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Question: For a reversible adiabatic expansion of an ideal gas \[\dfrac{dp}{p}\] equal to: A. \[\gamma \text...

For a reversible adiabatic expansion of an ideal gas dpp\dfrac{dp}{p} equal to:
A. γ dvv \gamma \text{ }\dfrac{dv}{v}\text{ }
B.  dv v ~\dfrac{dv}{~v\text{ }}
C. (γγ  1)dvv \left( \dfrac{\gamma }{\gamma \text{ }-\text{ }1} \right)\dfrac{dv}{v\text{ }}
D. γdvv -\gamma \dfrac{dv}{v\text{ }}

Explanation

Solution

Hint: We should know that adiabatic expansion is defined as an ideal behaviour for a closed system, in which the pressure is constant and the temperature is decreasing and it can be either reversible or irreversible.

Step by step answer:
So, first we should know about adiabatic processes. It is defined as: “The thermodynamic process in which there is no exchange of heat from the system to its surrounding neither during expansion nor during compression”.
We should know that adiabatic processes can be either reversible or irreversible. We should know about some essential conditions for adiabatic process to take place and they are as follows:
The system we are taking must be perfectly insulated from the surrounding.
The process must be carried out quickly so that there is sufficient amount of time for heat transfer to take place.
We should know about the adiabatic process equation.
PVγP{{V}^{\gamma }} = constant
In the above equation, we should know that, P is the pressure of the system, V is the volume of the system and γ is the adiabatic index and is defined as the ratio of heat capacity at constant pressure Cp to heat capacity at constant volume Cv.
Now, coming back to our question, in this question we have to find reversible adiabatic expansion of an ideal gasdpp\dfrac{dp}{p}. We know that:

& In\text{ }an\text{ }adiabatic\text{ }process: \\\ & P{{V}^{\gamma }}=K(constant) \\\ \end{aligned}$$ Now, if we differentiate the above equation we will have: $$\begin{aligned} & (1.dp){{v}^{\gamma }}+p.(\gamma {{v}^{\gamma -1}}dv)=0 \\\ & dp{{v}^{\gamma }}=-\gamma p{{v}^{\gamma -1}}dv \\\ \end{aligned}$$ $$\begin{aligned} & \dfrac{dp}{p}=\dfrac{-\gamma {{v}^{\gamma -1}}}{v\gamma }dv \\\ & \dfrac{dp}{p}=-\gamma \dfrac{dv}{v} \\\ \end{aligned}$$ We should note that, $\dfrac{dp}{p}$ and $\dfrac{dv}{v}$ are the fractional change in pressure and volume respectively. So, from the above calculation we came to know our answer that option D is correct. Note: Now at last we should know about some examples of adiabatic processes. There are several examples, some are stated below: We know that adiabatic process is a process where there is a gas compression and heat is generated. One of the simplest examples that we experience in daily life would be the release of air from a pneumatic tire. We should note that adiabatic Efficiency is applied to devices such as nozzles, compressors, and turbines. It is one of the good applications of the adiabatic process.