A rigid vessel of volume ` 0.50m^(3)` containing `H_(2) ` at `20.5^(@)C` and a pressure of `611 xx 10^(3)` Pa is connected to a second rigid vessel of volume 0.75`m^(3)` containing Ar at `31.2^(@)C` at a pressure of `433xx10^(3)` Pa. A value separating the two vessels is opened and both are cooled to a temperature of `14.5^(@)C`. What is the final pressure in the vessels?

Options

A. `2xx10^(5)`
B. `3.22xx10^(5)" Pa"`
C. 4840 pa
D. `4.84xx10^(5)" Pa"`

Explanation

To solve the problem, we need to find the final pressure in the two connected rigid vessels after they are cooled to a temperature of 14.5°C. We will use the ideal gas law, which states that \( PV = nRT \).

### Step 1: Convert Temperatures to Kelvin
First, we need to convert the given temperatures from Celsius to Kelvin.

- For \( H_2 \):
\[ T_1 = 20.5 + 273 = 293.5 \, K \]

- For \( Ar \):
\[ T_2 = 31.2 + 273 = 304.2 \, K \]

- Final temperature after cooling:
\[ T_f = 14.5 + 273 = 287.5 \, K \]

### Step 2: Calculate Number of Moles for Each Gas
Using the ideal gas equation \( n = \frac{PV}{RT} \), we can calculate the number of moles for each gas.

- For \( H_2 \):
\[ n_1 = \frac{P_1 V_1}{R T_1} = \frac{611 \times 10^3 \, Pa \times 0.50 \, m^3}{R \times 293.5} \]

- For \( Ar \):
\[ n_2 = \frac{P_2 V_2}{R T_2} = \frac{433 \times 10^3 \, Pa \times 0.75 \, m^3}{R \times 304.2} \]

### Step 3: Total Moles Calculation
The total number of moles \( n_{total} \) after the valve is opened is:
\[ n_{total} = n_1 + n_2 \]

### Step 4: Calculate Total Volume
The total volume \( V_{total} \) of the combined vessels is:
\[ V_{total} = V_1 + V_2 = 0.50 \, m^3 + 0.75 \, m^3 = 1.25 \, m^3 \]

### Step 5: Calculate Final Pressure Using Ideal Gas Law
The final pressure \( P_f \) can be calculated using the ideal gas law rearranged as:
\[ P_f = \frac{n_{total} R T_f}{V_{total}} \]

Substituting \( n_{total} \):
\[ P_f = \frac{(n_1 + n_2) R T_f}{V_{total}} \]

### Step 6: Substitute Values and Solve
Now we can substitute the values we calculated earlier into the equation for \( P_f \):
\[ P_f = \frac{\left(\frac{611 \times 10^3 \times 0.50}{R \times 293.5} + \frac{433 \times 10^3 \times 0.75}{R \times 304.2}\right) R \times 287.5}{1.25} \]

### Step 7: Simplify and Calculate Final Pressure
After simplifying, we can cancel \( R \) from the numerator and denominator:
\[ P_f = \frac{(611 \times 10^3 \times 0.50 \times 287.5 / 293.5) + (433 \times 10^3 \times 0.75 \times 287.5 / 304.2)}{1.25} \]

Calculating this will yield the final pressure \( P_f \).

### Final Result
After performing the calculations, we find:
\[ P_f \approx 4.84 \times 10^5 \, Pa \]

### Conclusion
The final pressure in the vessels after cooling to 14.5°C is approximately \( 4.84 \times 10^5 \, Pa \).

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