A diatomic gas obeys the law `pV^x=` constant. For what value of x, it has negative molar specific heat?

Options

A. (a) `xgt1.4`
B. (b) `xlt1.4`
C. (c) `1ltxlt1.4`
D. (d) `0ltxlt1`

Explanation

To solve the problem, we need to find the value of \( x \) for which the molar specific heat of a diatomic gas becomes negative when it obeys the law \( pV^x = \text{constant} \).

### Step-by-Step Solution:

1. **Understanding Molar Specific Heat**:
The molar specific heat \( C \) for a gas that follows the relation \( pV^x = \text{constant} \) is given by:
\[ C = C_v + \frac{R}{1 - x} \]
where \( C_v \) is the molar specific heat at constant volume and \( R \) is the universal gas constant.

2. **Identifying \( C_v \) for a Diatomic Gas**:
For a diatomic gas, the molar specific heat at constant volume \( C_v \) is:
\[ C_v = \frac{5}{2} R \]

3. **Substituting \( C_v \) into the Equation**:
Substitute \( C_v \) into the equation for \( C \):
\[ C = \frac{5}{2} R + \frac{R}{1 - x} \]

4. **Setting the Condition for Negative Molar Specific Heat**:
We want \( C < 0 \):
\[ \frac{5}{2} R + \frac{R}{1 - x} < 0 \]

5. **Rearranging the Inequality**:
Move \( \frac{R}{1 - x} \) to the other side:
\[ \frac{5}{2} R < -\frac{R}{1 - x} \]
Dividing both sides by \( R \) (assuming \( R > 0 \)):
\[ \frac{5}{2} < -\frac{1}{1 - x} \]

6. **Inverting the Inequality**:
Multiply both sides by \( -1 \) (which reverses the inequality):
\[ -\frac{5}{2} > \frac{1}{1 - x} \]

7. **Cross-Multiplying**:
Cross-multiply to eliminate the fraction:
\[ -5(1 - x) > 2 \]
Expanding gives:
\[ -5 + 5x > 2 \]
Rearranging gives:
\[ 5x > 7 \quad \Rightarrow \quad x > \frac{7}{5} = 1.4 \]

8. **Considering the Range of \( x \)**:
Since \( x \) must also satisfy \( x < 1 \) (from the earlier analysis), we find that:
\[ 1 < x < 1.4 \]

### Conclusion:
The value of \( x \) for which the diatomic gas has negative molar specific heat is:
\[ 1 < x < 1.4 \]