The pressure and density of a diatomic gas \((\gamma=7 / 5)\) change from \(\left(P_{1}, d_{1}\right)\) to \(\left(P_{2},\right.\), \(d_{2}\) ) adiabatically. If \(d_{2} / d_{1}=32\), then what is the value of \(P_{2} / P_{1}\) ? (a) 32 (b) 64 (c) 128 (d) 256

Short Answer

Expert verified
The value of \(P_2 / P_1\) is 128.

Step by step solution

01

Identify Relevant Adiabatic Process Formula

For an adiabatic process involving a diatomic gas, the pressure P and density d are related by the equation \(P d^{-\gamma} = \text{constant}\), where \(\gamma\) is the heat capacity ratio. We'll use this formula to relate \(P_1, d_1\) to \(P_2, d_2\).
02

Set Up the Equation

We start by writing the adiabatic condition for both the initial and final states, \(P_1 d_1^{-\gamma} = P_2 d_2^{-\gamma}\). We will find the ratio of \(P_2/P_1\) in terms of the density ratio \(d_2 / d_1\) and \(\gamma\).
03

Substitute the Given Values

We substitute the given value \(\gamma = 7/5\) and the density ratio \(d_2 / d_1 = 32\) into the adiabatic condition equation and solve for \(P_2 / P_1\).
04

Solve for the Pressure Ratio

Substituting the values, we have \(P_1 (d_1)^{-7/5} = P_2 (32d_1)^{-7/5}\). Simplifying the equation gives \(P_2/P_1 = 32^{7/5}\).
05

Calculate the Numerical Value of the Pressure Ratio

Calculating the value of \(32^{7/5}\), we get the pressure ratio \(P_2/P_1 = 128\), which corresponds to choice (c).

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Key Concepts

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

Adiabatic Process in Diatomic Gas
An adiabatic process is a thermodynamic change happening within a system where no heat is exchanged with the surroundings. This means any change in energy for a system undergoing an adiabatic process is due to work done by or on the system. For diatomic gases—those consisting of molecules with two atoms—the behavior during an adiabatic process is particularly interesting due to their energy absorption and release capabilities.

During such a process, the internal energy of a diatomic gas changes solely because of the work done, which affects the temperature, pressure, and volume of the gas. In mathematics, this is manifested through the pressure-density relationship and is described by the equation \( Pd^{-\gamma} = \text{constant} \) for a given quantity of gas. Here, \(\gamma\) is the heat capacity ratio, which is unique to the type of gas. For diatomic gases, such as oxygen or nitrogen, \(\gamma\) typically has a value of 7/5. By understanding how diatomic gases behave adiabatically, we can predict how parameters like pressure and volume will evolve without computing heat transfer, which is highly significant in areas such as aerodynamics and astrophysics.
Pressure-Density Relationship in Adiabatics
The pressure-density relationship is a cornerstone in understanding adiabatic processes. It shows how the pressure of a gas changes in response to a change in density when no heat is transferred to or from the gas. This relationship can be formulated using the equation for an adiabatic process, \( Pd^{-\gamma} = \text{constant} \) where \( P \) is the pressure, \( d \) is the density, and \( \gamma \) is the heat capacity ratio. For an ideal diatomic gas going through an adiabatic change, if the density increases, the pressure will also increase.

The math behind this relationship allows us to solve problems involving changes in gas conditions without dealing with the complexities of heat exchange. This proves especially useful in practical applications like understanding how the atmosphere behaves when air rises and expands or compresses as it falls. By applying the formula correctly and considering the initial and final states of the gas, we can find the resulting pressures after adiabatic processes, which is fundamental in thermodynamics and fluid dynamics.
Heat Capacity Ratio (\(\gamma\))
The heat capacity ratio, often denoted by \(\gamma\), is the ratio of the specific heat at constant pressure (\(C_p\)) to the specific heat at constant volume (\(C_v\)), that is, \(\gamma = C_p/C_v\). It is a crucial parameter in thermo-fluid dynamics as it influences the thermodynamic properties of a substance undergoing an adiabatic process.

For diatomic gases, the degree of freedom is more than in monatomic gases, leading to a \(\gamma\) value typically around 1.4 or 7/5. This specific ratio is essential when we deal with an adiabatic process because it determines how pressure and temperature will change when volume changes. A higher \(\gamma\) means the gas will react more dramatically to volume changes in terms of pressure and temperature. This property is utilized in a variety of engineering applications such as the design of nozzles and engines, as well as in understanding natural phenomena like sound speed propagation in the atmosphere.

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Most popular questions from this chapter

Which of the following pair does show the extensive properties? (a) temperature and pressure (b) viscosity and surface tension (c) refractive index and specific heat (d) volume and heat capacity

One mole of a certain ideal gas is contained under a weightless piston of a vertical cylinder at a temperature \(T\). The space over the piston opens into the atmosphere. What work has to be performed in order to increase isothermally the gas volume under the piston \(n\) times by slowly raising the piston? The friction of the piston against the cylinder walls is negligibly small. (a) \(R T(n-1-\ln n)\) (b) \(R T(1-n+\ln n)\) (c) \(R T \ln n\) (d) \(-R T \ln n\)

The equation of state for one mole of a gas is \(P V=R T+B P\), where \(B\) is a constant, independent of temperature. The internal energy of fixed amount of gas is function of temperature only. If one mole of the above gas is isothermally expanded from \(12 \mathrm{~L}\) to \(22 \mathrm{~L}\) at a constant external pressure of 1 bar at \(400 \mathrm{~K}\), then the change in enthalpy of the gas is approximately \((\mathrm{B}=2 \mathrm{~L} / \mathrm{mol})\) (a) 0 (b) \(-3.32 \mathrm{~J}\) (c) \(-332 \mathrm{~J}\) (d) \(-166 \mathrm{~J}\)

The latent heat of vaporization of a liquid at \(500 \mathrm{~K}\) and 1 atm pressure is \(10 \mathrm{kcal} / \mathrm{mol}\). What will be the change in internal energy if 3 moles of the liquid changes to vapour state at the same temperature and pressure? (a) \(27 \mathrm{kcal}\) (b) \(13 \mathrm{kcal}\) (c) \(-27 \mathrm{kcal}\) (d) \(-13\) kcal

An ideal gas \((\gamma=1.5)\) undergoes a change in state such that the magnitude of heat absorbed by the gas is equal to the magnitude of work done by the gas. The molar heat capacity of the gas in this process is (a) \(2 R\) (b) \(R\) (c) \(3 R\) (d) \(1.5 R\)

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