`rho`-T equation of a gas in adiabatic process is given by

Options

A. `T^(gamma-1)rho="constant"`
B. `rho^(gammaT)="constant"`
C. `Trho^(1-gamma)="constant"`
D. `T^(gamma)rho^(gamma-1)="constant"`

Answer

To derive the \(\rho\)-\(T\) equation of a gas in an adiabatic process, we will follow these steps:

### Step 1: Understand the Adiabatic Process
In an adiabatic process, there is no heat exchange with the surroundings. The relationship between pressure (P), volume (V), and temperature (T) can be described using the adiabatic condition, which states that:

\[ PV^\gamma = \text{constant} \]

where \(\gamma\) (gamma) is the heat capacity ratio \(C_p/C_v\).

### Step 2: Use the Ideal Gas Law
The ideal gas law is given by:

\[ PV = nRT \]

where \(n\) is the number of moles, \(R\) is the universal gas constant, and \(T\) is the temperature.

### Step 3: Relate Moles to Density
We can express the number of moles \(n\) in terms of mass \(m\) and molar mass \(M\):

\[ n = \frac{m}{M} \]

Substituting this into the ideal gas equation gives:

\[ PV = \frac{m}{M}RT \]

### Step 4: Express Density in Terms of Mass and Volume
The density \(\rho\) of the gas is defined as:

\[ \rho = \frac{m}{V} \]

From this, we can express mass \(m\) as:

\[ m = \rho V \]

### Step 5: Substitute Mass into the Ideal Gas Law
Substituting \(m = \rho V\) into the ideal gas equation gives:

\[ PV = \frac{\rho V}{M}RT \]

Cancelling \(V\) from both sides (assuming \(V \neq 0\)) leads to:

\[ P = \frac{\rho RT}{M} \]

### Step 6: Rearranging the Adiabatic Condition
Now, we can rearrange the adiabatic condition \(PV^\gamma = \text{constant}\) using our expression for \(P\):

\[ \left(\frac{\rho RT}{M}\right)V^\gamma = \text{constant} \]

### Step 7: Substitute for Volume
Using the relationship \(V = \frac{m}{\rho}\), we can express the volume in terms of density:

\[ \left(\frac{\rho RT}{M}\right)\left(\frac{m}{\rho}\right)^\gamma = \text{constant} \]

### Step 8: Simplifying the Equation
This can be simplified to:

\[ \frac{\rho RT}{M} \cdot \frac{m^\gamma}{\rho^\gamma} = \text{constant} \]

Rearranging gives:

\[ \rho^{1 - \gamma}T = \text{constant} \]

### Final Result
Thus, we arrive at the relationship:

\[ \rho^{1 - \gamma} T = C \]

where \(C\) is a constant.