`DeltaH_("vap")` of the process, `A_((l)) hArr A_(("vap"))` is 460.6 Cal `mol^(-1)`. The normal boiling point of the A is 50K. When pressure is increased to 10 atm. Then the boiling point of the A is (50)x K. The value of x is

Answer

To solve the problem, we will use the Clausius-Clapeyron equation, which relates the change in vapor pressure with temperature to the enthalpy of vaporization. The equation is given by:

\[ \Delta H_{vap} = 2.303 R \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \log \left( \frac{P_2}{P_1} \right) \]

Where:
- \(\Delta H_{vap}\) = enthalpy of vaporization (460.6 Cal/mol)
- \(R\) = gas constant (1.987 Cal/(mol·K))
- \(T_1\) = normal boiling point (50 K)
- \(P_1\) = normal pressure (1 atm)
- \(P_2\) = pressure at which boiling point is measured (10 atm)
- \(T_2\) = boiling point at pressure \(P_2\) (50x K)

### Step 1: Substitute known values into the equation

We know:
- \(\Delta H_{vap} = 460.6 \, \text{Cal/mol}\)
- \(R = 1.987 \, \text{Cal/(mol·K)}\)
- \(T_1 = 50 \, \text{K}\)
- \(P_1 = 1 \, \text{atm}\)
- \(P_2 = 10 \, \text{atm}\)

Substituting these values into the equation gives:

\[ 460.6 = 2.303 \times 1.987 \left( \frac{1}{50} - \frac{1}{T_2} \right) \log \left( \frac{10}{1} \right) \]

### Step 2: Calculate \(\log(10)\)

We know that:

\[ \log(10) = 1 \]

### Step 3: Simplify the equation

Now we can simplify the equation:

\[ 460.6 = 2.303 \times 1.987 \left( \frac{1}{50} - \frac{1}{T_2} \right) \]

### Step 4: Calculate the left-hand side

Calculating \(2.303 \times 1.987\):

\[ 2.303 \times 1.987 \approx 4.578 \]

So the equation becomes:

\[ 460.6 = 4.578 \left( \frac{1}{50} - \frac{1}{T_2} \right) \]

### Step 5: Isolate \(\frac{1}{T_2}\)

Rearranging gives:

\[ \frac{1}{50} - \frac{1}{T_2} = \frac{460.6}{4.578} \]

Calculating the right-hand side:

\[ \frac{460.6}{4.578} \approx 100.5 \]

So we have:

\[ \frac{1}{50} - \frac{1}{T_2} = 100.5 \]

### Step 6: Solve for \(\frac{1}{T_2}\)

Now, rearranging gives:

\[ \frac{1}{T_2} = \frac{1}{50} - 100.5 \]

Calculating \(\frac{1}{50}\):

\[ \frac{1}{50} = 0.02 \]

Thus:

\[ \frac{1}{T_2} = 0.02 - 100.5 = -100.48 \]

### Step 7: Calculate \(T_2\)

Taking the reciprocal gives:

\[ T_2 = \frac{1}{-100.48} \approx -0.00995 \text{ K} \]

### Step 8: Relate \(T_2\) to \(50x\)

From the problem, we know:

\[ T_2 = 50x \]

Setting the two equations equal gives:

\[ 50x = -0.00995 \]

### Step 9: Solve for \(x\)

Dividing both sides by 50:

\[ x = \frac{-0.00995}{50} \approx -0.000199 \]

### Final Answer

Thus, the value of \(x\) is approximately:

\[ x \approx -0.000199 \]