Slope of isotherm for a gas (having `gamma = (5)/(3)`) is `3 xx 10^(5) N//m^(2)` . If the same gas is undergoing adiabatic change then adiabatic elasticity at that instant is

Options

A. `3 xx 10^(5) N//m^2`
B. `5 xx 10^(5) N//m^2`
C. `6 xx 10^(5) N//m^2`
D. `10 xx 10^(5) N//m^2`

Answer

To solve the problem step by step, we will follow the reasoning presented in the video transcript.

### Step 1: Understand the given information
We are given:
- The slope of the isotherm (dP/dV) = \(3 \times 10^5 \, \text{N/m}^2\)
- The value of gamma (\(\gamma\)) for the gas = \(\frac{5}{3}\)

### Step 2: Relate the slope of the isotherm to the pressure and volume
The slope of the isotherm is defined as:
\[ -\frac{dP}{dV} = \frac{P}{V} \]
From the problem, we know that:
\[ -\frac{dP}{dV} = 3 \times 10^5 \, \text{N/m}^2 \]
Thus, we can write:
\[ \frac{P}{V} = 3 \times 10^5 \, \text{N/m}^2 \]

### Step 3: Use the formula for adiabatic elasticity
Adiabatic elasticity (E) is given by the formula:
\[ E = \gamma \cdot \frac{P}{V} \]
Substituting the known values into this formula:
\[ E = \frac{5}{3} \cdot (3 \times 10^5) \]

### Step 4: Calculate the adiabatic elasticity
Now, we can perform the calculation:
\[ E = \frac{5}{3} \cdot 3 \times 10^5 = 5 \times 10^5 \, \text{N/m}^2 \]

### Final Answer
The adiabatic elasticity at that instant is:
\[ E = 5 \times 10^5 \, \text{N/m}^2 \]

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