m grams of a gas of molecules weight M is flowing in an insulated tube whith velocity v. If the system is suddenly stopped then the rise in its temperature will be

Options

A. `(Mv^(2)(gamma-1))/(2RJ)`
B. `(mv^(2)gamma)/(2RJ)`
C. `(V^(2))/(2Js)`
D. `(Mv^(2)(gamma-1))/(2MRJ)`

Answer

To solve the problem, we need to determine the rise in temperature of a gas when it is suddenly stopped while flowing in an insulated tube. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the System
We have:
- Mass of the gas, \( m \) grams
- Molecular weight of the gas, \( M \)
- Velocity of the gas, \( v \)

### Step 2: Identify the Process
Since the gas is flowing in an insulated tube and is suddenly stopped, we can consider this as an adiabatic process. In an adiabatic process, there is no heat exchange with the surroundings.

### Step 3: Kinetic Energy of the Gas
When the gas is flowing, it possesses kinetic energy given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
This kinetic energy will be converted into internal energy when the gas is stopped.

### Step 4: Relate Kinetic Energy to Temperature Change
In an adiabatic process, the work done on the gas can be related to the change in internal energy, which in turn relates to the change in temperature. The work done can be expressed as:
\[ W = \Delta U = nC_v \Delta T \]
Where:
- \( n \) is the number of moles of the gas
- \( C_v \) is the molar specific heat at constant volume
- \( \Delta T \) is the change in temperature

### Step 5: Convert Mass to Moles
The number of moles \( n \) can be calculated from the mass \( m \) and molecular weight \( M \):
\[ n = \frac{m}{M} \]

### Step 6: Express Change in Internal Energy
The change in internal energy can also be expressed in terms of the kinetic energy:
\[ \frac{1}{2} mv^2 = nC_v \Delta T \]

### Step 7: Substitute for Moles
Substituting \( n \) into the equation gives:
\[ \frac{1}{2} mv^2 = \frac{m}{M} C_v \Delta T \]

### Step 8: Solve for Temperature Change
Rearranging the equation to solve for \( \Delta T \):
\[ \Delta T = \frac{Mv^2}{2C_v} \]

### Step 9: Use the Relation between \( C_v \) and \( R \)
For an ideal gas, we have the relation:
\[ C_v = \frac{R}{\gamma - 1} \]
where \( \gamma \) is the heat capacity ratio (specific heat at constant pressure \( C_p \) to specific heat at constant volume \( C_v \)).

### Step 10: Substitute \( C_v \) into the Temperature Change Equation
Substituting \( C_v \) into the equation for \( \Delta T \):
\[ \Delta T = \frac{Mv^2}{2 \cdot \frac{R}{\gamma - 1}} = \frac{Mv^2(\gamma - 1)}{2R} \]

### Final Result
Thus, the rise in temperature \( \Delta T \) when the gas is suddenly stopped is given by:
\[ \Delta T = \frac{Mv^2(\gamma - 1)}{2R} \]