \times 10^5 \)), \( T = 300 \; \text{K} \), and \( n = 2.0 \) moles, we use the ideal gas law: \( PV = nRT \). The gas constant \( R = 8.314 \; \text{J/(mol K)} \). Substitute and solve for \( V \): \[ V = \frac{nRT}{p} = \frac{2.0 \times 8.314 \times 300}{1.01325 \times 10^5} \text{ m}^3 \]
03

Calculate the Constant for the Adiabatic Process

With \( V \) calculated and \( p = 1.0 \; \text{atm} = 1.01325 \times 10^5 \; \text{Pa} \), evaluate \( pV^{\gamma} \): \[ \text{Constant} = p \times V^{1.4} = 1.01325 \times 10^5 \times V^{1.4} \] Use the volume from the previous step to substitute and find the constant.

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental principle in the study of gases. It provides a simple equation relating the four key physical properties of an ideal gas: pressure (\(p\)), volume (\(V\)), the number of moles (\(n\)), and temperature (\(T\)). This relationship is mathematically expressed as \(PV = nRT\), where \(R\) is the gas constant. This law is indispensable in calculating one property if the others are known. For instance, to find the volume of a gas, we rearrange the equation to \(V = \frac{nRT}{p}\). Each variable can be adjusted and substituted into the equation depending on the known conditions. Its simplicity and wide applicability make the Ideal Gas Law a cornerstone concept in thermodynamics and chemistry.

While the Ideal Gas Law assumes no interactions between gas particles and considers them to be point particles, it provides a good approximation for many gases under various conditions.
Diatomic Gas
Diatomic gases are those comprised of molecules containing two atoms. Common examples include hydrogen (\(H_2\)), nitrogen (\(N_2\)), and oxygen (\(O_2\)). Diatomic gases have unique physical properties compared to monatomic gases like helium or argon. In terms of thermodynamics, diatomic gases that are non-oscillating but can rotate have a specific heat capacity that reflects their additional degrees of freedom compared to monatomic gases. For a diatomic molecule that can rotate, the specific heat ratio \(\gamma\) is given as \(\frac{7}{5} = 1.4\). This indicates how energy is distributed between translational and rotational motion within the gas molecules. Understanding the behavior of diatomic gases is crucial in various scientific applications, especially in the study of atmospheric sciences and combustion processes.
Specific Heat Ratio
The specific heat ratio, \(\gamma\), also known as the adiabatic index, is the ratio of the specific heat capacity at constant pressure (\(C_p\)) to that at constant volume (\(C_v\)), thus \(\gamma = \frac{C_p}{C_v}\). This ratio is vital in various thermodynamic processes, including adiabatic processes where no heat is exchanged with the surroundings. In the case of an adiabatic process, \(pV^{\gamma} = \text{constant}\). For diatomic gases, \(\gamma\) often equals \(1.4\). This value impacts how gas behaves under compression or expansion without heat transfer with the environment. The specific heat ratio helps us predict temperature and pressure changes in a gas under different thermodynamic processes.

Understanding \(\gamma\) provides insights into the efficiency of engines and turbines, as well as atmospheric phenomena.
Gas Constant
The Gas Constant \(R\) is a crucial factor in the Ideal Gas Law, and it represents the relationship between pressure, volume, and temperature for a given amount of gas. Its value is \(8.314 \; \text{J/(mol K)}\), which relates energy per mole per degree Kelvin. The universality of \(R\) makes it a key component in calculations involving gases, allowing scientists and engineers to apply it across various conditions and gas types. In the context of the Ideal Gas Law, \(R\) helps link the macroscopic properties of gases to the amount of substance present.

Being a consistent factor, \(R\) facilitates transformations between different units in gaseous calculations and is foundational for deeper molecular kinetic theories.

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Most popular questions from this chapter

What is the average translational kinetic energy of nitrogen molecules at \(1600 \mathrm{~K}\) ?

At what frequency would the wavelength of sound in air be equal to the mean free path of oxygen molecules at \(1.0 \mathrm{~atm}\) pressure and \(0.00^{\circ} \mathrm{C} ?\) The molecular diameter is \(3.0 \times 10^{-8} \mathrm{~cm}\).

Air that initially occupies \(0.140 \mathrm{~m}^{3}\) at a gauge pressure of \(103.0 \mathrm{kPa}\) is expanded isothermally to a pressure of \(101.3 \mathrm{kPa}\) and then cooled at constant pressure until it reaches its initial volume. Compute the work done by the air. (Gauge pressure is the difference between the actual pressure and atmospheric pressure.)

At \(20^{\circ} \mathrm{C}\) and 750 torr pressure, the mean free paths for argon gas (Ar) and nitrogen gas \(\left(\mathrm{N}_{2}\right)\) are \(\lambda_{\mathrm{Ar}}=9.9 \times 10^{-6} \mathrm{~cm}\) and \(\lambda_{\mathrm{N}_{2}}=\) \(27.5 \times 10^{-6} \mathrm{~cm} .\) (a) Find the ratio of the diameter of an \(\mathrm{Ar}\) atom to that of an \(\mathrm{N}_{2}\) molecule. What is the mean free path of argon at (b) \(20^{\circ} \mathrm{C}\) and 150 torr, and \((\mathrm{c})-40^{\circ} \mathrm{C}\) and 750 torr?

Suppose \(1.80 \mathrm{~mol}\) of an ideal gas is taken from a volume of \(3.00 \mathrm{~m}^{3}\) to a volume of \(1.50 \mathrm{~m}^{3}\) via an isothermal compression at \(30^{\circ} \mathrm{C} .\) (a) How much energy is transferred as heat during the compression, and (b) is the transfer to or from the gas?

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