If the ratio of `PV_(m) & RT` for a real gas is `(x)/(24)` at a temp where `((delP)/(delV_(m))) =0`. The find value of `10x`.

Answer

To solve the problem, we need to find the value of \(10x\) given the ratio of \(\frac{PV_m}{RT} = \frac{x}{24}\) at a temperature where \(\frac{\partial P}{\partial V_m} = 0\). This condition indicates that we are at the critical point of the gas.

### Step-by-Step Solution:

1. **Understand the Critical Conditions**:
At the critical point, the following relationships hold:
- Critical temperature, \(T_c = \frac{8A}{27B}\)
- Critical pressure, \(P_c = \frac{A}{27B^2}\)
- Critical volume, \(V_c = 3B\)

2. **Set Up the Equation**:
We know that at the critical point:
\[ \frac{PV_m}{RT} = \frac{x}{24} \]
Here, we can substitute \(P\), \(V_m\), and \(T\) with their critical values:
\[ \frac{P_c \cdot V_c}{R \cdot T_c} = \frac{x}{24} \]

3. **Substitute the Critical Values**:
Substitute \(P_c\), \(V_c\), and \(T_c\) into the equation:
\[ \frac{\left(\frac{A}{27B^2}\right) \cdot (3B)}{R \cdot \left(\frac{8A}{27B}\right)} = \frac{x}{24} \]

4. **Simplify the Left Side**:
Simplifying the left side:
\[ \frac{\frac{3AB}{27B^2}}{\frac{8AR}{27B}} = \frac{3AB \cdot 27B}{27B^2 \cdot 8AR} = \frac{3}{8} \]

5. **Set the Equation Equal to \(\frac{x}{24}\)**:
Now we have:
\[ \frac{3}{8} = \frac{x}{24} \]

6. **Cross Multiply to Solve for \(x\)**:
Cross-multiplying gives:
\[ 3 \cdot 24 = 8x \implies 72 = 8x \implies x = \frac{72}{8} = 9 \]

7. **Calculate \(10x\)**:
Finally, we need to find \(10x\):
\[ 10x = 10 \cdot 9 = 90 \]

### Final Answer:
Thus, the value of \(10x\) is \(90\).

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